A series involves harmonic number
$begingroup$
How do we get a closed form for
$$sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}$$
real-analysis sequences-and-series riemann-zeta harmonic-numbers
edited Sep 20 '15 at 12:48
Martin Sleziak
44.8k9118272
asked Feb 26 '13 at 3:31
RyanRyan
1,7941025
$endgroup$
|
show 3 more comments
$begingroup$
How do we get a closed form for
$$sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}$$
real-analysis sequences-and-series riemann-zeta harmonic-numbers
edited Sep 20 '15 at 12:48
Martin Sleziak
44.8k9118272
asked Feb 26 '13 at 3:31
RyanRyan
1,7941025
$endgroup$
$begingroup$
no idea. Why do you think there is one?
$endgroup$
– Will Jagy
Feb 26 '13 at 3:34
2
$begingroup$
Mathematica says the answer is $frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$ which probably suggests it knows a way how to reduce it to known series. One attempt may be at trying to go backwards from Mathematica's answer and see if you can rediscover what it did.
$endgroup$
– muzzlator
Feb 26 '13 at 3:40
2
$begingroup$
It would have been better to put that information into the question, both for motivation and to help others with their thoughts. So post it as "Mathematica says $sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}=frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$. Is there an easy to demonstrate this?" If you have information on why you were motivated to ask Mathematica, that might be useful, too.
$endgroup$
– Ross Millikan
Feb 26 '13 at 3:44
1
$begingroup$
Some similar series are on this guy's blog, the same guy's blog, and this other guy's blog, but they omit the identity you're after.
$endgroup$
– Aeolian
Feb 26 '13 at 4:25
1
$begingroup$
@joriki: you are right. Then OP came back with some motivation. There is progress.
$endgroup$
– Ross Millikan
Feb 26 '13 at 4:35
|
show 3 more comments
$begingroup$
How do we get a closed form for
$$sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}$$
real-analysis sequences-and-series riemann-zeta harmonic-numbers
edited Sep 20 '15 at 12:48
Martin Sleziak
44.8k9118272
asked Feb 26 '13 at 3:31
RyanRyan
1,7941025
$endgroup$
How do we get a closed form for
$$sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}$$
real-analysis sequences-and-series riemann-zeta harmonic-numbers
real-analysis sequences-and-series riemann-zeta harmonic-numbers
edited Sep 20 '15 at 12:48
Martin Sleziak
44.8k9118272
asked Feb 26 '13 at 3:31
RyanRyan
1,7941025
edited Sep 20 '15 at 12:48
Martin Sleziak
44.8k9118272
asked Feb 26 '13 at 3:31
RyanRyan
1,7941025
edited Sep 20 '15 at 12:48
Martin Sleziak
44.8k9118272
edited Sep 20 '15 at 12:48
Martin Sleziak
44.8k9118272
edited Sep 20 '15 at 12:48
Martin Sleziak
44.8k9118272
44.8k9118272
asked Feb 26 '13 at 3:31
RyanRyan
1,7941025
asked Feb 26 '13 at 3:31
RyanRyan
1,7941025
asked Feb 26 '13 at 3:31
RyanRyan
1,7941025
1,7941025
$begingroup$
no idea. Why do you think there is one?
$endgroup$
– Will Jagy
Feb 26 '13 at 3:34
2
$begingroup$
Mathematica says the answer is $frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$ which probably suggests it knows a way how to reduce it to known series. One attempt may be at trying to go backwards from Mathematica's answer and see if you can rediscover what it did.
$endgroup$
– muzzlator
Feb 26 '13 at 3:40
2
$begingroup$
It would have been better to put that information into the question, both for motivation and to help others with their thoughts. So post it as "Mathematica says $sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}=frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$. Is there an easy to demonstrate this?" If you have information on why you were motivated to ask Mathematica, that might be useful, too.
$endgroup$
– Ross Millikan
Feb 26 '13 at 3:44
1
$begingroup$
Some similar series are on this guy's blog, the same guy's blog, and this other guy's blog, but they omit the identity you're after.
$endgroup$
– Aeolian
Feb 26 '13 at 4:25
1
$begingroup$
@joriki: you are right. Then OP came back with some motivation. There is progress.
$endgroup$
– Ross Millikan
Feb 26 '13 at 4:35
|
show 3 more comments
$begingroup$
no idea. Why do you think there is one?
$endgroup$
– Will Jagy
Feb 26 '13 at 3:34
2
$begingroup$
Mathematica says the answer is $frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$ which probably suggests it knows a way how to reduce it to known series. One attempt may be at trying to go backwards from Mathematica's answer and see if you can rediscover what it did.
$endgroup$
– muzzlator
Feb 26 '13 at 3:40
2
$begingroup$
It would have been better to put that information into the question, both for motivation and to help others with their thoughts. So post it as "Mathematica says $sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}=frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$. Is there an easy to demonstrate this?" If you have information on why you were motivated to ask Mathematica, that might be useful, too.
$endgroup$
– Ross Millikan
Feb 26 '13 at 3:44
1
$begingroup$
Some similar series are on this guy's blog, the same guy's blog, and this other guy's blog, but they omit the identity you're after.
$endgroup$
– Aeolian
Feb 26 '13 at 4:25
1
$begingroup$
@joriki: you are right. Then OP came back with some motivation. There is progress.
$endgroup$
– Ross Millikan
Feb 26 '13 at 4:35
$begingroup$
no idea. Why do you think there is one?
$endgroup$
– Will Jagy
Feb 26 '13 at 3:34
$begingroup$
no idea. Why do you think there is one?
$endgroup$
– Will Jagy
Feb 26 '13 at 3:34
2
2
$begingroup$
Mathematica says the answer is $frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$ which probably suggests it knows a way how to reduce it to known series. One attempt may be at trying to go backwards from Mathematica's answer and see if you can rediscover what it did.
$endgroup$
– muzzlator
Feb 26 '13 at 3:40
$begingroup$
Mathematica says the answer is $frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$ which probably suggests it knows a way how to reduce it to known series. One attempt may be at trying to go backwards from Mathematica's answer and see if you can rediscover what it did.
$endgroup$
– muzzlator
Feb 26 '13 at 3:40
2
2
$begingroup$
It would have been better to put that information into the question, both for motivation and to help others with their thoughts. So post it as "Mathematica says $sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}=frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$. Is there an easy to demonstrate this?" If you have information on why you were motivated to ask Mathematica, that might be useful, too.
$endgroup$
– Ross Millikan
Feb 26 '13 at 3:44
$begingroup$
It would have been better to put that information into the question, both for motivation and to help others with their thoughts. So post it as "Mathematica says $sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}=frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$. Is there an easy to demonstrate this?" If you have information on why you were motivated to ask Mathematica, that might be useful, too.
$endgroup$
– Ross Millikan
Feb 26 '13 at 3:44
1
1
$begingroup$
Some similar series are on this guy's blog, the same guy's blog, and this other guy's blog, but they omit the identity you're after.
$endgroup$
– Aeolian
Feb 26 '13 at 4:25
$begingroup$
Some similar series are on this guy's blog, the same guy's blog, and this other guy's blog, but they omit the identity you're after.
$endgroup$
– Aeolian
Feb 26 '13 at 4:25
1
1
$begingroup$
@joriki: you are right. Then OP came back with some motivation. There is progress.
$endgroup$
– Ross Millikan
Feb 26 '13 at 4:35
$begingroup$
@joriki: you are right. Then OP came back with some motivation. There is progress.
$endgroup$
– Ross Millikan
Feb 26 '13 at 4:35
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Here's another solution. I'll denote various versions of the sum
$$
sum_{k=1}^inftysum_{j=1}^kfrac1jfrac1{k^2}
$$
by an $S$ with two subscripts indicating which parities are included, the first subscript referring to the parity of $j$ and the second to the parity of $k$, with '$mathrm e$' denoting only the even terms, '$mathrm o$' denoting only the odd terms, '$+$' denoting the sum of the even and odd terms, i.e. the regular sum, and '$-$' denoting the difference between the even and the odd terms, i.e. the alternating sum. Then
$$
begin{align}
sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}
&=
2sum_{n=1}^inftysum_{i=1}^nfrac1{2i}frac1{(2n+1)^2}
\
&=
2S_{mathrm{eo}}
\
&=
2(S_{++}-S_{mathrm o+}-S_{mathrm{ee}})
\
&=
2left(S_{++}-S_{mathrm o+}-frac18S_{++}right)
\
&=
2left(frac38S_{++}+left(frac12S_{++}-S_{mathrm o+}right)right)
\
&=
frac34S_{++}+S_{-+}
\
&=
frac32zeta(3)+sum_{k=1}^inftysum_{j=1}^kfrac{(-1)^j}jfrac1{k^2};,
end{align}
$$
where I used the result $sum_nH_n/n^2=2zeta(3)$ from the blog post Aeolian linked to and reduced the present problem to finding the analogue of that result with the sign alternating with $j$, which we can rewrite as
$$
begin{align}
sum_{k=1}^inftysum_{j=1}^kfrac{(-1)^j}jfrac1{k^2}
&=
sum_{k=1}^inftysum_{j=1}^inftyfrac{(-1)^j}jfrac1{k^2}-sum_{k=1}^inftysum_{j=k+1}^inftyfrac{(-1)^j}jfrac1{k^2}
\
&=
-zeta(2)log2+sum_{j=1}^inftyfrac{(-1)^j}{j+1}sum_{k=1}^jfrac1{k^2};.
end{align}
$$
This last double sum can be evaluated by the method applied in the blog post, making use of the fact that summing the coefficients of a power series in $x$ corresponds to dividing it by $1-x$:
$$
begin{align}
sum_{j=1}^infty x^jsum_{k=1}^jfrac1{k^2}=defLi{operatorname{Li}}frac{Li_2(x)}{1-x};,
end{align}
$$
where $Li_2$ is the dilogarithm. Thus
$$
begin{align}
sum_{j=1}^inftyfrac{(-1)^j}{j+1}sum_{k=1}^jfrac1{k^2}
&=
int_0^1sum_{j=1}^infty (-x)^jsum_{k=1}^jfrac1{k^2}mathrm dx
\
&=
int_0^1frac{Li_2(-x)}{1+x}mathrm dx
\
&=
left[Li_2(-x)log(1+x)right]_0^1+int_0^1frac{log^2(1+x)}xmathrm dx
\
&=-frac{zeta(2)}2log2+frac{zeta(3)}4;,
end{align}
$$
where the boundary term is evaluated using $Li_2(-1)=-eta(2)=-zeta(2)+2zeta(2)/4=-zeta(2)/2$ and the integral in the second term is evaluated in this separate question. Putting it all together, we have
$$
begin{align}
sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}
&=
frac74zeta(3)-frac32zeta(2)log2
\
&=
frac74zeta(3)-frac{pi^2}4log2;.
end{align}
$$
I believe all the rearrangements can be justified, despite the series being only conditionally convergent in $j$, by considering the partial sums with $j$ and $k$ both going up to $M$; then all the rearrangements can be carried out within that finite square of the grid, and the sums of the remaining terms go to zero with $Mtoinfty$.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 28 '13 at 10:21
jorikijoriki
170k10184343
$endgroup$
add a comment |
$begingroup$
I gave some integral representations for sums close to this one before. Here is an integral representation for your sum
$$sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}= frac{1}{4},int_{0}^{1}!{frac {ln left( 1-z right) ln left( zright) }{zsqrt {1-z}}}{dz}= frac{1}{4}(7,zeta left( 3 right) -{pi }^{2}ln left( 2 right))sim 0.393327464. $$
The above integral can be evaluated through beta function. Here is the technique from previous problems. Basically, you need to consider the integral
$$ int_{0}^{1} z^s (1-z)^{w-1/2} dz. $$
edited Apr 13 '17 at 12:19
Community♦
1
answered Feb 26 '13 at 14:57
Mhenni BenghorbalMhenni Benghorbal
43.2k63675
$endgroup$
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}: {large ?}}$.
Lets consider
$ds{fermipars{x}equiv
sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}},x^{2n + 1}.
qquadfermipars{1}={large ?},,quad fermipars{0} = 0}$.
begin{align}
fermi'pars{x}&=sum_{n = 1}^{infty}{H_{n} over 2n + 1},x^{2n}
imp
bracks{xfermi'pars{x}}'=sum_{n = 1}^{infty}H_{n},x^{2n}
=-,{lnpars{1 - x^{2}} over 1 - x^{2}},,qquadfermi'pars{0} = 0
end{align}
where we used the
Harmonic Number Generating Function.
Then
begin{align}
&xfermi'pars{x}=-int_{0}^{x}{lnpars{1 - t^{2}} over 1 - t^{2}},dd t
\[3mm]&imp
fermipars{1}=-int_{0}^{1}{dd x over x}int_{0}^{x}
{lnpars{1 - t^{2}} over 1 - t^{2}},dd t
=-int_{0}^{1}{lnpars{1 - t^{2}} over 1 - t^{2}}int_{t}^{1}{dd x over x}
,dd t
end{align}
$$begin{array}{|c|}hline\
quadsum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}
=int_{0}^{1}{lnpars{t}lnpars{1 - t^{2}} over 1 - t^{2}},dd tquad
\ \ hline
end{array}
$$
begin{align}
&color{#c00000}{sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}}
=int_{0}^{1}{lnpars{t^{1/2}}lnpars{1 - t} over 1 - t},half,t^{-1/2}
,dd t
={1 over 4}int_{0}^{1}{t^{-1/2}lnpars{t}lnpars{1 - t} over 1 - t},dd t
\[3mm]&={1 over 4}lim_{mu to 0 atop{vphantom{LARGE A}nu to 0}}
partiald{}{mu}partiald{}{nu}int_{0}^{1}t^{mu - 1/2}
pars{1 - t}^{nu - 1},dd t
={1 over 4}lim_{mu to 0 atop{vphantom{LARGE A}nu to 0}}
partiald{}{nu}Gammapars{nu}partiald{}{mu}
bracks{Gammapars{mu + 1/2} over Gammapars{mu + nu + 1/2}}
\[3mm]&={1 over 4}lim_{nu to 0}
partiald{}{nu}braces{%
Gammapars{nu},{Gammapars{1/2} over Gammapars{nu + 1/2}}
bracks{Psipars{half} - Psipars{nu + half}}}
\[3mm]&=-,{1 over 4},Gammapars{half}lim_{nu to 0}
partiald{}{nu}bracks{%
{Gammapars{nu + 1} over Gammapars{nu + 1/2}},
{Psipars{1/2 + nu} - Psipars{1/2} over nu}}
\[3mm]&=-,{1 over 4},Gammapars{half}lim_{nu to 0}
partiald{}{nu}braces{%
{Gammapars{nu + 1} over Gammapars{nu + 1/2}},
bracks{Psi'pars{half} + half,Psi''pars{half}nu}}
\[3mm]&={pi^{2}gamma + pi^{2}Psipars{1/2} + 14zetapars{3} over 8}
quadmbox{where we used}quadPsipars{1} = -gamma,,quad
Psi''pars{half} = -14zetapars{3}.
end{align}
With $ds{Psipars{half} = -2lnpars{2} - gamma}$:
$$
color{#66f}{largesum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}
={1 over 4},bracks{7zetapars{3} - pi^{2}lnpars{2}}}
approx {tt 0.3933}
$$
answered Jul 20 '14 at 16:01
Felix MarinFelix Marin
67.9k7107142
$endgroup$
add a comment |
$begingroup$
$$displaystyle I=int_0^1 dfrac{ln xln(1-x^2)}{1-x^2}dx$$
Define the function $R$ on $[0;1]$,
$$R(x)=int_0^xdfrac{ln t}{1-t^2}dt=int_0^1dfrac{xln(tx)}{1-t^2x^2}dt$$
Let $epsilon$, real, such that $0<epsilon<1$.
begin{align}
J(epsilon)&=Big[left(R(x)-R(1)right)ln(1-x^2)Big]_0^{1-epsilon}+int_0^{1-epsilon} dfrac{2xleft(R(x)-R(1)right)}{1-x^2}dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+int_0^{1-epsilon} dfrac{2xR(x)}{1-x^2}dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&int_0^{1-epsilon}left(int_0^1dfrac{2x^2ln(tx)}{(1-x^2)(1-t^2x^2)}dtright)dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&int_0^{1-epsilon}left(int_0^1dfrac{2x^2ln x}{(1-x^2)(1-t^2x^2)}dtright)dx+int_0^1left(int_0^{1-epsilon}dfrac{2x^2ln t}{(1-x^2)(1-t^2x^2)}dxright)dt\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&displaystyleint_0^{1-epsilon}left[dfrac{xln xlnleft(tfrac{1+tx}{1-tx}right)}{1-x^2}right]_{t=0}^{t=1}dx+int_0^1 left[dfrac{ln tlnleft(tfrac{1-x}{1+x}right)}{t^2-1}+dfrac{ln tlnleft(tfrac{1-tx}{1+tx}right)}{t}-dfrac{tln tlnleft(tfrac{1+tx}{1-tx}right)}{1-t^2}right]_{x=0}^{x=1-epsilon}dt\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&displaystyleint_0^{1-epsilon}dfrac{xln xlnleft(tfrac{1+x}{1-x}right)}{1-x^2}dx-lnleft(dfrac{epsilon}{2+epsilon}right)R(1)+int_0^1dfrac{ln tlnleft(tfrac{1-t(1-epsilon)}{1+t(1-epsilon)}right)}{t}dt-\
&int_0^1dfrac{tln tlnleft(tfrac{1+t(1-epsilon)}{1-t(1-epsilon)}right)}{1-t^2}dt
end{align}
Since,
$$lim_{epsilonrightarrow 0}left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)=0$$
and,
$$lim_{epsilonrightarrow 0}R(1)lnleft(tfrac{1-(1-epsilon)^2}{epsilon}right)=R(1)ln 2$$
then,
$$boxed{lim_{epsilonrightarrow 0}J(epsilon)=2R(1)ln 2+int_0^1dfrac{ln tlnleft(tfrac{1-t}{1+t}right)}{t}dt}$$
and then,
begin{align}
int_0^1dfrac{ln tlnleft(tfrac{1-t}{1+t}right)}{t}dt&=int_0^1dfrac{ln tleft(ln(1-t)-ln(1+t)right)}{t}dt\
&=-2int_0^1left(sum_{n=0}^{infty}dfrac{t^{2n}}{2n+1}right)ln tdt\
&=-2sum_{n=0}^{infty}left(dfrac{1}{2n+1}int_0^1 t^{2n}ln tdtright)\
&=2sum_{n=0}^{infty}dfrac{1}{(2n+1)^3}\
&=2left(sum_{n=1}^{infty}dfrac{1}{n^3}-sum_{n=1}^{infty}dfrac{1}{(2n)^3}right)\
&=2left(zeta(3)-dfrac{1}{8}zeta(2)right)\
&=dfrac{7}{4}zeta(3)\
end{align}
and,
begin{align}
displaystyle R(1)&=int_0^1dfrac{ln x}{1-x^2}dx\
&=int_0^1 left(sum_{n=0}^{infty}x^{2n}right)ln xdx\
&=sum_{n=0}^{infty}left(int_0^1 x^{2n}ln x dxright)\
&=-sum_{n=0}^{infty}dfrac{1}{(2n+1)^2}\
&=sum_{n=1}^{infty}dfrac{1}{(2n)^2}-sum_{n=0}^{infty}dfrac{1}{n^2}\
&=dfrac{1}{4}zeta(2)-zeta(2)\
&=-dfrac{3}{4}zeta(2)\
&=-dfrac{pi^2}{8}
end{align}
Therefore,
$$boxed{I=dfrac{7}{4}zeta(3)--dfrac{1}{4}pi^2ln 2}$$
answered Aug 21 '16 at 23:22
FDPFDP
5,33211524
$endgroup$
add a comment |
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4 Answers
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$begingroup$
Here's another solution. I'll denote various versions of the sum
$$
sum_{k=1}^inftysum_{j=1}^kfrac1jfrac1{k^2}
$$
by an $S$ with two subscripts indicating which parities are included, the first subscript referring to the parity of $j$ and the second to the parity of $k$, with '$mathrm e$' denoting only the even terms, '$mathrm o$' denoting only the odd terms, '$+$' denoting the sum of the even and odd terms, i.e. the regular sum, and '$-$' denoting the difference between the even and the odd terms, i.e. the alternating sum. Then
$$
begin{align}
sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}
&=
2sum_{n=1}^inftysum_{i=1}^nfrac1{2i}frac1{(2n+1)^2}
\
&=
2S_{mathrm{eo}}
\
&=
2(S_{++}-S_{mathrm o+}-S_{mathrm{ee}})
\
&=
2left(S_{++}-S_{mathrm o+}-frac18S_{++}right)
\
&=
2left(frac38S_{++}+left(frac12S_{++}-S_{mathrm o+}right)right)
\
&=
frac34S_{++}+S_{-+}
\
&=
frac32zeta(3)+sum_{k=1}^inftysum_{j=1}^kfrac{(-1)^j}jfrac1{k^2};,
end{align}
$$
where I used the result $sum_nH_n/n^2=2zeta(3)$ from the blog post Aeolian linked to and reduced the present problem to finding the analogue of that result with the sign alternating with $j$, which we can rewrite as
$$
begin{align}
sum_{k=1}^inftysum_{j=1}^kfrac{(-1)^j}jfrac1{k^2}
&=
sum_{k=1}^inftysum_{j=1}^inftyfrac{(-1)^j}jfrac1{k^2}-sum_{k=1}^inftysum_{j=k+1}^inftyfrac{(-1)^j}jfrac1{k^2}
\
&=
-zeta(2)log2+sum_{j=1}^inftyfrac{(-1)^j}{j+1}sum_{k=1}^jfrac1{k^2};.
end{align}
$$
This last double sum can be evaluated by the method applied in the blog post, making use of the fact that summing the coefficients of a power series in $x$ corresponds to dividing it by $1-x$:
$$
begin{align}
sum_{j=1}^infty x^jsum_{k=1}^jfrac1{k^2}=defLi{operatorname{Li}}frac{Li_2(x)}{1-x};,
end{align}
$$
where $Li_2$ is the dilogarithm. Thus
$$
begin{align}
sum_{j=1}^inftyfrac{(-1)^j}{j+1}sum_{k=1}^jfrac1{k^2}
&=
int_0^1sum_{j=1}^infty (-x)^jsum_{k=1}^jfrac1{k^2}mathrm dx
\
&=
int_0^1frac{Li_2(-x)}{1+x}mathrm dx
\
&=
left[Li_2(-x)log(1+x)right]_0^1+int_0^1frac{log^2(1+x)}xmathrm dx
\
&=-frac{zeta(2)}2log2+frac{zeta(3)}4;,
end{align}
$$
where the boundary term is evaluated using $Li_2(-1)=-eta(2)=-zeta(2)+2zeta(2)/4=-zeta(2)/2$ and the integral in the second term is evaluated in this separate question. Putting it all together, we have
$$
begin{align}
sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}
&=
frac74zeta(3)-frac32zeta(2)log2
\
&=
frac74zeta(3)-frac{pi^2}4log2;.
end{align}
$$
I believe all the rearrangements can be justified, despite the series being only conditionally convergent in $j$, by considering the partial sums with $j$ and $k$ both going up to $M$; then all the rearrangements can be carried out within that finite square of the grid, and the sums of the remaining terms go to zero with $Mtoinfty$.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 28 '13 at 10:21
jorikijoriki
170k10184343
$endgroup$
add a comment |
$begingroup$
Here's another solution. I'll denote various versions of the sum
$$
sum_{k=1}^inftysum_{j=1}^kfrac1jfrac1{k^2}
$$
by an $S$ with two subscripts indicating which parities are included, the first subscript referring to the parity of $j$ and the second to the parity of $k$, with '$mathrm e$' denoting only the even terms, '$mathrm o$' denoting only the odd terms, '$+$' denoting the sum of the even and odd terms, i.e. the regular sum, and '$-$' denoting the difference between the even and the odd terms, i.e. the alternating sum. Then
$$
begin{align}
sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}
&=
2sum_{n=1}^inftysum_{i=1}^nfrac1{2i}frac1{(2n+1)^2}
\
&=
2S_{mathrm{eo}}
\
&=
2(S_{++}-S_{mathrm o+}-S_{mathrm{ee}})
\
&=
2left(S_{++}-S_{mathrm o+}-frac18S_{++}right)
\
&=
2left(frac38S_{++}+left(frac12S_{++}-S_{mathrm o+}right)right)
\
&=
frac34S_{++}+S_{-+}
\
&=
frac32zeta(3)+sum_{k=1}^inftysum_{j=1}^kfrac{(-1)^j}jfrac1{k^2};,
end{align}
$$
where I used the result $sum_nH_n/n^2=2zeta(3)$ from the blog post Aeolian linked to and reduced the present problem to finding the analogue of that result with the sign alternating with $j$, which we can rewrite as
$$
begin{align}
sum_{k=1}^inftysum_{j=1}^kfrac{(-1)^j}jfrac1{k^2}
&=
sum_{k=1}^inftysum_{j=1}^inftyfrac{(-1)^j}jfrac1{k^2}-sum_{k=1}^inftysum_{j=k+1}^inftyfrac{(-1)^j}jfrac1{k^2}
\
&=
-zeta(2)log2+sum_{j=1}^inftyfrac{(-1)^j}{j+1}sum_{k=1}^jfrac1{k^2};.
end{align}
$$
This last double sum can be evaluated by the method applied in the blog post, making use of the fact that summing the coefficients of a power series in $x$ corresponds to dividing it by $1-x$:
$$
begin{align}
sum_{j=1}^infty x^jsum_{k=1}^jfrac1{k^2}=defLi{operatorname{Li}}frac{Li_2(x)}{1-x};,
end{align}
$$
where $Li_2$ is the dilogarithm. Thus
$$
begin{align}
sum_{j=1}^inftyfrac{(-1)^j}{j+1}sum_{k=1}^jfrac1{k^2}
&=
int_0^1sum_{j=1}^infty (-x)^jsum_{k=1}^jfrac1{k^2}mathrm dx
\
&=
int_0^1frac{Li_2(-x)}{1+x}mathrm dx
\
&=
left[Li_2(-x)log(1+x)right]_0^1+int_0^1frac{log^2(1+x)}xmathrm dx
\
&=-frac{zeta(2)}2log2+frac{zeta(3)}4;,
end{align}
$$
where the boundary term is evaluated using $Li_2(-1)=-eta(2)=-zeta(2)+2zeta(2)/4=-zeta(2)/2$ and the integral in the second term is evaluated in this separate question. Putting it all together, we have
$$
begin{align}
sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}
&=
frac74zeta(3)-frac32zeta(2)log2
\
&=
frac74zeta(3)-frac{pi^2}4log2;.
end{align}
$$
I believe all the rearrangements can be justified, despite the series being only conditionally convergent in $j$, by considering the partial sums with $j$ and $k$ both going up to $M$; then all the rearrangements can be carried out within that finite square of the grid, and the sums of the remaining terms go to zero with $Mtoinfty$.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 28 '13 at 10:21
jorikijoriki
170k10184343
$endgroup$
add a comment |
$begingroup$
Here's another solution. I'll denote various versions of the sum
$$
sum_{k=1}^inftysum_{j=1}^kfrac1jfrac1{k^2}
$$
by an $S$ with two subscripts indicating which parities are included, the first subscript referring to the parity of $j$ and the second to the parity of $k$, with '$mathrm e$' denoting only the even terms, '$mathrm o$' denoting only the odd terms, '$+$' denoting the sum of the even and odd terms, i.e. the regular sum, and '$-$' denoting the difference between the even and the odd terms, i.e. the alternating sum. Then
$$
begin{align}
sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}
&=
2sum_{n=1}^inftysum_{i=1}^nfrac1{2i}frac1{(2n+1)^2}
\
&=
2S_{mathrm{eo}}
\
&=
2(S_{++}-S_{mathrm o+}-S_{mathrm{ee}})
\
&=
2left(S_{++}-S_{mathrm o+}-frac18S_{++}right)
\
&=
2left(frac38S_{++}+left(frac12S_{++}-S_{mathrm o+}right)right)
\
&=
frac34S_{++}+S_{-+}
\
&=
frac32zeta(3)+sum_{k=1}^inftysum_{j=1}^kfrac{(-1)^j}jfrac1{k^2};,
end{align}
$$
where I used the result $sum_nH_n/n^2=2zeta(3)$ from the blog post Aeolian linked to and reduced the present problem to finding the analogue of that result with the sign alternating with $j$, which we can rewrite as
$$
begin{align}
sum_{k=1}^inftysum_{j=1}^kfrac{(-1)^j}jfrac1{k^2}
&=
sum_{k=1}^inftysum_{j=1}^inftyfrac{(-1)^j}jfrac1{k^2}-sum_{k=1}^inftysum_{j=k+1}^inftyfrac{(-1)^j}jfrac1{k^2}
\
&=
-zeta(2)log2+sum_{j=1}^inftyfrac{(-1)^j}{j+1}sum_{k=1}^jfrac1{k^2};.
end{align}
$$
This last double sum can be evaluated by the method applied in the blog post, making use of the fact that summing the coefficients of a power series in $x$ corresponds to dividing it by $1-x$:
$$
begin{align}
sum_{j=1}^infty x^jsum_{k=1}^jfrac1{k^2}=defLi{operatorname{Li}}frac{Li_2(x)}{1-x};,
end{align}
$$
where $Li_2$ is the dilogarithm. Thus
$$
begin{align}
sum_{j=1}^inftyfrac{(-1)^j}{j+1}sum_{k=1}^jfrac1{k^2}
&=
int_0^1sum_{j=1}^infty (-x)^jsum_{k=1}^jfrac1{k^2}mathrm dx
\
&=
int_0^1frac{Li_2(-x)}{1+x}mathrm dx
\
&=
left[Li_2(-x)log(1+x)right]_0^1+int_0^1frac{log^2(1+x)}xmathrm dx
\
&=-frac{zeta(2)}2log2+frac{zeta(3)}4;,
end{align}
$$
where the boundary term is evaluated using $Li_2(-1)=-eta(2)=-zeta(2)+2zeta(2)/4=-zeta(2)/2$ and the integral in the second term is evaluated in this separate question. Putting it all together, we have
$$
begin{align}
sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}
&=
frac74zeta(3)-frac32zeta(2)log2
\
&=
frac74zeta(3)-frac{pi^2}4log2;.
end{align}
$$
I believe all the rearrangements can be justified, despite the series being only conditionally convergent in $j$, by considering the partial sums with $j$ and $k$ both going up to $M$; then all the rearrangements can be carried out within that finite square of the grid, and the sums of the remaining terms go to zero with $Mtoinfty$.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 28 '13 at 10:21
jorikijoriki
170k10184343
$endgroup$
Here's another solution. I'll denote various versions of the sum
$$
sum_{k=1}^inftysum_{j=1}^kfrac1jfrac1{k^2}
$$
by an $S$ with two subscripts indicating which parities are included, the first subscript referring to the parity of $j$ and the second to the parity of $k$, with '$mathrm e$' denoting only the even terms, '$mathrm o$' denoting only the odd terms, '$+$' denoting the sum of the even and odd terms, i.e. the regular sum, and '$-$' denoting the difference between the even and the odd terms, i.e. the alternating sum. Then
$$
begin{align}
sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}
&=
2sum_{n=1}^inftysum_{i=1}^nfrac1{2i}frac1{(2n+1)^2}
\
&=
2S_{mathrm{eo}}
\
&=
2(S_{++}-S_{mathrm o+}-S_{mathrm{ee}})
\
&=
2left(S_{++}-S_{mathrm o+}-frac18S_{++}right)
\
&=
2left(frac38S_{++}+left(frac12S_{++}-S_{mathrm o+}right)right)
\
&=
frac34S_{++}+S_{-+}
\
&=
frac32zeta(3)+sum_{k=1}^inftysum_{j=1}^kfrac{(-1)^j}jfrac1{k^2};,
end{align}
$$
where I used the result $sum_nH_n/n^2=2zeta(3)$ from the blog post Aeolian linked to and reduced the present problem to finding the analogue of that result with the sign alternating with $j$, which we can rewrite as
$$
begin{align}
sum_{k=1}^inftysum_{j=1}^kfrac{(-1)^j}jfrac1{k^2}
&=
sum_{k=1}^inftysum_{j=1}^inftyfrac{(-1)^j}jfrac1{k^2}-sum_{k=1}^inftysum_{j=k+1}^inftyfrac{(-1)^j}jfrac1{k^2}
\
&=
-zeta(2)log2+sum_{j=1}^inftyfrac{(-1)^j}{j+1}sum_{k=1}^jfrac1{k^2};.
end{align}
$$
This last double sum can be evaluated by the method applied in the blog post, making use of the fact that summing the coefficients of a power series in $x$ corresponds to dividing it by $1-x$:
$$
begin{align}
sum_{j=1}^infty x^jsum_{k=1}^jfrac1{k^2}=defLi{operatorname{Li}}frac{Li_2(x)}{1-x};,
end{align}
$$
where $Li_2$ is the dilogarithm. Thus
$$
begin{align}
sum_{j=1}^inftyfrac{(-1)^j}{j+1}sum_{k=1}^jfrac1{k^2}
&=
int_0^1sum_{j=1}^infty (-x)^jsum_{k=1}^jfrac1{k^2}mathrm dx
\
&=
int_0^1frac{Li_2(-x)}{1+x}mathrm dx
\
&=
left[Li_2(-x)log(1+x)right]_0^1+int_0^1frac{log^2(1+x)}xmathrm dx
\
&=-frac{zeta(2)}2log2+frac{zeta(3)}4;,
end{align}
$$
where the boundary term is evaluated using $Li_2(-1)=-eta(2)=-zeta(2)+2zeta(2)/4=-zeta(2)/2$ and the integral in the second term is evaluated in this separate question. Putting it all together, we have
$$
begin{align}
sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}
&=
frac74zeta(3)-frac32zeta(2)log2
\
&=
frac74zeta(3)-frac{pi^2}4log2;.
end{align}
$$
I believe all the rearrangements can be justified, despite the series being only conditionally convergent in $j$, by considering the partial sums with $j$ and $k$ both going up to $M$; then all the rearrangements can be carried out within that finite square of the grid, and the sums of the remaining terms go to zero with $Mtoinfty$.
edited Apr 13 '17 at 12:21
Community♦
1
answered Feb 28 '13 at 10:21
jorikijoriki
170k10184343
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Feb 28 '13 at 10:21
jorikijoriki
170k10184343
answered Feb 28 '13 at 10:21
jorikijoriki
170k10184343
answered Feb 28 '13 at 10:21
jorikijoriki
170k10184343
170k10184343
add a comment |
add a comment |
$begingroup$
I gave some integral representations for sums close to this one before. Here is an integral representation for your sum
$$sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}= frac{1}{4},int_{0}^{1}!{frac {ln left( 1-z right) ln left( zright) }{zsqrt {1-z}}}{dz}= frac{1}{4}(7,zeta left( 3 right) -{pi }^{2}ln left( 2 right))sim 0.393327464. $$
The above integral can be evaluated through beta function. Here is the technique from previous problems. Basically, you need to consider the integral
$$ int_{0}^{1} z^s (1-z)^{w-1/2} dz. $$
edited Apr 13 '17 at 12:19
Community♦
1
answered Feb 26 '13 at 14:57
Mhenni BenghorbalMhenni Benghorbal
43.2k63675
$endgroup$
add a comment |
$begingroup$
I gave some integral representations for sums close to this one before. Here is an integral representation for your sum
$$sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}= frac{1}{4},int_{0}^{1}!{frac {ln left( 1-z right) ln left( zright) }{zsqrt {1-z}}}{dz}= frac{1}{4}(7,zeta left( 3 right) -{pi }^{2}ln left( 2 right))sim 0.393327464. $$
The above integral can be evaluated through beta function. Here is the technique from previous problems. Basically, you need to consider the integral
$$ int_{0}^{1} z^s (1-z)^{w-1/2} dz. $$
edited Apr 13 '17 at 12:19
Community♦
1
answered Feb 26 '13 at 14:57
Mhenni BenghorbalMhenni Benghorbal
43.2k63675
$endgroup$
add a comment |
$begingroup$
I gave some integral representations for sums close to this one before. Here is an integral representation for your sum
$$sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}= frac{1}{4},int_{0}^{1}!{frac {ln left( 1-z right) ln left( zright) }{zsqrt {1-z}}}{dz}= frac{1}{4}(7,zeta left( 3 right) -{pi }^{2}ln left( 2 right))sim 0.393327464. $$
The above integral can be evaluated through beta function. Here is the technique from previous problems. Basically, you need to consider the integral
$$ int_{0}^{1} z^s (1-z)^{w-1/2} dz. $$
edited Apr 13 '17 at 12:19
Community♦
1
answered Feb 26 '13 at 14:57
Mhenni BenghorbalMhenni Benghorbal
43.2k63675
$endgroup$
I gave some integral representations for sums close to this one before. Here is an integral representation for your sum
$$sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}= frac{1}{4},int_{0}^{1}!{frac {ln left( 1-z right) ln left( zright) }{zsqrt {1-z}}}{dz}= frac{1}{4}(7,zeta left( 3 right) -{pi }^{2}ln left( 2 right))sim 0.393327464. $$
The above integral can be evaluated through beta function. Here is the technique from previous problems. Basically, you need to consider the integral
$$ int_{0}^{1} z^s (1-z)^{w-1/2} dz. $$
edited Apr 13 '17 at 12:19
Community♦
1
answered Feb 26 '13 at 14:57
Mhenni BenghorbalMhenni Benghorbal
43.2k63675
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
answered Feb 26 '13 at 14:57
Mhenni BenghorbalMhenni Benghorbal
43.2k63675
answered Feb 26 '13 at 14:57
Mhenni BenghorbalMhenni Benghorbal
43.2k63675
answered Feb 26 '13 at 14:57
Mhenni BenghorbalMhenni Benghorbal
43.2k63675
43.2k63675
add a comment |
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
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newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
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newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
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newcommand{wt}[1]{widetilde{#1}}$
$ds{sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}: {large ?}}$.
Lets consider
$ds{fermipars{x}equiv
sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}},x^{2n + 1}.
qquadfermipars{1}={large ?},,quad fermipars{0} = 0}$.
begin{align}
fermi'pars{x}&=sum_{n = 1}^{infty}{H_{n} over 2n + 1},x^{2n}
imp
bracks{xfermi'pars{x}}'=sum_{n = 1}^{infty}H_{n},x^{2n}
=-,{lnpars{1 - x^{2}} over 1 - x^{2}},,qquadfermi'pars{0} = 0
end{align}
where we used the
Harmonic Number Generating Function.
Then
begin{align}
&xfermi'pars{x}=-int_{0}^{x}{lnpars{1 - t^{2}} over 1 - t^{2}},dd t
\[3mm]&imp
fermipars{1}=-int_{0}^{1}{dd x over x}int_{0}^{x}
{lnpars{1 - t^{2}} over 1 - t^{2}},dd t
=-int_{0}^{1}{lnpars{1 - t^{2}} over 1 - t^{2}}int_{t}^{1}{dd x over x}
,dd t
end{align}
$$begin{array}{|c|}hline\
quadsum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}
=int_{0}^{1}{lnpars{t}lnpars{1 - t^{2}} over 1 - t^{2}},dd tquad
\ \ hline
end{array}
$$
begin{align}
&color{#c00000}{sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}}
=int_{0}^{1}{lnpars{t^{1/2}}lnpars{1 - t} over 1 - t},half,t^{-1/2}
,dd t
={1 over 4}int_{0}^{1}{t^{-1/2}lnpars{t}lnpars{1 - t} over 1 - t},dd t
\[3mm]&={1 over 4}lim_{mu to 0 atop{vphantom{LARGE A}nu to 0}}
partiald{}{mu}partiald{}{nu}int_{0}^{1}t^{mu - 1/2}
pars{1 - t}^{nu - 1},dd t
={1 over 4}lim_{mu to 0 atop{vphantom{LARGE A}nu to 0}}
partiald{}{nu}Gammapars{nu}partiald{}{mu}
bracks{Gammapars{mu + 1/2} over Gammapars{mu + nu + 1/2}}
\[3mm]&={1 over 4}lim_{nu to 0}
partiald{}{nu}braces{%
Gammapars{nu},{Gammapars{1/2} over Gammapars{nu + 1/2}}
bracks{Psipars{half} - Psipars{nu + half}}}
\[3mm]&=-,{1 over 4},Gammapars{half}lim_{nu to 0}
partiald{}{nu}bracks{%
{Gammapars{nu + 1} over Gammapars{nu + 1/2}},
{Psipars{1/2 + nu} - Psipars{1/2} over nu}}
\[3mm]&=-,{1 over 4},Gammapars{half}lim_{nu to 0}
partiald{}{nu}braces{%
{Gammapars{nu + 1} over Gammapars{nu + 1/2}},
bracks{Psi'pars{half} + half,Psi''pars{half}nu}}
\[3mm]&={pi^{2}gamma + pi^{2}Psipars{1/2} + 14zetapars{3} over 8}
quadmbox{where we used}quadPsipars{1} = -gamma,,quad
Psi''pars{half} = -14zetapars{3}.
end{align}
With $ds{Psipars{half} = -2lnpars{2} - gamma}$:
$$
color{#66f}{largesum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}
={1 over 4},bracks{7zetapars{3} - pi^{2}lnpars{2}}}
approx {tt 0.3933}
$$
answered Jul 20 '14 at 16:01
Felix MarinFelix Marin
67.9k7107142
$endgroup$
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}
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newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}: {large ?}}$.
Lets consider
$ds{fermipars{x}equiv
sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}},x^{2n + 1}.
qquadfermipars{1}={large ?},,quad fermipars{0} = 0}$.
begin{align}
fermi'pars{x}&=sum_{n = 1}^{infty}{H_{n} over 2n + 1},x^{2n}
imp
bracks{xfermi'pars{x}}'=sum_{n = 1}^{infty}H_{n},x^{2n}
=-,{lnpars{1 - x^{2}} over 1 - x^{2}},,qquadfermi'pars{0} = 0
end{align}
where we used the
Harmonic Number Generating Function.
Then
begin{align}
&xfermi'pars{x}=-int_{0}^{x}{lnpars{1 - t^{2}} over 1 - t^{2}},dd t
\[3mm]&imp
fermipars{1}=-int_{0}^{1}{dd x over x}int_{0}^{x}
{lnpars{1 - t^{2}} over 1 - t^{2}},dd t
=-int_{0}^{1}{lnpars{1 - t^{2}} over 1 - t^{2}}int_{t}^{1}{dd x over x}
,dd t
end{align}
$$begin{array}{|c|}hline\
quadsum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}
=int_{0}^{1}{lnpars{t}lnpars{1 - t^{2}} over 1 - t^{2}},dd tquad
\ \ hline
end{array}
$$
begin{align}
&color{#c00000}{sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}}
=int_{0}^{1}{lnpars{t^{1/2}}lnpars{1 - t} over 1 - t},half,t^{-1/2}
,dd t
={1 over 4}int_{0}^{1}{t^{-1/2}lnpars{t}lnpars{1 - t} over 1 - t},dd t
\[3mm]&={1 over 4}lim_{mu to 0 atop{vphantom{LARGE A}nu to 0}}
partiald{}{mu}partiald{}{nu}int_{0}^{1}t^{mu - 1/2}
pars{1 - t}^{nu - 1},dd t
={1 over 4}lim_{mu to 0 atop{vphantom{LARGE A}nu to 0}}
partiald{}{nu}Gammapars{nu}partiald{}{mu}
bracks{Gammapars{mu + 1/2} over Gammapars{mu + nu + 1/2}}
\[3mm]&={1 over 4}lim_{nu to 0}
partiald{}{nu}braces{%
Gammapars{nu},{Gammapars{1/2} over Gammapars{nu + 1/2}}
bracks{Psipars{half} - Psipars{nu + half}}}
\[3mm]&=-,{1 over 4},Gammapars{half}lim_{nu to 0}
partiald{}{nu}bracks{%
{Gammapars{nu + 1} over Gammapars{nu + 1/2}},
{Psipars{1/2 + nu} - Psipars{1/2} over nu}}
\[3mm]&=-,{1 over 4},Gammapars{half}lim_{nu to 0}
partiald{}{nu}braces{%
{Gammapars{nu + 1} over Gammapars{nu + 1/2}},
bracks{Psi'pars{half} + half,Psi''pars{half}nu}}
\[3mm]&={pi^{2}gamma + pi^{2}Psipars{1/2} + 14zetapars{3} over 8}
quadmbox{where we used}quadPsipars{1} = -gamma,,quad
Psi''pars{half} = -14zetapars{3}.
end{align}
With $ds{Psipars{half} = -2lnpars{2} - gamma}$:
$$
color{#66f}{largesum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}
={1 over 4},bracks{7zetapars{3} - pi^{2}lnpars{2}}}
approx {tt 0.3933}
$$
answered Jul 20 '14 at 16:01
Felix MarinFelix Marin
67.9k7107142
$endgroup$
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
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newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
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newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
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newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
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newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}: {large ?}}$.
Lets consider
$ds{fermipars{x}equiv
sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}},x^{2n + 1}.
qquadfermipars{1}={large ?},,quad fermipars{0} = 0}$.
begin{align}
fermi'pars{x}&=sum_{n = 1}^{infty}{H_{n} over 2n + 1},x^{2n}
imp
bracks{xfermi'pars{x}}'=sum_{n = 1}^{infty}H_{n},x^{2n}
=-,{lnpars{1 - x^{2}} over 1 - x^{2}},,qquadfermi'pars{0} = 0
end{align}
where we used the
Harmonic Number Generating Function.
Then
begin{align}
&xfermi'pars{x}=-int_{0}^{x}{lnpars{1 - t^{2}} over 1 - t^{2}},dd t
\[3mm]&imp
fermipars{1}=-int_{0}^{1}{dd x over x}int_{0}^{x}
{lnpars{1 - t^{2}} over 1 - t^{2}},dd t
=-int_{0}^{1}{lnpars{1 - t^{2}} over 1 - t^{2}}int_{t}^{1}{dd x over x}
,dd t
end{align}
$$begin{array}{|c|}hline\
quadsum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}
=int_{0}^{1}{lnpars{t}lnpars{1 - t^{2}} over 1 - t^{2}},dd tquad
\ \ hline
end{array}
$$
begin{align}
&color{#c00000}{sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}}
=int_{0}^{1}{lnpars{t^{1/2}}lnpars{1 - t} over 1 - t},half,t^{-1/2}
,dd t
={1 over 4}int_{0}^{1}{t^{-1/2}lnpars{t}lnpars{1 - t} over 1 - t},dd t
\[3mm]&={1 over 4}lim_{mu to 0 atop{vphantom{LARGE A}nu to 0}}
partiald{}{mu}partiald{}{nu}int_{0}^{1}t^{mu - 1/2}
pars{1 - t}^{nu - 1},dd t
={1 over 4}lim_{mu to 0 atop{vphantom{LARGE A}nu to 0}}
partiald{}{nu}Gammapars{nu}partiald{}{mu}
bracks{Gammapars{mu + 1/2} over Gammapars{mu + nu + 1/2}}
\[3mm]&={1 over 4}lim_{nu to 0}
partiald{}{nu}braces{%
Gammapars{nu},{Gammapars{1/2} over Gammapars{nu + 1/2}}
bracks{Psipars{half} - Psipars{nu + half}}}
\[3mm]&=-,{1 over 4},Gammapars{half}lim_{nu to 0}
partiald{}{nu}bracks{%
{Gammapars{nu + 1} over Gammapars{nu + 1/2}},
{Psipars{1/2 + nu} - Psipars{1/2} over nu}}
\[3mm]&=-,{1 over 4},Gammapars{half}lim_{nu to 0}
partiald{}{nu}braces{%
{Gammapars{nu + 1} over Gammapars{nu + 1/2}},
bracks{Psi'pars{half} + half,Psi''pars{half}nu}}
\[3mm]&={pi^{2}gamma + pi^{2}Psipars{1/2} + 14zetapars{3} over 8}
quadmbox{where we used}quadPsipars{1} = -gamma,,quad
Psi''pars{half} = -14zetapars{3}.
end{align}
With $ds{Psipars{half} = -2lnpars{2} - gamma}$:
$$
color{#66f}{largesum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}
={1 over 4},bracks{7zetapars{3} - pi^{2}lnpars{2}}}
approx {tt 0.3933}
$$
answered Jul 20 '14 at 16:01
Felix MarinFelix Marin
67.9k7107142
$endgroup$
$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
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newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
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newcommand{half}{{1 over 2}}
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newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}: {large ?}}$.
Lets consider
$ds{fermipars{x}equiv
sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}},x^{2n + 1}.
qquadfermipars{1}={large ?},,quad fermipars{0} = 0}$.
begin{align}
fermi'pars{x}&=sum_{n = 1}^{infty}{H_{n} over 2n + 1},x^{2n}
imp
bracks{xfermi'pars{x}}'=sum_{n = 1}^{infty}H_{n},x^{2n}
=-,{lnpars{1 - x^{2}} over 1 - x^{2}},,qquadfermi'pars{0} = 0
end{align}
where we used the
Harmonic Number Generating Function.
Then
begin{align}
&xfermi'pars{x}=-int_{0}^{x}{lnpars{1 - t^{2}} over 1 - t^{2}},dd t
\[3mm]&imp
fermipars{1}=-int_{0}^{1}{dd x over x}int_{0}^{x}
{lnpars{1 - t^{2}} over 1 - t^{2}},dd t
=-int_{0}^{1}{lnpars{1 - t^{2}} over 1 - t^{2}}int_{t}^{1}{dd x over x}
,dd t
end{align}
$$begin{array}{|c|}hline\
quadsum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}
=int_{0}^{1}{lnpars{t}lnpars{1 - t^{2}} over 1 - t^{2}},dd tquad
\ \ hline
end{array}
$$
begin{align}
&color{#c00000}{sum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}}
=int_{0}^{1}{lnpars{t^{1/2}}lnpars{1 - t} over 1 - t},half,t^{-1/2}
,dd t
={1 over 4}int_{0}^{1}{t^{-1/2}lnpars{t}lnpars{1 - t} over 1 - t},dd t
\[3mm]&={1 over 4}lim_{mu to 0 atop{vphantom{LARGE A}nu to 0}}
partiald{}{mu}partiald{}{nu}int_{0}^{1}t^{mu - 1/2}
pars{1 - t}^{nu - 1},dd t
={1 over 4}lim_{mu to 0 atop{vphantom{LARGE A}nu to 0}}
partiald{}{nu}Gammapars{nu}partiald{}{mu}
bracks{Gammapars{mu + 1/2} over Gammapars{mu + nu + 1/2}}
\[3mm]&={1 over 4}lim_{nu to 0}
partiald{}{nu}braces{%
Gammapars{nu},{Gammapars{1/2} over Gammapars{nu + 1/2}}
bracks{Psipars{half} - Psipars{nu + half}}}
\[3mm]&=-,{1 over 4},Gammapars{half}lim_{nu to 0}
partiald{}{nu}bracks{%
{Gammapars{nu + 1} over Gammapars{nu + 1/2}},
{Psipars{1/2 + nu} - Psipars{1/2} over nu}}
\[3mm]&=-,{1 over 4},Gammapars{half}lim_{nu to 0}
partiald{}{nu}braces{%
{Gammapars{nu + 1} over Gammapars{nu + 1/2}},
bracks{Psi'pars{half} + half,Psi''pars{half}nu}}
\[3mm]&={pi^{2}gamma + pi^{2}Psipars{1/2} + 14zetapars{3} over 8}
quadmbox{where we used}quadPsipars{1} = -gamma,,quad
Psi''pars{half} = -14zetapars{3}.
end{align}
With $ds{Psipars{half} = -2lnpars{2} - gamma}$:
$$
color{#66f}{largesum_{n = 1}^{infty}{H_{n} over pars{2n + 1}^{2}}
={1 over 4},bracks{7zetapars{3} - pi^{2}lnpars{2}}}
approx {tt 0.3933}
$$
answered Jul 20 '14 at 16:01
Felix MarinFelix Marin
67.9k7107142
edited Jul 29 '14 at 0:02
answered Jul 20 '14 at 16:01
Felix MarinFelix Marin
67.9k7107142
answered Jul 20 '14 at 16:01
Felix MarinFelix Marin
67.9k7107142
answered Jul 20 '14 at 16:01
Felix MarinFelix Marin
67.9k7107142
67.9k7107142
add a comment |
add a comment |
$begingroup$
$$displaystyle I=int_0^1 dfrac{ln xln(1-x^2)}{1-x^2}dx$$
Define the function $R$ on $[0;1]$,
$$R(x)=int_0^xdfrac{ln t}{1-t^2}dt=int_0^1dfrac{xln(tx)}{1-t^2x^2}dt$$
Let $epsilon$, real, such that $0<epsilon<1$.
begin{align}
J(epsilon)&=Big[left(R(x)-R(1)right)ln(1-x^2)Big]_0^{1-epsilon}+int_0^{1-epsilon} dfrac{2xleft(R(x)-R(1)right)}{1-x^2}dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+int_0^{1-epsilon} dfrac{2xR(x)}{1-x^2}dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&int_0^{1-epsilon}left(int_0^1dfrac{2x^2ln(tx)}{(1-x^2)(1-t^2x^2)}dtright)dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&int_0^{1-epsilon}left(int_0^1dfrac{2x^2ln x}{(1-x^2)(1-t^2x^2)}dtright)dx+int_0^1left(int_0^{1-epsilon}dfrac{2x^2ln t}{(1-x^2)(1-t^2x^2)}dxright)dt\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&displaystyleint_0^{1-epsilon}left[dfrac{xln xlnleft(tfrac{1+tx}{1-tx}right)}{1-x^2}right]_{t=0}^{t=1}dx+int_0^1 left[dfrac{ln tlnleft(tfrac{1-x}{1+x}right)}{t^2-1}+dfrac{ln tlnleft(tfrac{1-tx}{1+tx}right)}{t}-dfrac{tln tlnleft(tfrac{1+tx}{1-tx}right)}{1-t^2}right]_{x=0}^{x=1-epsilon}dt\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&displaystyleint_0^{1-epsilon}dfrac{xln xlnleft(tfrac{1+x}{1-x}right)}{1-x^2}dx-lnleft(dfrac{epsilon}{2+epsilon}right)R(1)+int_0^1dfrac{ln tlnleft(tfrac{1-t(1-epsilon)}{1+t(1-epsilon)}right)}{t}dt-\
&int_0^1dfrac{tln tlnleft(tfrac{1+t(1-epsilon)}{1-t(1-epsilon)}right)}{1-t^2}dt
end{align}
Since,
$$lim_{epsilonrightarrow 0}left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)=0$$
and,
$$lim_{epsilonrightarrow 0}R(1)lnleft(tfrac{1-(1-epsilon)^2}{epsilon}right)=R(1)ln 2$$
then,
$$boxed{lim_{epsilonrightarrow 0}J(epsilon)=2R(1)ln 2+int_0^1dfrac{ln tlnleft(tfrac{1-t}{1+t}right)}{t}dt}$$
and then,
begin{align}
int_0^1dfrac{ln tlnleft(tfrac{1-t}{1+t}right)}{t}dt&=int_0^1dfrac{ln tleft(ln(1-t)-ln(1+t)right)}{t}dt\
&=-2int_0^1left(sum_{n=0}^{infty}dfrac{t^{2n}}{2n+1}right)ln tdt\
&=-2sum_{n=0}^{infty}left(dfrac{1}{2n+1}int_0^1 t^{2n}ln tdtright)\
&=2sum_{n=0}^{infty}dfrac{1}{(2n+1)^3}\
&=2left(sum_{n=1}^{infty}dfrac{1}{n^3}-sum_{n=1}^{infty}dfrac{1}{(2n)^3}right)\
&=2left(zeta(3)-dfrac{1}{8}zeta(2)right)\
&=dfrac{7}{4}zeta(3)\
end{align}
and,
begin{align}
displaystyle R(1)&=int_0^1dfrac{ln x}{1-x^2}dx\
&=int_0^1 left(sum_{n=0}^{infty}x^{2n}right)ln xdx\
&=sum_{n=0}^{infty}left(int_0^1 x^{2n}ln x dxright)\
&=-sum_{n=0}^{infty}dfrac{1}{(2n+1)^2}\
&=sum_{n=1}^{infty}dfrac{1}{(2n)^2}-sum_{n=0}^{infty}dfrac{1}{n^2}\
&=dfrac{1}{4}zeta(2)-zeta(2)\
&=-dfrac{3}{4}zeta(2)\
&=-dfrac{pi^2}{8}
end{align}
Therefore,
$$boxed{I=dfrac{7}{4}zeta(3)--dfrac{1}{4}pi^2ln 2}$$
answered Aug 21 '16 at 23:22
FDPFDP
5,33211524
$endgroup$
add a comment |
$begingroup$
$$displaystyle I=int_0^1 dfrac{ln xln(1-x^2)}{1-x^2}dx$$
Define the function $R$ on $[0;1]$,
$$R(x)=int_0^xdfrac{ln t}{1-t^2}dt=int_0^1dfrac{xln(tx)}{1-t^2x^2}dt$$
Let $epsilon$, real, such that $0<epsilon<1$.
begin{align}
J(epsilon)&=Big[left(R(x)-R(1)right)ln(1-x^2)Big]_0^{1-epsilon}+int_0^{1-epsilon} dfrac{2xleft(R(x)-R(1)right)}{1-x^2}dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+int_0^{1-epsilon} dfrac{2xR(x)}{1-x^2}dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&int_0^{1-epsilon}left(int_0^1dfrac{2x^2ln(tx)}{(1-x^2)(1-t^2x^2)}dtright)dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&int_0^{1-epsilon}left(int_0^1dfrac{2x^2ln x}{(1-x^2)(1-t^2x^2)}dtright)dx+int_0^1left(int_0^{1-epsilon}dfrac{2x^2ln t}{(1-x^2)(1-t^2x^2)}dxright)dt\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&displaystyleint_0^{1-epsilon}left[dfrac{xln xlnleft(tfrac{1+tx}{1-tx}right)}{1-x^2}right]_{t=0}^{t=1}dx+int_0^1 left[dfrac{ln tlnleft(tfrac{1-x}{1+x}right)}{t^2-1}+dfrac{ln tlnleft(tfrac{1-tx}{1+tx}right)}{t}-dfrac{tln tlnleft(tfrac{1+tx}{1-tx}right)}{1-t^2}right]_{x=0}^{x=1-epsilon}dt\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&displaystyleint_0^{1-epsilon}dfrac{xln xlnleft(tfrac{1+x}{1-x}right)}{1-x^2}dx-lnleft(dfrac{epsilon}{2+epsilon}right)R(1)+int_0^1dfrac{ln tlnleft(tfrac{1-t(1-epsilon)}{1+t(1-epsilon)}right)}{t}dt-\
&int_0^1dfrac{tln tlnleft(tfrac{1+t(1-epsilon)}{1-t(1-epsilon)}right)}{1-t^2}dt
end{align}
Since,
$$lim_{epsilonrightarrow 0}left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)=0$$
and,
$$lim_{epsilonrightarrow 0}R(1)lnleft(tfrac{1-(1-epsilon)^2}{epsilon}right)=R(1)ln 2$$
then,
$$boxed{lim_{epsilonrightarrow 0}J(epsilon)=2R(1)ln 2+int_0^1dfrac{ln tlnleft(tfrac{1-t}{1+t}right)}{t}dt}$$
and then,
begin{align}
int_0^1dfrac{ln tlnleft(tfrac{1-t}{1+t}right)}{t}dt&=int_0^1dfrac{ln tleft(ln(1-t)-ln(1+t)right)}{t}dt\
&=-2int_0^1left(sum_{n=0}^{infty}dfrac{t^{2n}}{2n+1}right)ln tdt\
&=-2sum_{n=0}^{infty}left(dfrac{1}{2n+1}int_0^1 t^{2n}ln tdtright)\
&=2sum_{n=0}^{infty}dfrac{1}{(2n+1)^3}\
&=2left(sum_{n=1}^{infty}dfrac{1}{n^3}-sum_{n=1}^{infty}dfrac{1}{(2n)^3}right)\
&=2left(zeta(3)-dfrac{1}{8}zeta(2)right)\
&=dfrac{7}{4}zeta(3)\
end{align}
and,
begin{align}
displaystyle R(1)&=int_0^1dfrac{ln x}{1-x^2}dx\
&=int_0^1 left(sum_{n=0}^{infty}x^{2n}right)ln xdx\
&=sum_{n=0}^{infty}left(int_0^1 x^{2n}ln x dxright)\
&=-sum_{n=0}^{infty}dfrac{1}{(2n+1)^2}\
&=sum_{n=1}^{infty}dfrac{1}{(2n)^2}-sum_{n=0}^{infty}dfrac{1}{n^2}\
&=dfrac{1}{4}zeta(2)-zeta(2)\
&=-dfrac{3}{4}zeta(2)\
&=-dfrac{pi^2}{8}
end{align}
Therefore,
$$boxed{I=dfrac{7}{4}zeta(3)--dfrac{1}{4}pi^2ln 2}$$
answered Aug 21 '16 at 23:22
FDPFDP
5,33211524
$endgroup$
add a comment |
$begingroup$
$$displaystyle I=int_0^1 dfrac{ln xln(1-x^2)}{1-x^2}dx$$
Define the function $R$ on $[0;1]$,
$$R(x)=int_0^xdfrac{ln t}{1-t^2}dt=int_0^1dfrac{xln(tx)}{1-t^2x^2}dt$$
Let $epsilon$, real, such that $0<epsilon<1$.
begin{align}
J(epsilon)&=Big[left(R(x)-R(1)right)ln(1-x^2)Big]_0^{1-epsilon}+int_0^{1-epsilon} dfrac{2xleft(R(x)-R(1)right)}{1-x^2}dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+int_0^{1-epsilon} dfrac{2xR(x)}{1-x^2}dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&int_0^{1-epsilon}left(int_0^1dfrac{2x^2ln(tx)}{(1-x^2)(1-t^2x^2)}dtright)dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&int_0^{1-epsilon}left(int_0^1dfrac{2x^2ln x}{(1-x^2)(1-t^2x^2)}dtright)dx+int_0^1left(int_0^{1-epsilon}dfrac{2x^2ln t}{(1-x^2)(1-t^2x^2)}dxright)dt\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&displaystyleint_0^{1-epsilon}left[dfrac{xln xlnleft(tfrac{1+tx}{1-tx}right)}{1-x^2}right]_{t=0}^{t=1}dx+int_0^1 left[dfrac{ln tlnleft(tfrac{1-x}{1+x}right)}{t^2-1}+dfrac{ln tlnleft(tfrac{1-tx}{1+tx}right)}{t}-dfrac{tln tlnleft(tfrac{1+tx}{1-tx}right)}{1-t^2}right]_{x=0}^{x=1-epsilon}dt\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&displaystyleint_0^{1-epsilon}dfrac{xln xlnleft(tfrac{1+x}{1-x}right)}{1-x^2}dx-lnleft(dfrac{epsilon}{2+epsilon}right)R(1)+int_0^1dfrac{ln tlnleft(tfrac{1-t(1-epsilon)}{1+t(1-epsilon)}right)}{t}dt-\
&int_0^1dfrac{tln tlnleft(tfrac{1+t(1-epsilon)}{1-t(1-epsilon)}right)}{1-t^2}dt
end{align}
Since,
$$lim_{epsilonrightarrow 0}left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)=0$$
and,
$$lim_{epsilonrightarrow 0}R(1)lnleft(tfrac{1-(1-epsilon)^2}{epsilon}right)=R(1)ln 2$$
then,
$$boxed{lim_{epsilonrightarrow 0}J(epsilon)=2R(1)ln 2+int_0^1dfrac{ln tlnleft(tfrac{1-t}{1+t}right)}{t}dt}$$
and then,
begin{align}
int_0^1dfrac{ln tlnleft(tfrac{1-t}{1+t}right)}{t}dt&=int_0^1dfrac{ln tleft(ln(1-t)-ln(1+t)right)}{t}dt\
&=-2int_0^1left(sum_{n=0}^{infty}dfrac{t^{2n}}{2n+1}right)ln tdt\
&=-2sum_{n=0}^{infty}left(dfrac{1}{2n+1}int_0^1 t^{2n}ln tdtright)\
&=2sum_{n=0}^{infty}dfrac{1}{(2n+1)^3}\
&=2left(sum_{n=1}^{infty}dfrac{1}{n^3}-sum_{n=1}^{infty}dfrac{1}{(2n)^3}right)\
&=2left(zeta(3)-dfrac{1}{8}zeta(2)right)\
&=dfrac{7}{4}zeta(3)\
end{align}
and,
begin{align}
displaystyle R(1)&=int_0^1dfrac{ln x}{1-x^2}dx\
&=int_0^1 left(sum_{n=0}^{infty}x^{2n}right)ln xdx\
&=sum_{n=0}^{infty}left(int_0^1 x^{2n}ln x dxright)\
&=-sum_{n=0}^{infty}dfrac{1}{(2n+1)^2}\
&=sum_{n=1}^{infty}dfrac{1}{(2n)^2}-sum_{n=0}^{infty}dfrac{1}{n^2}\
&=dfrac{1}{4}zeta(2)-zeta(2)\
&=-dfrac{3}{4}zeta(2)\
&=-dfrac{pi^2}{8}
end{align}
Therefore,
$$boxed{I=dfrac{7}{4}zeta(3)--dfrac{1}{4}pi^2ln 2}$$
answered Aug 21 '16 at 23:22
FDPFDP
5,33211524
$endgroup$
$$displaystyle I=int_0^1 dfrac{ln xln(1-x^2)}{1-x^2}dx$$
Define the function $R$ on $[0;1]$,
$$R(x)=int_0^xdfrac{ln t}{1-t^2}dt=int_0^1dfrac{xln(tx)}{1-t^2x^2}dt$$
Let $epsilon$, real, such that $0<epsilon<1$.
begin{align}
J(epsilon)&=Big[left(R(x)-R(1)right)ln(1-x^2)Big]_0^{1-epsilon}+int_0^{1-epsilon} dfrac{2xleft(R(x)-R(1)right)}{1-x^2}dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+int_0^{1-epsilon} dfrac{2xR(x)}{1-x^2}dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&int_0^{1-epsilon}left(int_0^1dfrac{2x^2ln(tx)}{(1-x^2)(1-t^2x^2)}dtright)dx\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&int_0^{1-epsilon}left(int_0^1dfrac{2x^2ln x}{(1-x^2)(1-t^2x^2)}dtright)dx+int_0^1left(int_0^{1-epsilon}dfrac{2x^2ln t}{(1-x^2)(1-t^2x^2)}dxright)dt\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&displaystyleint_0^{1-epsilon}left[dfrac{xln xlnleft(tfrac{1+tx}{1-tx}right)}{1-x^2}right]_{t=0}^{t=1}dx+int_0^1 left[dfrac{ln tlnleft(tfrac{1-x}{1+x}right)}{t^2-1}+dfrac{ln tlnleft(tfrac{1-tx}{1+tx}right)}{t}-dfrac{tln tlnleft(tfrac{1+tx}{1-tx}right)}{1-t^2}right]_{x=0}^{x=1-epsilon}dt\
&=left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)+R(1)lnleft(1-(1-epsilon)^2right)+\
&displaystyleint_0^{1-epsilon}dfrac{xln xlnleft(tfrac{1+x}{1-x}right)}{1-x^2}dx-lnleft(dfrac{epsilon}{2+epsilon}right)R(1)+int_0^1dfrac{ln tlnleft(tfrac{1-t(1-epsilon)}{1+t(1-epsilon)}right)}{t}dt-\
&int_0^1dfrac{tln tlnleft(tfrac{1+t(1-epsilon)}{1-t(1-epsilon)}right)}{1-t^2}dt
end{align}
Since,
$$lim_{epsilonrightarrow 0}left(R(1-epsilon)-R(1)right)ln(1-(1-epsilon)^2)=0$$
and,
$$lim_{epsilonrightarrow 0}R(1)lnleft(tfrac{1-(1-epsilon)^2}{epsilon}right)=R(1)ln 2$$
then,
$$boxed{lim_{epsilonrightarrow 0}J(epsilon)=2R(1)ln 2+int_0^1dfrac{ln tlnleft(tfrac{1-t}{1+t}right)}{t}dt}$$
and then,
begin{align}
int_0^1dfrac{ln tlnleft(tfrac{1-t}{1+t}right)}{t}dt&=int_0^1dfrac{ln tleft(ln(1-t)-ln(1+t)right)}{t}dt\
&=-2int_0^1left(sum_{n=0}^{infty}dfrac{t^{2n}}{2n+1}right)ln tdt\
&=-2sum_{n=0}^{infty}left(dfrac{1}{2n+1}int_0^1 t^{2n}ln tdtright)\
&=2sum_{n=0}^{infty}dfrac{1}{(2n+1)^3}\
&=2left(sum_{n=1}^{infty}dfrac{1}{n^3}-sum_{n=1}^{infty}dfrac{1}{(2n)^3}right)\
&=2left(zeta(3)-dfrac{1}{8}zeta(2)right)\
&=dfrac{7}{4}zeta(3)\
end{align}
and,
begin{align}
displaystyle R(1)&=int_0^1dfrac{ln x}{1-x^2}dx\
&=int_0^1 left(sum_{n=0}^{infty}x^{2n}right)ln xdx\
&=sum_{n=0}^{infty}left(int_0^1 x^{2n}ln x dxright)\
&=-sum_{n=0}^{infty}dfrac{1}{(2n+1)^2}\
&=sum_{n=1}^{infty}dfrac{1}{(2n)^2}-sum_{n=0}^{infty}dfrac{1}{n^2}\
&=dfrac{1}{4}zeta(2)-zeta(2)\
&=-dfrac{3}{4}zeta(2)\
&=-dfrac{pi^2}{8}
end{align}
Therefore,
$$boxed{I=dfrac{7}{4}zeta(3)--dfrac{1}{4}pi^2ln 2}$$
answered Aug 21 '16 at 23:22
FDPFDP
5,33211524
answered Aug 21 '16 at 23:22
FDPFDP
5,33211524
answered Aug 21 '16 at 23:22
FDPFDP
5,33211524
answered Aug 21 '16 at 23:22
FDPFDP
5,33211524
5,33211524
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$begingroup$
no idea. Why do you think there is one?
$endgroup$
– Will Jagy
Feb 26 '13 at 3:34
2
$begingroup$
Mathematica says the answer is $frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$ which probably suggests it knows a way how to reduce it to known series. One attempt may be at trying to go backwards from Mathematica's answer and see if you can rediscover what it did.
$endgroup$
– muzzlator
Feb 26 '13 at 3:40
2
$begingroup$
It would have been better to put that information into the question, both for motivation and to help others with their thoughts. So post it as "Mathematica says $sum_{n=1}^inftyfrac{H_n}{(2n+1)^2}=frac{1}{4} (-pi^2 ln{2} + 7 zeta(3))$. Is there an easy to demonstrate this?" If you have information on why you were motivated to ask Mathematica, that might be useful, too.
$endgroup$
– Ross Millikan
Feb 26 '13 at 3:44
1
$begingroup$
Some similar series are on this guy's blog, the same guy's blog, and this other guy's blog, but they omit the identity you're after.
$endgroup$
– Aeolian
Feb 26 '13 at 4:25
1
$begingroup$
@joriki: you are right. Then OP came back with some motivation. There is progress.
$endgroup$
– Ross Millikan
Feb 26 '13 at 4:35