TITLE: Interesting Integral $\int_{-\infty}^{\infty}\frac{e^{i nx}}{\Gamma(\alpha+x) \Gamma(\beta -x)}dx$
QUESTION [24 upvotes]: 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?

REPLY [15 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\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}}$

$\ds{\verts{n} < \pi\,,\ \Re\pars{\alpha + \beta} > 1}$.

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}!}
\\[3mm] & ={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}
\\[5mm] = &\ {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}
\\[3mm] & =
\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 $\pars{1}$:
  \begin{align}
&\color{#66f}{\large%
\int_{-\infty}^{\infty}\!\!%
{\expo{\ic nx} \over \Gamma\pars{\alpha + x}\Gamma\pars{\beta - x}}\,\dd x}
 \\[5mm] = &\
{\bracks{2\cos\pars{n/2}}^{\alpha + \beta - 2}\over
   \Gamma\pars{\alpha + \beta - 1}}\,\expo{\ic n\pars{\beta - \alpha}/2}
\end{align}