prove $sum_{n=0}^{infty}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$
$begingroup$
I am seeking alternate proofs for
$$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
Here's mine:
Recall that, for $xin(0,2)$,
$$frac1x=sum_{ngeq0}(1-x)^n$$
Hence we have that, for $frac{1-sqrt5}2<x<frac{1+sqrt5}2$,
$$frac1{x^2-x+1}=sum_{ngeq0}x^n(1-x)^n$$
Which gives
$$
begin{align}
int_0^1frac{mathrm dx}{x^2-x+1}=&int_0^1sum_{ngeq0}x^n(1-x)^nmathrm dx\
=&sum_{ngeq0}int_0^1x^n(1-x)^nmathrm dx\
=&sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}
end{align}
$$
That last step was from the definition of the Beta function:
$$B(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
Setting $I=int_0^1frac{mathrm dx}{x^2-x+1}$, we complete the square:
$$I=int_0^1frac{mathrm dx}{(x-frac12)^2+frac34}$$
Preforming the substitution $x=frac{1+3^{1/2}tan u}2$, we have
$$I=frac{3^{1/2}}2int_{-pi/6}^{pi/6}frac{sec^2u mathrm du}{frac34+frac34tan^2u}$$
And with a little simplification, we arrive at
$$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
Which brings me to my questions. How else can we prove this? What other series can be evaluated using similar tricks? Have fun!
sequences-and-series soft-question special-functions alternative-proof big-list
asked Dec 2 '18 at 21:29
clathratusclathratus
3,862333
$endgroup$
add a comment |
$begingroup$
I am seeking alternate proofs for
$$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
Here's mine:
Recall that, for $xin(0,2)$,
$$frac1x=sum_{ngeq0}(1-x)^n$$
Hence we have that, for $frac{1-sqrt5}2<x<frac{1+sqrt5}2$,
$$frac1{x^2-x+1}=sum_{ngeq0}x^n(1-x)^n$$
Which gives
$$
begin{align}
int_0^1frac{mathrm dx}{x^2-x+1}=&int_0^1sum_{ngeq0}x^n(1-x)^nmathrm dx\
=&sum_{ngeq0}int_0^1x^n(1-x)^nmathrm dx\
=&sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}
end{align}
$$
That last step was from the definition of the Beta function:
$$B(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
Setting $I=int_0^1frac{mathrm dx}{x^2-x+1}$, we complete the square:
$$I=int_0^1frac{mathrm dx}{(x-frac12)^2+frac34}$$
Preforming the substitution $x=frac{1+3^{1/2}tan u}2$, we have
$$I=frac{3^{1/2}}2int_{-pi/6}^{pi/6}frac{sec^2u mathrm du}{frac34+frac34tan^2u}$$
And with a little simplification, we arrive at
$$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
Which brings me to my questions. How else can we prove this? What other series can be evaluated using similar tricks? Have fun!
sequences-and-series soft-question special-functions alternative-proof big-list
asked Dec 2 '18 at 21:29
clathratusclathratus
3,862333
$endgroup$
add a comment |
$begingroup$
I am seeking alternate proofs for
$$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
Here's mine:
Recall that, for $xin(0,2)$,
$$frac1x=sum_{ngeq0}(1-x)^n$$
Hence we have that, for $frac{1-sqrt5}2<x<frac{1+sqrt5}2$,
$$frac1{x^2-x+1}=sum_{ngeq0}x^n(1-x)^n$$
Which gives
$$
begin{align}
int_0^1frac{mathrm dx}{x^2-x+1}=&int_0^1sum_{ngeq0}x^n(1-x)^nmathrm dx\
=&sum_{ngeq0}int_0^1x^n(1-x)^nmathrm dx\
=&sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}
end{align}
$$
That last step was from the definition of the Beta function:
$$B(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
Setting $I=int_0^1frac{mathrm dx}{x^2-x+1}$, we complete the square:
$$I=int_0^1frac{mathrm dx}{(x-frac12)^2+frac34}$$
Preforming the substitution $x=frac{1+3^{1/2}tan u}2$, we have
$$I=frac{3^{1/2}}2int_{-pi/6}^{pi/6}frac{sec^2u mathrm du}{frac34+frac34tan^2u}$$
And with a little simplification, we arrive at
$$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
Which brings me to my questions. How else can we prove this? What other series can be evaluated using similar tricks? Have fun!
sequences-and-series soft-question special-functions alternative-proof big-list
asked Dec 2 '18 at 21:29
clathratusclathratus
3,862333
$endgroup$
I am seeking alternate proofs for
$$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
Here's mine:
Recall that, for $xin(0,2)$,
$$frac1x=sum_{ngeq0}(1-x)^n$$
Hence we have that, for $frac{1-sqrt5}2<x<frac{1+sqrt5}2$,
$$frac1{x^2-x+1}=sum_{ngeq0}x^n(1-x)^n$$
Which gives
$$
begin{align}
int_0^1frac{mathrm dx}{x^2-x+1}=&int_0^1sum_{ngeq0}x^n(1-x)^nmathrm dx\
=&sum_{ngeq0}int_0^1x^n(1-x)^nmathrm dx\
=&sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}
end{align}
$$
That last step was from the definition of the Beta function:
$$B(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
Setting $I=int_0^1frac{mathrm dx}{x^2-x+1}$, we complete the square:
$$I=int_0^1frac{mathrm dx}{(x-frac12)^2+frac34}$$
Preforming the substitution $x=frac{1+3^{1/2}tan u}2$, we have
$$I=frac{3^{1/2}}2int_{-pi/6}^{pi/6}frac{sec^2u mathrm du}{frac34+frac34tan^2u}$$
And with a little simplification, we arrive at
$$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
Which brings me to my questions. How else can we prove this? What other series can be evaluated using similar tricks? Have fun!
sequences-and-series soft-question special-functions alternative-proof big-list
sequences-and-series soft-question special-functions alternative-proof big-list
asked Dec 2 '18 at 21:29
clathratusclathratus
3,862333
asked Dec 2 '18 at 21:29
clathratusclathratus
3,862333
edited Dec 3 '18 at 3:47
clathratus
asked Dec 2 '18 at 21:29
clathratusclathratus
3,862333
asked Dec 2 '18 at 21:29
clathratusclathratus
3,862333
asked Dec 2 '18 at 21:29
clathratusclathratus
3,862333
3,862333
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A not-really-alternative approach:
$$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=sum_{ngeq 0}frac{1}{(2n+1)binom{2n}{n}}=sum_{ngeq 1}frac{2}{nbinom{2n}{n}} $$
and since $sum_{ngeq 1}frac{z^{2n}}{n^2binom{2n}{n}}$ equals $2arcsin^2tfrac{z}{2}$ (by the Lagrange inversion theorem or equivalent approaches), we simply have
$$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=frac{d}{dz}left.2arcsin^2tfrac{z}{2}right|_{z=1}=left.frac{4arcsin(z/2)}{sqrt{4-z^2}}right|_{z=1}=frac{2pi}{3sqrt{3}}.$$
answered Dec 2 '18 at 21:41
Jack D'AurizioJack D'Aurizio
289k33281660
$endgroup$
$begingroup$
different enough for me! Thanks Jack (+1)
$endgroup$
– clathratus
Dec 2 '18 at 21:42
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
Hereafter, $ds{Psi}$ is the
Digamma Function: $ds{Psipars{z} equiv totald{lnpars{Gammapars{z}}}{z}}$ where $ds{Gamma}$ is the
Gamma Function.
begin{align}
&bbox[10px,#ffd]{int_{0}^{1}{dd x over x^{2} - x + 1}} =
int_{0}^{1}{1 + x over 1 + x^{3}},dd x
\[5mm] = &
int_{0}^{1}{1 + x - x^{3} - x^{4}over 1 - x^{6}},dd x
\[5mm] = &
{1 over 6}int_{0}^{1}{x^{-5/6} + x^{-2/3} - x^{-1/3} - x^{-1/6} over 1 - x},dd x
\[5mm] = &
{1 over 6}left{bracks{%
int_{0}^{1}{1 - x^{-1/3} over 1 - x},dd x -
int_{0}^{1}{1 - x^{-2/3} over 1 - x},dd x}
right.
\[2mm] & +
left.phantom{left{,right.}
bracks{int_{0}^{1}{1 - x^{-1/6} over 1 - x},dd x -
int_{0}^{1}{1 - x^{-5/6} over 1 - x},dd x}
label{1}tag{1}
right}
\[5mm] = &
{1 over 6}braces{!%
bracks{Psipars{2 over 3} -
Psipars{1 over 3}} +
bracks{Psipars{5 over 6} -
Psipars{1 over 6}}!}label{2}tag{2}
\[5mm] = &
{1 over 6}bracks{picotpars{pi over 3} +
picotpars{pi over 6}} =
{1 over 6}pars{{root{3} over 3} + root{3}}pi
label{3}tag{3}
\[5mm] = &
bbx{{2root{3} over 9},pi} approx 1.2092
end{align}
eqref{1} is evaluated with
$ds{mathbf{color{black}{6.3.22}}}$ A & S Identity. In eqref{2} and eqref{3}, I used the
Euler Reflection Formula.
answered Dec 4 '18 at 5:57
Felix MarinFelix Marin
67.5k7107141
$endgroup$
$begingroup$
Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
$endgroup$
– clathratus
Dec 4 '18 at 17:41
1
$begingroup$
@clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
$endgroup$
– Felix Marin
Dec 4 '18 at 19:52
$begingroup$
I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
$endgroup$
– clathratus
Dec 8 '18 at 21:26
$begingroup$
Thanks @clathratus
$endgroup$
– Felix Marin
Dec 8 '18 at 23:55
add a comment |
$begingroup$
Still another way is through the Hypergeometric function.
Since the term of the sum is
$$
t_{,n} = {{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2n + 2} right)}}
$$
and
$$
eqalign{
& t_{,0} = {{Gamma left( 1 right)^{,2} } over {Gamma left( 2 right)}} = 1 cr
& {{t_{,n + 1} } over {t_{,n} }} = {{Gamma left( {n + 2} right)^{,2} } over {Gamma left( {2n + 4} right)}}{{Gamma left( {2n + 2} right)}
over {Gamma left( {n + 1} right)^{,2} }} = {{left( {n + 1} right)^{,2} } over {left( {2n + 3} right)left( {2n + 2} right)}} cr
& = {{left( {n + 1} right)left( {n + 1} right)} over {left( {n + 3/2} right)}}{{1/4} over {left( {n + 1} right)}} cr}
$$
then
$$
eqalign{
& sumlimits_{n = 0}^infty {{{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2left( {n + 1} right)} right)}}}
= {}_2F_{,1} left( {left. {matrix{ {1,;1} cr {3/2} cr } ;} right|;1/4} right) = cr
& = {{arcsin left( {sqrt {1/4} } right)} over {sqrt {1 - 1/4} sqrt {1/4} }} = {{4pi } over {6sqrt 3 }} = {{2pi } over {3sqrt 3 }} cr}
$$
answered Dec 4 '18 at 23:19
G CabG Cab
18.6k31238
$endgroup$
$begingroup$
Cool! I really like any approach that involves hypergeometric functions.
$endgroup$
– clathratus
Dec 5 '18 at 0:33
$begingroup$
@clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
$endgroup$
– G Cab
Dec 5 '18 at 1:13
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A not-really-alternative approach:
$$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=sum_{ngeq 0}frac{1}{(2n+1)binom{2n}{n}}=sum_{ngeq 1}frac{2}{nbinom{2n}{n}} $$
and since $sum_{ngeq 1}frac{z^{2n}}{n^2binom{2n}{n}}$ equals $2arcsin^2tfrac{z}{2}$ (by the Lagrange inversion theorem or equivalent approaches), we simply have
$$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=frac{d}{dz}left.2arcsin^2tfrac{z}{2}right|_{z=1}=left.frac{4arcsin(z/2)}{sqrt{4-z^2}}right|_{z=1}=frac{2pi}{3sqrt{3}}.$$
answered Dec 2 '18 at 21:41
Jack D'AurizioJack D'Aurizio
289k33281660
$endgroup$
$begingroup$
different enough for me! Thanks Jack (+1)
$endgroup$
– clathratus
Dec 2 '18 at 21:42
add a comment |
$begingroup$
A not-really-alternative approach:
$$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=sum_{ngeq 0}frac{1}{(2n+1)binom{2n}{n}}=sum_{ngeq 1}frac{2}{nbinom{2n}{n}} $$
and since $sum_{ngeq 1}frac{z^{2n}}{n^2binom{2n}{n}}$ equals $2arcsin^2tfrac{z}{2}$ (by the Lagrange inversion theorem or equivalent approaches), we simply have
$$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=frac{d}{dz}left.2arcsin^2tfrac{z}{2}right|_{z=1}=left.frac{4arcsin(z/2)}{sqrt{4-z^2}}right|_{z=1}=frac{2pi}{3sqrt{3}}.$$
answered Dec 2 '18 at 21:41
Jack D'AurizioJack D'Aurizio
289k33281660
$endgroup$
$begingroup$
different enough for me! Thanks Jack (+1)
$endgroup$
– clathratus
Dec 2 '18 at 21:42
add a comment |
$begingroup$
A not-really-alternative approach:
$$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=sum_{ngeq 0}frac{1}{(2n+1)binom{2n}{n}}=sum_{ngeq 1}frac{2}{nbinom{2n}{n}} $$
and since $sum_{ngeq 1}frac{z^{2n}}{n^2binom{2n}{n}}$ equals $2arcsin^2tfrac{z}{2}$ (by the Lagrange inversion theorem or equivalent approaches), we simply have
$$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=frac{d}{dz}left.2arcsin^2tfrac{z}{2}right|_{z=1}=left.frac{4arcsin(z/2)}{sqrt{4-z^2}}right|_{z=1}=frac{2pi}{3sqrt{3}}.$$
answered Dec 2 '18 at 21:41
Jack D'AurizioJack D'Aurizio
289k33281660
$endgroup$
A not-really-alternative approach:
$$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=sum_{ngeq 0}frac{1}{(2n+1)binom{2n}{n}}=sum_{ngeq 1}frac{2}{nbinom{2n}{n}} $$
and since $sum_{ngeq 1}frac{z^{2n}}{n^2binom{2n}{n}}$ equals $2arcsin^2tfrac{z}{2}$ (by the Lagrange inversion theorem or equivalent approaches), we simply have
$$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=frac{d}{dz}left.2arcsin^2tfrac{z}{2}right|_{z=1}=left.frac{4arcsin(z/2)}{sqrt{4-z^2}}right|_{z=1}=frac{2pi}{3sqrt{3}}.$$
answered Dec 2 '18 at 21:41
Jack D'AurizioJack D'Aurizio
289k33281660
answered Dec 2 '18 at 21:41
Jack D'AurizioJack D'Aurizio
289k33281660
answered Dec 2 '18 at 21:41
Jack D'AurizioJack D'Aurizio
289k33281660
answered Dec 2 '18 at 21:41
Jack D'AurizioJack D'Aurizio
289k33281660
289k33281660
$begingroup$
different enough for me! Thanks Jack (+1)
$endgroup$
– clathratus
Dec 2 '18 at 21:42
add a comment |
$begingroup$
different enough for me! Thanks Jack (+1)
$endgroup$
– clathratus
Dec 2 '18 at 21:42
$begingroup$
different enough for me! Thanks Jack (+1)
$endgroup$
– clathratus
Dec 2 '18 at 21:42
$begingroup$
different enough for me! Thanks Jack (+1)
$endgroup$
– clathratus
Dec 2 '18 at 21:42
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
Hereafter, $ds{Psi}$ is the
Digamma Function: $ds{Psipars{z} equiv totald{lnpars{Gammapars{z}}}{z}}$ where $ds{Gamma}$ is the
Gamma Function.
begin{align}
&bbox[10px,#ffd]{int_{0}^{1}{dd x over x^{2} - x + 1}} =
int_{0}^{1}{1 + x over 1 + x^{3}},dd x
\[5mm] = &
int_{0}^{1}{1 + x - x^{3} - x^{4}over 1 - x^{6}},dd x
\[5mm] = &
{1 over 6}int_{0}^{1}{x^{-5/6} + x^{-2/3} - x^{-1/3} - x^{-1/6} over 1 - x},dd x
\[5mm] = &
{1 over 6}left{bracks{%
int_{0}^{1}{1 - x^{-1/3} over 1 - x},dd x -
int_{0}^{1}{1 - x^{-2/3} over 1 - x},dd x}
right.
\[2mm] & +
left.phantom{left{,right.}
bracks{int_{0}^{1}{1 - x^{-1/6} over 1 - x},dd x -
int_{0}^{1}{1 - x^{-5/6} over 1 - x},dd x}
label{1}tag{1}
right}
\[5mm] = &
{1 over 6}braces{!%
bracks{Psipars{2 over 3} -
Psipars{1 over 3}} +
bracks{Psipars{5 over 6} -
Psipars{1 over 6}}!}label{2}tag{2}
\[5mm] = &
{1 over 6}bracks{picotpars{pi over 3} +
picotpars{pi over 6}} =
{1 over 6}pars{{root{3} over 3} + root{3}}pi
label{3}tag{3}
\[5mm] = &
bbx{{2root{3} over 9},pi} approx 1.2092
end{align}
eqref{1} is evaluated with
$ds{mathbf{color{black}{6.3.22}}}$ A & S Identity. In eqref{2} and eqref{3}, I used the
Euler Reflection Formula.
answered Dec 4 '18 at 5:57
Felix MarinFelix Marin
67.5k7107141
$endgroup$
$begingroup$
Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
$endgroup$
– clathratus
Dec 4 '18 at 17:41
1
$begingroup$
@clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
$endgroup$
– Felix Marin
Dec 4 '18 at 19:52
$begingroup$
I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
$endgroup$
– clathratus
Dec 8 '18 at 21:26
$begingroup$
Thanks @clathratus
$endgroup$
– Felix Marin
Dec 8 '18 at 23:55
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
Hereafter, $ds{Psi}$ is the
Digamma Function: $ds{Psipars{z} equiv totald{lnpars{Gammapars{z}}}{z}}$ where $ds{Gamma}$ is the
Gamma Function.
begin{align}
&bbox[10px,#ffd]{int_{0}^{1}{dd x over x^{2} - x + 1}} =
int_{0}^{1}{1 + x over 1 + x^{3}},dd x
\[5mm] = &
int_{0}^{1}{1 + x - x^{3} - x^{4}over 1 - x^{6}},dd x
\[5mm] = &
{1 over 6}int_{0}^{1}{x^{-5/6} + x^{-2/3} - x^{-1/3} - x^{-1/6} over 1 - x},dd x
\[5mm] = &
{1 over 6}left{bracks{%
int_{0}^{1}{1 - x^{-1/3} over 1 - x},dd x -
int_{0}^{1}{1 - x^{-2/3} over 1 - x},dd x}
right.
\[2mm] & +
left.phantom{left{,right.}
bracks{int_{0}^{1}{1 - x^{-1/6} over 1 - x},dd x -
int_{0}^{1}{1 - x^{-5/6} over 1 - x},dd x}
label{1}tag{1}
right}
\[5mm] = &
{1 over 6}braces{!%
bracks{Psipars{2 over 3} -
Psipars{1 over 3}} +
bracks{Psipars{5 over 6} -
Psipars{1 over 6}}!}label{2}tag{2}
\[5mm] = &
{1 over 6}bracks{picotpars{pi over 3} +
picotpars{pi over 6}} =
{1 over 6}pars{{root{3} over 3} + root{3}}pi
label{3}tag{3}
\[5mm] = &
bbx{{2root{3} over 9},pi} approx 1.2092
end{align}
eqref{1} is evaluated with
$ds{mathbf{color{black}{6.3.22}}}$ A & S Identity. In eqref{2} and eqref{3}, I used the
Euler Reflection Formula.
answered Dec 4 '18 at 5:57
Felix MarinFelix Marin
67.5k7107141
$endgroup$
$begingroup$
Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
$endgroup$
– clathratus
Dec 4 '18 at 17:41
1
$begingroup$
@clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
$endgroup$
– Felix Marin
Dec 4 '18 at 19:52
$begingroup$
I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
$endgroup$
– clathratus
Dec 8 '18 at 21:26
$begingroup$
Thanks @clathratus
$endgroup$
– Felix Marin
Dec 8 '18 at 23:55
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
Hereafter, $ds{Psi}$ is the
Digamma Function: $ds{Psipars{z} equiv totald{lnpars{Gammapars{z}}}{z}}$ where $ds{Gamma}$ is the
Gamma Function.
begin{align}
&bbox[10px,#ffd]{int_{0}^{1}{dd x over x^{2} - x + 1}} =
int_{0}^{1}{1 + x over 1 + x^{3}},dd x
\[5mm] = &
int_{0}^{1}{1 + x - x^{3} - x^{4}over 1 - x^{6}},dd x
\[5mm] = &
{1 over 6}int_{0}^{1}{x^{-5/6} + x^{-2/3} - x^{-1/3} - x^{-1/6} over 1 - x},dd x
\[5mm] = &
{1 over 6}left{bracks{%
int_{0}^{1}{1 - x^{-1/3} over 1 - x},dd x -
int_{0}^{1}{1 - x^{-2/3} over 1 - x},dd x}
right.
\[2mm] & +
left.phantom{left{,right.}
bracks{int_{0}^{1}{1 - x^{-1/6} over 1 - x},dd x -
int_{0}^{1}{1 - x^{-5/6} over 1 - x},dd x}
label{1}tag{1}
right}
\[5mm] = &
{1 over 6}braces{!%
bracks{Psipars{2 over 3} -
Psipars{1 over 3}} +
bracks{Psipars{5 over 6} -
Psipars{1 over 6}}!}label{2}tag{2}
\[5mm] = &
{1 over 6}bracks{picotpars{pi over 3} +
picotpars{pi over 6}} =
{1 over 6}pars{{root{3} over 3} + root{3}}pi
label{3}tag{3}
\[5mm] = &
bbx{{2root{3} over 9},pi} approx 1.2092
end{align}
eqref{1} is evaluated with
$ds{mathbf{color{black}{6.3.22}}}$ A & S Identity. In eqref{2} and eqref{3}, I used the
Euler Reflection Formula.
answered Dec 4 '18 at 5:57
Felix MarinFelix Marin
67.5k7107141
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
Hereafter, $ds{Psi}$ is the
Digamma Function: $ds{Psipars{z} equiv totald{lnpars{Gammapars{z}}}{z}}$ where $ds{Gamma}$ is the
Gamma Function.
begin{align}
&bbox[10px,#ffd]{int_{0}^{1}{dd x over x^{2} - x + 1}} =
int_{0}^{1}{1 + x over 1 + x^{3}},dd x
\[5mm] = &
int_{0}^{1}{1 + x - x^{3} - x^{4}over 1 - x^{6}},dd x
\[5mm] = &
{1 over 6}int_{0}^{1}{x^{-5/6} + x^{-2/3} - x^{-1/3} - x^{-1/6} over 1 - x},dd x
\[5mm] = &
{1 over 6}left{bracks{%
int_{0}^{1}{1 - x^{-1/3} over 1 - x},dd x -
int_{0}^{1}{1 - x^{-2/3} over 1 - x},dd x}
right.
\[2mm] & +
left.phantom{left{,right.}
bracks{int_{0}^{1}{1 - x^{-1/6} over 1 - x},dd x -
int_{0}^{1}{1 - x^{-5/6} over 1 - x},dd x}
label{1}tag{1}
right}
\[5mm] = &
{1 over 6}braces{!%
bracks{Psipars{2 over 3} -
Psipars{1 over 3}} +
bracks{Psipars{5 over 6} -
Psipars{1 over 6}}!}label{2}tag{2}
\[5mm] = &
{1 over 6}bracks{picotpars{pi over 3} +
picotpars{pi over 6}} =
{1 over 6}pars{{root{3} over 3} + root{3}}pi
label{3}tag{3}
\[5mm] = &
bbx{{2root{3} over 9},pi} approx 1.2092
end{align}
eqref{1} is evaluated with
$ds{mathbf{color{black}{6.3.22}}}$ A & S Identity. In eqref{2} and eqref{3}, I used the
Euler Reflection Formula.
answered Dec 4 '18 at 5:57
Felix MarinFelix Marin
67.5k7107141
edited Dec 4 '18 at 19:56
answered Dec 4 '18 at 5:57
Felix MarinFelix Marin
67.5k7107141
answered Dec 4 '18 at 5:57
Felix MarinFelix Marin
67.5k7107141
answered Dec 4 '18 at 5:57
Felix MarinFelix Marin
67.5k7107141
67.5k7107141
$begingroup$
Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
$endgroup$
– clathratus
Dec 4 '18 at 17:41
1
$begingroup$
@clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
$endgroup$
– Felix Marin
Dec 4 '18 at 19:52
$begingroup$
I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
$endgroup$
– clathratus
Dec 8 '18 at 21:26
$begingroup$
Thanks @clathratus
$endgroup$
– Felix Marin
Dec 8 '18 at 23:55
add a comment |
$begingroup$
Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
$endgroup$
– clathratus
Dec 4 '18 at 17:41
1
$begingroup$
@clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
$endgroup$
– Felix Marin
Dec 4 '18 at 19:52
$begingroup$
I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
$endgroup$
– clathratus
Dec 8 '18 at 21:26
$begingroup$
Thanks @clathratus
$endgroup$
– Felix Marin
Dec 8 '18 at 23:55
$begingroup$
Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
$endgroup$
– clathratus
Dec 4 '18 at 17:41
$begingroup$
Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
$endgroup$
– clathratus
Dec 4 '18 at 17:41
1
1
$begingroup$
@clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
$endgroup$
– Felix Marin
Dec 4 '18 at 19:52
$begingroup$
@clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
$endgroup$
– Felix Marin
Dec 4 '18 at 19:52
$begingroup$
I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
$endgroup$
– clathratus
Dec 8 '18 at 21:26
$begingroup$
I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
$endgroup$
– clathratus
Dec 8 '18 at 21:26
$begingroup$
Thanks @clathratus
$endgroup$
– Felix Marin
Dec 8 '18 at 23:55
$begingroup$
Thanks @clathratus
$endgroup$
– Felix Marin
Dec 8 '18 at 23:55
add a comment |
$begingroup$
Still another way is through the Hypergeometric function.
Since the term of the sum is
$$
t_{,n} = {{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2n + 2} right)}}
$$
and
$$
eqalign{
& t_{,0} = {{Gamma left( 1 right)^{,2} } over {Gamma left( 2 right)}} = 1 cr
& {{t_{,n + 1} } over {t_{,n} }} = {{Gamma left( {n + 2} right)^{,2} } over {Gamma left( {2n + 4} right)}}{{Gamma left( {2n + 2} right)}
over {Gamma left( {n + 1} right)^{,2} }} = {{left( {n + 1} right)^{,2} } over {left( {2n + 3} right)left( {2n + 2} right)}} cr
& = {{left( {n + 1} right)left( {n + 1} right)} over {left( {n + 3/2} right)}}{{1/4} over {left( {n + 1} right)}} cr}
$$
then
$$
eqalign{
& sumlimits_{n = 0}^infty {{{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2left( {n + 1} right)} right)}}}
= {}_2F_{,1} left( {left. {matrix{ {1,;1} cr {3/2} cr } ;} right|;1/4} right) = cr
& = {{arcsin left( {sqrt {1/4} } right)} over {sqrt {1 - 1/4} sqrt {1/4} }} = {{4pi } over {6sqrt 3 }} = {{2pi } over {3sqrt 3 }} cr}
$$
answered Dec 4 '18 at 23:19
G CabG Cab
18.6k31238
$endgroup$
$begingroup$
Cool! I really like any approach that involves hypergeometric functions.
$endgroup$
– clathratus
Dec 5 '18 at 0:33
$begingroup$
@clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
$endgroup$
– G Cab
Dec 5 '18 at 1:13
add a comment |
$begingroup$
Still another way is through the Hypergeometric function.
Since the term of the sum is
$$
t_{,n} = {{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2n + 2} right)}}
$$
and
$$
eqalign{
& t_{,0} = {{Gamma left( 1 right)^{,2} } over {Gamma left( 2 right)}} = 1 cr
& {{t_{,n + 1} } over {t_{,n} }} = {{Gamma left( {n + 2} right)^{,2} } over {Gamma left( {2n + 4} right)}}{{Gamma left( {2n + 2} right)}
over {Gamma left( {n + 1} right)^{,2} }} = {{left( {n + 1} right)^{,2} } over {left( {2n + 3} right)left( {2n + 2} right)}} cr
& = {{left( {n + 1} right)left( {n + 1} right)} over {left( {n + 3/2} right)}}{{1/4} over {left( {n + 1} right)}} cr}
$$
then
$$
eqalign{
& sumlimits_{n = 0}^infty {{{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2left( {n + 1} right)} right)}}}
= {}_2F_{,1} left( {left. {matrix{ {1,;1} cr {3/2} cr } ;} right|;1/4} right) = cr
& = {{arcsin left( {sqrt {1/4} } right)} over {sqrt {1 - 1/4} sqrt {1/4} }} = {{4pi } over {6sqrt 3 }} = {{2pi } over {3sqrt 3 }} cr}
$$
answered Dec 4 '18 at 23:19
G CabG Cab
18.6k31238
$endgroup$
$begingroup$
Cool! I really like any approach that involves hypergeometric functions.
$endgroup$
– clathratus
Dec 5 '18 at 0:33
$begingroup$
@clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
$endgroup$
– G Cab
Dec 5 '18 at 1:13
add a comment |
$begingroup$
Still another way is through the Hypergeometric function.
Since the term of the sum is
$$
t_{,n} = {{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2n + 2} right)}}
$$
and
$$
eqalign{
& t_{,0} = {{Gamma left( 1 right)^{,2} } over {Gamma left( 2 right)}} = 1 cr
& {{t_{,n + 1} } over {t_{,n} }} = {{Gamma left( {n + 2} right)^{,2} } over {Gamma left( {2n + 4} right)}}{{Gamma left( {2n + 2} right)}
over {Gamma left( {n + 1} right)^{,2} }} = {{left( {n + 1} right)^{,2} } over {left( {2n + 3} right)left( {2n + 2} right)}} cr
& = {{left( {n + 1} right)left( {n + 1} right)} over {left( {n + 3/2} right)}}{{1/4} over {left( {n + 1} right)}} cr}
$$
then
$$
eqalign{
& sumlimits_{n = 0}^infty {{{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2left( {n + 1} right)} right)}}}
= {}_2F_{,1} left( {left. {matrix{ {1,;1} cr {3/2} cr } ;} right|;1/4} right) = cr
& = {{arcsin left( {sqrt {1/4} } right)} over {sqrt {1 - 1/4} sqrt {1/4} }} = {{4pi } over {6sqrt 3 }} = {{2pi } over {3sqrt 3 }} cr}
$$
answered Dec 4 '18 at 23:19
G CabG Cab
18.6k31238
$endgroup$
Still another way is through the Hypergeometric function.
Since the term of the sum is
$$
t_{,n} = {{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2n + 2} right)}}
$$
and
$$
eqalign{
& t_{,0} = {{Gamma left( 1 right)^{,2} } over {Gamma left( 2 right)}} = 1 cr
& {{t_{,n + 1} } over {t_{,n} }} = {{Gamma left( {n + 2} right)^{,2} } over {Gamma left( {2n + 4} right)}}{{Gamma left( {2n + 2} right)}
over {Gamma left( {n + 1} right)^{,2} }} = {{left( {n + 1} right)^{,2} } over {left( {2n + 3} right)left( {2n + 2} right)}} cr
& = {{left( {n + 1} right)left( {n + 1} right)} over {left( {n + 3/2} right)}}{{1/4} over {left( {n + 1} right)}} cr}
$$
then
$$
eqalign{
& sumlimits_{n = 0}^infty {{{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2left( {n + 1} right)} right)}}}
= {}_2F_{,1} left( {left. {matrix{ {1,;1} cr {3/2} cr } ;} right|;1/4} right) = cr
& = {{arcsin left( {sqrt {1/4} } right)} over {sqrt {1 - 1/4} sqrt {1/4} }} = {{4pi } over {6sqrt 3 }} = {{2pi } over {3sqrt 3 }} cr}
$$
answered Dec 4 '18 at 23:19
G CabG Cab
18.6k31238
answered Dec 4 '18 at 23:19
G CabG Cab
18.6k31238
answered Dec 4 '18 at 23:19
G CabG Cab
18.6k31238
answered Dec 4 '18 at 23:19
G CabG Cab
18.6k31238
18.6k31238
$begingroup$
Cool! I really like any approach that involves hypergeometric functions.
$endgroup$
– clathratus
Dec 5 '18 at 0:33
$begingroup$
@clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
$endgroup$
– G Cab
Dec 5 '18 at 1:13
add a comment |
$begingroup$
Cool! I really like any approach that involves hypergeometric functions.
$endgroup$
– clathratus
Dec 5 '18 at 0:33
$begingroup$
@clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
$endgroup$
– G Cab
Dec 5 '18 at 1:13
$begingroup$
Cool! I really like any approach that involves hypergeometric functions.
$endgroup$
– clathratus
Dec 5 '18 at 0:33
$begingroup$
Cool! I really like any approach that involves hypergeometric functions.
$endgroup$
– clathratus
Dec 5 '18 at 0:33
$begingroup$
@clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
$endgroup$
– G Cab
Dec 5 '18 at 1:13
$begingroup$
@clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
$endgroup$
– G Cab
Dec 5 '18 at 1:13
add a comment |
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