Interesting Integral $int_{-infty}^{infty}frac{e^{i nx}}{Gamma(alpha+x) Gamma(beta -x)}dx$
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I am asking this question out of curiosity.
$$int_{-infty}^{infty}frac{e^{i nx}}{Gamma(alpha+x) Gamma(beta -x)}dx = frac{ left(2cos frac{n}{2} right)^{alpha +beta-2}}{Gamma(alpha+beta-1)}e^{frac{in}{2}(beta - alpha)} quad |n|<pi quad text{and} quad Re(alpha+beta)>1$$
- How did Ramanujan derive this formula?
- I have noticed that Ramanujan has discovered many integrals involving gamma function. Is there a general method to deal with such integrals?
integration special-functions improper-integrals gamma-function
asked Mar 15 '13 at 11:28
Shobhit BhatnagarShobhit Bhatnagar
6,32053274
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add a comment |
$begingroup$
I am asking this question out of curiosity.
$$int_{-infty}^{infty}frac{e^{i nx}}{Gamma(alpha+x) Gamma(beta -x)}dx = frac{ left(2cos frac{n}{2} right)^{alpha +beta-2}}{Gamma(alpha+beta-1)}e^{frac{in}{2}(beta - alpha)} quad |n|<pi quad text{and} quad Re(alpha+beta)>1$$
- How did Ramanujan derive this formula?
- I have noticed that Ramanujan has discovered many integrals involving gamma function. Is there a general method to deal with such integrals?
integration special-functions improper-integrals gamma-function
asked Mar 15 '13 at 11:28
Shobhit BhatnagarShobhit Bhatnagar
6,32053274
$endgroup$
1
$begingroup$
Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
$endgroup$
– daniel
Mar 15 '13 at 13:44
$begingroup$
@daniel: I will try to find the book you suggested.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:12
add a comment |
$begingroup$
I am asking this question out of curiosity.
$$int_{-infty}^{infty}frac{e^{i nx}}{Gamma(alpha+x) Gamma(beta -x)}dx = frac{ left(2cos frac{n}{2} right)^{alpha +beta-2}}{Gamma(alpha+beta-1)}e^{frac{in}{2}(beta - alpha)} quad |n|<pi quad text{and} quad Re(alpha+beta)>1$$
- How did Ramanujan derive this formula?
- I have noticed that Ramanujan has discovered many integrals involving gamma function. Is there a general method to deal with such integrals?
integration special-functions improper-integrals gamma-function
asked Mar 15 '13 at 11:28
Shobhit BhatnagarShobhit Bhatnagar
6,32053274
$endgroup$
I am asking this question out of curiosity.
$$int_{-infty}^{infty}frac{e^{i nx}}{Gamma(alpha+x) Gamma(beta -x)}dx = frac{ left(2cos frac{n}{2} right)^{alpha +beta-2}}{Gamma(alpha+beta-1)}e^{frac{in}{2}(beta - alpha)} quad |n|<pi quad text{and} quad Re(alpha+beta)>1$$
- How did Ramanujan derive this formula?
- I have noticed that Ramanujan has discovered many integrals involving gamma function. Is there a general method to deal with such integrals?
integration special-functions improper-integrals gamma-function
integration special-functions improper-integrals gamma-function
asked Mar 15 '13 at 11:28
Shobhit BhatnagarShobhit Bhatnagar
6,32053274
asked Mar 15 '13 at 11:28
Shobhit BhatnagarShobhit Bhatnagar
6,32053274
edited Mar 15 '13 at 14:46
Shobhit Bhatnagar
asked Mar 15 '13 at 11:28
Shobhit BhatnagarShobhit Bhatnagar
6,32053274
asked Mar 15 '13 at 11:28
Shobhit BhatnagarShobhit Bhatnagar
6,32053274
asked Mar 15 '13 at 11:28
Shobhit BhatnagarShobhit Bhatnagar
6,32053274
6,32053274
1
$begingroup$
Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
$endgroup$
– daniel
Mar 15 '13 at 13:44
$begingroup$
@daniel: I will try to find the book you suggested.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:12
add a comment |
1
$begingroup$
Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
$endgroup$
– daniel
Mar 15 '13 at 13:44
$begingroup$
@daniel: I will try to find the book you suggested.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:12
1
1
$begingroup$
Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
$endgroup$
– daniel
Mar 15 '13 at 13:44
$begingroup$
Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
$endgroup$
– daniel
Mar 15 '13 at 13:44
$begingroup$
@daniel: I will try to find the book you suggested.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:12
$begingroup$
@daniel: I will try to find the book you suggested.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is not a complete answer but maybe it will save a trip to the library.
In Ramanujan's Collected Works at ch. 27 he says it is "well known" that
$$ intlimits_{-pi/2}^{pi/2}(cos x)^m e^{inx}dx = frac{pi}{2^m}frac{Gamma(1+m)}{Gammaleft(1+ frac{1}{2}(m+n)right)Gammaleft(1+frac{1}{2}(m-n)right)}. tag{1.1} $$
He says that from this and Fourier's Theorem we can derive the relation in your question. He may work out some details in the chapter (I haven't looked). If you accept or can prove (1.1) this might be enough to solve the problem.
Edit in response to comment: All he is saying is that a function can sometimes be represented as a Fourier series. I suspect the proof is in an earlier paper and ch. 27 is building on your relation. But he clearly says that "it follows" from (1.1) so it's a start. There are 4 papers cited in the notes...
Edit: Here is one way to do this proof. I thought we could take advantage of the (visually obvious) F-transform aspect of the relation, but this proof is very direct and maybe there is no simpler way...
edited Apr 13 '17 at 12:21
Community♦
1
answered Mar 15 '13 at 13:25
danieldaniel
6,29222259
$endgroup$
2
$begingroup$
You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
$endgroup$
– Sangchul Lee
Mar 15 '13 at 15:42
$begingroup$
@daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:53
$begingroup$
I think I understood it!
$endgroup$
– Shobhit Bhatnagar
Mar 17 '13 at 3:20
2
$begingroup$
Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
$endgroup$
– J. M. is not a mathematician
Apr 9 '13 at 1:41
$begingroup$
Using (1.1) implies $alpha=beta$
$endgroup$
– Anastasiya-Romanova 秀
Sep 26 '14 at 9:58
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
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newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x
={bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}}
,expo{ic npars{beta - alpha}/2}}$
$ds{verts{n} < pi,, Repars{alpha + beta} > 1}$.
Note that
begin{align}
&{1 over Gammapars{alpha + x}Gammapars{beta - x}}
={1 over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm]&={1 over Gammapars{alpha + beta - 1}},
{pars{alpha + beta - 2}! over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm] & ={1 over Gammapars{alpha + beta - 1}},
{alpha + beta - 2 choose alpha + x - 1}
end{align}
begin{align}&color{#c00000}{%
int_{-infty}^{infty}%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = & {1 over Gammapars{alpha + beta - 1}}
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}tag{1}
end{align}
begin{align}&color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=int_{-infty}^{infty}bracks{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha + x}},{dd z over 2piic}}expo{ic n x},dd x
\[3mm]&=-icoint_{verts{z} = 1}
{pars{1 + z}^{alpha + beta - 2} over z^{alpha}}braces{%
int_{-infty}^{infty}expo{icbracks{n - {rm Arg}pars{z}}x}
,{dd x over 2pi}},dd z
\[3mm]&=-ic overbrace{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha}},deltapars{n - {rm Arg}pars{z}},dd z}
^{ds{mbox{Set} z equiv expo{ic theta},,quadverts{theta} < pi}}
\[3mm]&=-icint_{-pi}^{pi}
{pars{1 + expo{ictheta}}^{alpha + beta - 2} over
expo{icalphatheta}},deltapars{n - theta},expo{ictheta}ic,ddtheta
=pars{1 + expo{ic n}}^{alpha + beta - 2}expo{icpars{1 - alpha}n}
\[3mm]&=expo{icpars{alpha + beta - 2}n/2}
pars{expo{-ic n/2} + expo{ic n/2}}^{alpha + beta - 2}
expo{icpars{1 - alpha}n}
\[3mm] & =
bracks{2cospars{{n over 2}}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
end{align}
$$
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=bracks{2cospars{n over 2}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
$$
Replace this result in $pars{1}$:
begin{align}
&color{#66f}{large%
int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = &
{bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}},expo{ic npars{beta - alpha}/2}
end{align}
answered Jul 12 '14 at 7:50
Felix MarinFelix Marin
68.6k7109145
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$begingroup$
The manipulations with the delta function were really helpful to me.
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– daniel
Sep 25 '14 at 7:54
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@daniel It's nice it was helpful for you. Thanks.
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– Felix Marin
Sep 25 '14 at 21:25
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@daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
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– Did
Dec 17 '18 at 0:22
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@Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
$endgroup$
– daniel
Dec 17 '18 at 7:06
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is not a complete answer but maybe it will save a trip to the library.
In Ramanujan's Collected Works at ch. 27 he says it is "well known" that
$$ intlimits_{-pi/2}^{pi/2}(cos x)^m e^{inx}dx = frac{pi}{2^m}frac{Gamma(1+m)}{Gammaleft(1+ frac{1}{2}(m+n)right)Gammaleft(1+frac{1}{2}(m-n)right)}. tag{1.1} $$
He says that from this and Fourier's Theorem we can derive the relation in your question. He may work out some details in the chapter (I haven't looked). If you accept or can prove (1.1) this might be enough to solve the problem.
Edit in response to comment: All he is saying is that a function can sometimes be represented as a Fourier series. I suspect the proof is in an earlier paper and ch. 27 is building on your relation. But he clearly says that "it follows" from (1.1) so it's a start. There are 4 papers cited in the notes...
Edit: Here is one way to do this proof. I thought we could take advantage of the (visually obvious) F-transform aspect of the relation, but this proof is very direct and maybe there is no simpler way...
edited Apr 13 '17 at 12:21
Community♦
1
answered Mar 15 '13 at 13:25
danieldaniel
6,29222259
$endgroup$
2
$begingroup$
You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
$endgroup$
– Sangchul Lee
Mar 15 '13 at 15:42
$begingroup$
@daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:53
$begingroup$
I think I understood it!
$endgroup$
– Shobhit Bhatnagar
Mar 17 '13 at 3:20
2
$begingroup$
Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
$endgroup$
– J. M. is not a mathematician
Apr 9 '13 at 1:41
$begingroup$
Using (1.1) implies $alpha=beta$
$endgroup$
– Anastasiya-Romanova 秀
Sep 26 '14 at 9:58
add a comment |
$begingroup$
This is not a complete answer but maybe it will save a trip to the library.
In Ramanujan's Collected Works at ch. 27 he says it is "well known" that
$$ intlimits_{-pi/2}^{pi/2}(cos x)^m e^{inx}dx = frac{pi}{2^m}frac{Gamma(1+m)}{Gammaleft(1+ frac{1}{2}(m+n)right)Gammaleft(1+frac{1}{2}(m-n)right)}. tag{1.1} $$
He says that from this and Fourier's Theorem we can derive the relation in your question. He may work out some details in the chapter (I haven't looked). If you accept or can prove (1.1) this might be enough to solve the problem.
Edit in response to comment: All he is saying is that a function can sometimes be represented as a Fourier series. I suspect the proof is in an earlier paper and ch. 27 is building on your relation. But he clearly says that "it follows" from (1.1) so it's a start. There are 4 papers cited in the notes...
Edit: Here is one way to do this proof. I thought we could take advantage of the (visually obvious) F-transform aspect of the relation, but this proof is very direct and maybe there is no simpler way...
edited Apr 13 '17 at 12:21
Community♦
1
answered Mar 15 '13 at 13:25
danieldaniel
6,29222259
$endgroup$
2
$begingroup$
You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
$endgroup$
– Sangchul Lee
Mar 15 '13 at 15:42
$begingroup$
@daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:53
$begingroup$
I think I understood it!
$endgroup$
– Shobhit Bhatnagar
Mar 17 '13 at 3:20
2
$begingroup$
Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
$endgroup$
– J. M. is not a mathematician
Apr 9 '13 at 1:41
$begingroup$
Using (1.1) implies $alpha=beta$
$endgroup$
– Anastasiya-Romanova 秀
Sep 26 '14 at 9:58
add a comment |
$begingroup$
This is not a complete answer but maybe it will save a trip to the library.
In Ramanujan's Collected Works at ch. 27 he says it is "well known" that
$$ intlimits_{-pi/2}^{pi/2}(cos x)^m e^{inx}dx = frac{pi}{2^m}frac{Gamma(1+m)}{Gammaleft(1+ frac{1}{2}(m+n)right)Gammaleft(1+frac{1}{2}(m-n)right)}. tag{1.1} $$
He says that from this and Fourier's Theorem we can derive the relation in your question. He may work out some details in the chapter (I haven't looked). If you accept or can prove (1.1) this might be enough to solve the problem.
Edit in response to comment: All he is saying is that a function can sometimes be represented as a Fourier series. I suspect the proof is in an earlier paper and ch. 27 is building on your relation. But he clearly says that "it follows" from (1.1) so it's a start. There are 4 papers cited in the notes...
Edit: Here is one way to do this proof. I thought we could take advantage of the (visually obvious) F-transform aspect of the relation, but this proof is very direct and maybe there is no simpler way...
edited Apr 13 '17 at 12:21
Community♦
1
answered Mar 15 '13 at 13:25
danieldaniel
6,29222259
$endgroup$
This is not a complete answer but maybe it will save a trip to the library.
In Ramanujan's Collected Works at ch. 27 he says it is "well known" that
$$ intlimits_{-pi/2}^{pi/2}(cos x)^m e^{inx}dx = frac{pi}{2^m}frac{Gamma(1+m)}{Gammaleft(1+ frac{1}{2}(m+n)right)Gammaleft(1+frac{1}{2}(m-n)right)}. tag{1.1} $$
He says that from this and Fourier's Theorem we can derive the relation in your question. He may work out some details in the chapter (I haven't looked). If you accept or can prove (1.1) this might be enough to solve the problem.
Edit in response to comment: All he is saying is that a function can sometimes be represented as a Fourier series. I suspect the proof is in an earlier paper and ch. 27 is building on your relation. But he clearly says that "it follows" from (1.1) so it's a start. There are 4 papers cited in the notes...
Edit: Here is one way to do this proof. I thought we could take advantage of the (visually obvious) F-transform aspect of the relation, but this proof is very direct and maybe there is no simpler way...
edited Apr 13 '17 at 12:21
Community♦
1
answered Mar 15 '13 at 13:25
danieldaniel
6,29222259
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 Mar 15 '13 at 13:25
danieldaniel
6,29222259
answered Mar 15 '13 at 13:25
danieldaniel
6,29222259
answered Mar 15 '13 at 13:25
danieldaniel
6,29222259
6,29222259
2
$begingroup$
You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
$endgroup$
– Sangchul Lee
Mar 15 '13 at 15:42
$begingroup$
@daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:53
$begingroup$
I think I understood it!
$endgroup$
– Shobhit Bhatnagar
Mar 17 '13 at 3:20
2
$begingroup$
Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
$endgroup$
– J. M. is not a mathematician
Apr 9 '13 at 1:41
$begingroup$
Using (1.1) implies $alpha=beta$
$endgroup$
– Anastasiya-Romanova 秀
Sep 26 '14 at 9:58
add a comment |
2
$begingroup$
You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
$endgroup$
– Sangchul Lee
Mar 15 '13 at 15:42
$begingroup$
@daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:53
$begingroup$
I think I understood it!
$endgroup$
– Shobhit Bhatnagar
Mar 17 '13 at 3:20
2
$begingroup$
Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
$endgroup$
– J. M. is not a mathematician
Apr 9 '13 at 1:41
$begingroup$
Using (1.1) implies $alpha=beta$
$endgroup$
– Anastasiya-Romanova 秀
Sep 26 '14 at 9:58
2
2
$begingroup$
You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
$endgroup$
– Sangchul Lee
Mar 15 '13 at 15:42
$begingroup$
You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
$endgroup$
– Sangchul Lee
Mar 15 '13 at 15:42
$begingroup$
@daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:53
$begingroup$
@daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:53
$begingroup$
I think I understood it!
$endgroup$
– Shobhit Bhatnagar
Mar 17 '13 at 3:20
$begingroup$
I think I understood it!
$endgroup$
– Shobhit Bhatnagar
Mar 17 '13 at 3:20
2
2
$begingroup$
Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
$endgroup$
– J. M. is not a mathematician
Apr 9 '13 at 1:41
$begingroup$
Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
$endgroup$
– J. M. is not a mathematician
Apr 9 '13 at 1:41
$begingroup$
Using (1.1) implies $alpha=beta$
$endgroup$
– Anastasiya-Romanova 秀
Sep 26 '14 at 9:58
$begingroup$
Using (1.1) implies $alpha=beta$
$endgroup$
– Anastasiya-Romanova 秀
Sep 26 '14 at 9:58
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x
={bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}}
,expo{ic npars{beta - alpha}/2}}$
$ds{verts{n} < pi,, Repars{alpha + beta} > 1}$.
Note that
begin{align}
&{1 over Gammapars{alpha + x}Gammapars{beta - x}}
={1 over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm]&={1 over Gammapars{alpha + beta - 1}},
{pars{alpha + beta - 2}! over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm] & ={1 over Gammapars{alpha + beta - 1}},
{alpha + beta - 2 choose alpha + x - 1}
end{align}
begin{align}&color{#c00000}{%
int_{-infty}^{infty}%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = & {1 over Gammapars{alpha + beta - 1}}
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}tag{1}
end{align}
begin{align}&color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=int_{-infty}^{infty}bracks{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha + x}},{dd z over 2piic}}expo{ic n x},dd x
\[3mm]&=-icoint_{verts{z} = 1}
{pars{1 + z}^{alpha + beta - 2} over z^{alpha}}braces{%
int_{-infty}^{infty}expo{icbracks{n - {rm Arg}pars{z}}x}
,{dd x over 2pi}},dd z
\[3mm]&=-ic overbrace{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha}},deltapars{n - {rm Arg}pars{z}},dd z}
^{ds{mbox{Set} z equiv expo{ic theta},,quadverts{theta} < pi}}
\[3mm]&=-icint_{-pi}^{pi}
{pars{1 + expo{ictheta}}^{alpha + beta - 2} over
expo{icalphatheta}},deltapars{n - theta},expo{ictheta}ic,ddtheta
=pars{1 + expo{ic n}}^{alpha + beta - 2}expo{icpars{1 - alpha}n}
\[3mm]&=expo{icpars{alpha + beta - 2}n/2}
pars{expo{-ic n/2} + expo{ic n/2}}^{alpha + beta - 2}
expo{icpars{1 - alpha}n}
\[3mm] & =
bracks{2cospars{{n over 2}}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
end{align}
$$
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=bracks{2cospars{n over 2}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
$$
Replace this result in $pars{1}$:
begin{align}
&color{#66f}{large%
int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = &
{bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}},expo{ic npars{beta - alpha}/2}
end{align}
answered Jul 12 '14 at 7:50
Felix MarinFelix Marin
68.6k7109145
$endgroup$
$begingroup$
The manipulations with the delta function were really helpful to me.
$endgroup$
– daniel
Sep 25 '14 at 7:54
$begingroup$
@daniel It's nice it was helpful for you. Thanks.
$endgroup$
– Felix Marin
Sep 25 '14 at 21:25
$begingroup$
@daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
$endgroup$
– Did
Dec 17 '18 at 0:22
$begingroup$
@Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
$endgroup$
– daniel
Dec 17 '18 at 7:06
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x
={bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}}
,expo{ic npars{beta - alpha}/2}}$
$ds{verts{n} < pi,, Repars{alpha + beta} > 1}$.
Note that
begin{align}
&{1 over Gammapars{alpha + x}Gammapars{beta - x}}
={1 over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm]&={1 over Gammapars{alpha + beta - 1}},
{pars{alpha + beta - 2}! over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm] & ={1 over Gammapars{alpha + beta - 1}},
{alpha + beta - 2 choose alpha + x - 1}
end{align}
begin{align}&color{#c00000}{%
int_{-infty}^{infty}%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = & {1 over Gammapars{alpha + beta - 1}}
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}tag{1}
end{align}
begin{align}&color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=int_{-infty}^{infty}bracks{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha + x}},{dd z over 2piic}}expo{ic n x},dd x
\[3mm]&=-icoint_{verts{z} = 1}
{pars{1 + z}^{alpha + beta - 2} over z^{alpha}}braces{%
int_{-infty}^{infty}expo{icbracks{n - {rm Arg}pars{z}}x}
,{dd x over 2pi}},dd z
\[3mm]&=-ic overbrace{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha}},deltapars{n - {rm Arg}pars{z}},dd z}
^{ds{mbox{Set} z equiv expo{ic theta},,quadverts{theta} < pi}}
\[3mm]&=-icint_{-pi}^{pi}
{pars{1 + expo{ictheta}}^{alpha + beta - 2} over
expo{icalphatheta}},deltapars{n - theta},expo{ictheta}ic,ddtheta
=pars{1 + expo{ic n}}^{alpha + beta - 2}expo{icpars{1 - alpha}n}
\[3mm]&=expo{icpars{alpha + beta - 2}n/2}
pars{expo{-ic n/2} + expo{ic n/2}}^{alpha + beta - 2}
expo{icpars{1 - alpha}n}
\[3mm] & =
bracks{2cospars{{n over 2}}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
end{align}
$$
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=bracks{2cospars{n over 2}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
$$
Replace this result in $pars{1}$:
begin{align}
&color{#66f}{large%
int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = &
{bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}},expo{ic npars{beta - alpha}/2}
end{align}
answered Jul 12 '14 at 7:50
Felix MarinFelix Marin
68.6k7109145
$endgroup$
$begingroup$
The manipulations with the delta function were really helpful to me.
$endgroup$
– daniel
Sep 25 '14 at 7:54
$begingroup$
@daniel It's nice it was helpful for you. Thanks.
$endgroup$
– Felix Marin
Sep 25 '14 at 21:25
$begingroup$
@daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
$endgroup$
– Did
Dec 17 '18 at 0:22
$begingroup$
@Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
$endgroup$
– daniel
Dec 17 '18 at 7:06
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x
={bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}}
,expo{ic npars{beta - alpha}/2}}$
$ds{verts{n} < pi,, Repars{alpha + beta} > 1}$.
Note that
begin{align}
&{1 over Gammapars{alpha + x}Gammapars{beta - x}}
={1 over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm]&={1 over Gammapars{alpha + beta - 1}},
{pars{alpha + beta - 2}! over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm] & ={1 over Gammapars{alpha + beta - 1}},
{alpha + beta - 2 choose alpha + x - 1}
end{align}
begin{align}&color{#c00000}{%
int_{-infty}^{infty}%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = & {1 over Gammapars{alpha + beta - 1}}
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}tag{1}
end{align}
begin{align}&color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=int_{-infty}^{infty}bracks{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha + x}},{dd z over 2piic}}expo{ic n x},dd x
\[3mm]&=-icoint_{verts{z} = 1}
{pars{1 + z}^{alpha + beta - 2} over z^{alpha}}braces{%
int_{-infty}^{infty}expo{icbracks{n - {rm Arg}pars{z}}x}
,{dd x over 2pi}},dd z
\[3mm]&=-ic overbrace{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha}},deltapars{n - {rm Arg}pars{z}},dd z}
^{ds{mbox{Set} z equiv expo{ic theta},,quadverts{theta} < pi}}
\[3mm]&=-icint_{-pi}^{pi}
{pars{1 + expo{ictheta}}^{alpha + beta - 2} over
expo{icalphatheta}},deltapars{n - theta},expo{ictheta}ic,ddtheta
=pars{1 + expo{ic n}}^{alpha + beta - 2}expo{icpars{1 - alpha}n}
\[3mm]&=expo{icpars{alpha + beta - 2}n/2}
pars{expo{-ic n/2} + expo{ic n/2}}^{alpha + beta - 2}
expo{icpars{1 - alpha}n}
\[3mm] & =
bracks{2cospars{{n over 2}}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
end{align}
$$
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=bracks{2cospars{n over 2}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
$$
Replace this result in $pars{1}$:
begin{align}
&color{#66f}{large%
int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = &
{bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}},expo{ic npars{beta - alpha}/2}
end{align}
answered Jul 12 '14 at 7:50
Felix MarinFelix Marin
68.6k7109145
$endgroup$
$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x
={bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}}
,expo{ic npars{beta - alpha}/2}}$
$ds{verts{n} < pi,, Repars{alpha + beta} > 1}$.
Note that
begin{align}
&{1 over Gammapars{alpha + x}Gammapars{beta - x}}
={1 over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm]&={1 over Gammapars{alpha + beta - 1}},
{pars{alpha + beta - 2}! over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm] & ={1 over Gammapars{alpha + beta - 1}},
{alpha + beta - 2 choose alpha + x - 1}
end{align}
begin{align}&color{#c00000}{%
int_{-infty}^{infty}%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = & {1 over Gammapars{alpha + beta - 1}}
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}tag{1}
end{align}
begin{align}&color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=int_{-infty}^{infty}bracks{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha + x}},{dd z over 2piic}}expo{ic n x},dd x
\[3mm]&=-icoint_{verts{z} = 1}
{pars{1 + z}^{alpha + beta - 2} over z^{alpha}}braces{%
int_{-infty}^{infty}expo{icbracks{n - {rm Arg}pars{z}}x}
,{dd x over 2pi}},dd z
\[3mm]&=-ic overbrace{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha}},deltapars{n - {rm Arg}pars{z}},dd z}
^{ds{mbox{Set} z equiv expo{ic theta},,quadverts{theta} < pi}}
\[3mm]&=-icint_{-pi}^{pi}
{pars{1 + expo{ictheta}}^{alpha + beta - 2} over
expo{icalphatheta}},deltapars{n - theta},expo{ictheta}ic,ddtheta
=pars{1 + expo{ic n}}^{alpha + beta - 2}expo{icpars{1 - alpha}n}
\[3mm]&=expo{icpars{alpha + beta - 2}n/2}
pars{expo{-ic n/2} + expo{ic n/2}}^{alpha + beta - 2}
expo{icpars{1 - alpha}n}
\[3mm] & =
bracks{2cospars{{n over 2}}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
end{align}
$$
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=bracks{2cospars{n over 2}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
$$
Replace this result in $pars{1}$:
begin{align}
&color{#66f}{large%
int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = &
{bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}},expo{ic npars{beta - alpha}/2}
end{align}
answered Jul 12 '14 at 7:50
Felix MarinFelix Marin
68.6k7109145
edited Dec 17 '18 at 0:16
answered Jul 12 '14 at 7:50
Felix MarinFelix Marin
68.6k7109145
answered Jul 12 '14 at 7:50
Felix MarinFelix Marin
68.6k7109145
answered Jul 12 '14 at 7:50
Felix MarinFelix Marin
68.6k7109145
68.6k7109145
$begingroup$
The manipulations with the delta function were really helpful to me.
$endgroup$
– daniel
Sep 25 '14 at 7:54
$begingroup$
@daniel It's nice it was helpful for you. Thanks.
$endgroup$
– Felix Marin
Sep 25 '14 at 21:25
$begingroup$
@daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
$endgroup$
– Did
Dec 17 '18 at 0:22
$begingroup$
@Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
$endgroup$
– daniel
Dec 17 '18 at 7:06
add a comment |
$begingroup$
The manipulations with the delta function were really helpful to me.
$endgroup$
– daniel
Sep 25 '14 at 7:54
$begingroup$
@daniel It's nice it was helpful for you. Thanks.
$endgroup$
– Felix Marin
Sep 25 '14 at 21:25
$begingroup$
@daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
$endgroup$
– Did
Dec 17 '18 at 0:22
$begingroup$
@Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
$endgroup$
– daniel
Dec 17 '18 at 7:06
$begingroup$
The manipulations with the delta function were really helpful to me.
$endgroup$
– daniel
Sep 25 '14 at 7:54
$begingroup$
The manipulations with the delta function were really helpful to me.
$endgroup$
– daniel
Sep 25 '14 at 7:54
$begingroup$
@daniel It's nice it was helpful for you. Thanks.
$endgroup$
– Felix Marin
Sep 25 '14 at 21:25
$begingroup$
@daniel It's nice it was helpful for you. Thanks.
$endgroup$
– Felix Marin
Sep 25 '14 at 21:25
$begingroup$
@daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
$endgroup$
– Did
Dec 17 '18 at 0:22
$begingroup$
@daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
$endgroup$
– Did
Dec 17 '18 at 0:22
$begingroup$
@Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
$endgroup$
– daniel
Dec 17 '18 at 7:06
$begingroup$
@Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
$endgroup$
– daniel
Dec 17 '18 at 7:06
add a comment |
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Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
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– daniel
Mar 15 '13 at 13:44
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@daniel: I will try to find the book you suggested.
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– Shobhit Bhatnagar
Mar 15 '13 at 15:12