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At the beginning of his paper Ramanujan makes the following statement:

It is well known that, $$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \cos x \right)^m e^{inx} dx = \frac{\pi}{2^m} \frac{\Gamma(1+m)}{\Gamma\left(1+\frac{m+n}{2}\right)\Gamma\left(1+\frac{m-n}{2}\right)} $$ if $\Re[m]>-1$.

Unfortunately I do not know that. Can someone explain how the l.h.s. produces the r.h.s. (or give a reference that explains this)?

Then Ramanujan makes another statement:

It follows from this and Fourier's theorem that, if $n$ is any real number except $\pm \pi$ and $\Re[\alpha + \beta]>1$, or if $n=\pm \pi$ and $\Re[\alpha+\beta]>2$ then $$ \int_{-\infty}^{\infty} \frac{e^{inx}}{\Gamma(\alpha+x)\Gamma(\beta-x)} = \frac{\left(2 \cos\left(\frac{n}{2}\right)\right)^{\alpha+\beta-2}}{\Gamma(\alpha+\beta-1)} e^{\frac{in(\beta-\alpha)}{2}}\quad \mathrm{or} \quad 0, $$ according as $\vert n \vert < \pi$ or $\vert n \vert \geq \pi$.

This is another statement that is not trivial for me. It would be great if someone could provide proof or a reference.

Ted Black
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    They are not trivial for most of the people – Vanessa Oct 08 '22 at 20:15
  • The paper in question is available at http://ramanujan.sirinudi.org/Volumes/published/ram27.html – Paramanand Singh Oct 10 '22 at 14:15
  • $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \cos x \right)^m e^{inx} dx=2\int_0^{\frac{\pi}{2}} \left( \cos x \right)^m \cos (nx) dx$$ the evaluation is here - https://math.stackexchange.com/questions/3419834/integrals-related-to-the-reciprocal-beta-function?noredirect=1 – Svyatoslav Oct 11 '22 at 11:33
  • Sorry @Svyatoslav, I cannot follow the proof in the link and also I do not think he is looking at the integral you wrote. – Ted Black Oct 12 '22 at 22:37
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    The method in the link is applicable to this integral as well. You can find other solutions here - https://approach0.xyz/search/?q=OR%20content%3A%24%5Cint_0%5E%7B%5Cpi%2F2%7D%5Ccos%5Em(x)%5Ccos(nx)dx%24&p=1 – Svyatoslav Oct 14 '22 at 04:23

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