# Interesting Integral $\int_{-\infty}^{\infty}\frac{e^{i nx}}{\Gamma(\alpha+x) \Gamma(\beta -x)}dx$

I am asking this question out of curiosity.

$$\int_{-\infty}^{\infty}\frac{e^{i nx}}{\Gamma(\alpha+x) \Gamma(\beta -x)}dx = \frac{ \left(2\cos \frac{n}{2} \right)^{\alpha +\beta-2}}{\Gamma(\alpha+\beta-1)}e^{\frac{in}{2}(\beta - \alpha)} \quad |n|<\pi \quad \text{and} \quad \Re(\alpha+\beta)>1$$

• 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?
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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. –  daniel Mar 15 '13 at 13:44
@daniel: I will try to find the book you suggested. –  Integrals and Series Mar 15 '13 at 15:12

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

$$\int\limits_{-\pi/2}^{\pi/2}(\cos x)^m e^{inx}dx = \frac{\pi}{2^m}\frac{\Gamma(1+m)}{\Gamma\left(1+ \frac{1}{2}(m+n)\right)\Gamma\left(1+\frac{1}{2}(m-n)\right)}. \tag{1.1}$$

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...

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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. –  sos440 Mar 15 '13 at 15:42
@daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result. –  Integrals and Series Mar 15 '13 at 15:53
I think I understood it! –  Integrals and Series Mar 17 '13 at 3:20
Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae... –  Ｊ. Ｍ. Apr 9 '13 at 1:41
Using (1.1) implies $\alpha=\beta$ –  Anastasiya-Romanova Sep 26 at 9:58


Note that \begin{align} &{1 \over \Gamma\pars{\alpha + x}\Gamma\pars{\beta - x}} ={1 \over \pars{\alpha + x - 1}!\pars{\beta - x - 1}!} \\[3mm]&={1 \over \Gamma\pars{\alpha + \beta - 1}}\, {\pars{\alpha + \beta - 2}! \over \pars{\alpha + x - 1}!\pars{\beta - x - 1}!} ={1 \over \Gamma\pars{\alpha + \beta - 1}}\, {\alpha + \beta - 2 \choose \alpha + x - 1} \end{align}

\begin{align}&\color{#c00000}{% \int_{-\infty}^{\infty}% {\expo{\ic nx} \over \Gamma\pars{\alpha + x}\Gamma\pars{\beta - x}}\,\dd x} ={1 \over \Gamma\pars{\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 2\pi\ic}}\expo{\ic n x}\,\dd x \\[3mm]&=-\ic\oint_{\verts{z}\ =\ 1} {\pars{1 + z}^{\alpha + \beta - 2} \over z^{\alpha}}\braces{% \int_{-\infty}^{\infty}\expo{\ic\bracks{n - {\rm Arg}\pars{z}}x} \,{\dd x \over 2\pi}}\,\dd z \\[3mm]&=-\ic\ \overbrace{\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{\alpha + \beta - 2} \over z^{\alpha}}\,\delta\pars{n - {\rm Arg}\pars{z}}\,\dd z} ^{\ds{\mbox{Set}\ z \equiv \expo{\ic \theta}\,,\quad\verts{\theta} < \pi}} \\[3mm]&=-\ic\int_{-\pi}^{\pi} {\pars{1 + \expo{\ic\theta}}^{\alpha + \beta - 2} \over \expo{\ic\alpha\theta}}\,\delta\pars{n - \theta}\,\expo{\ic\theta}\ic\,\dd\theta =\pars{1 + \expo{\ic n}}^{\alpha + \beta - 2}\expo{\ic\pars{1 - \alpha}n} \\[3mm]&=\expo{\ic\pars{\alpha + \beta - 2}n/2} \pars{\expo{-\ic n/2} + \expo{\ic n/2}}^{\alpha + \beta - 2} \expo{\ic\pars{1 - \alpha}n} =\bracks{2\cos\pars{n \over 2}}^{\alpha + \beta - 2} \expo{\ic\pars{\beta - \alpha}n/2} \end{align}

$$\color{#00f}{\int_{-\infty}^{\infty} {\alpha + \beta - 2 \choose \alpha + x - 1}\expo{\ic nx}\,\dd x} =\bracks{2\cos\pars{n \over 2}}^{\alpha + \beta - 2} \expo{\ic\pars{\beta - \alpha}n/2}$$

Replace this result in $\pars{1}$: $$\color{#66f}{\large% \int_{-\infty}^{\infty}\!\!% {\expo{\ic nx} \over \Gamma\pars{\alpha + x}\Gamma\pars{\beta - x}}\,\dd x ={\bracks{2\cos\pars{n/2}}^{\alpha + \beta - 2}\over \Gamma\pars{\alpha + \beta - 1}}\,\expo{\ic n\pars{\beta - \alpha}/2}}$$

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The manipulations with the delta function were really helpful to me. –  daniel Sep 25 at 7:54
@daniel It's nice it was helpful for you. Thanks. –  Felix Marin Sep 25 at 21:25