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?

 A: $\newcommand{\+}{^{\dagger}}
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$\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}

A: 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...
