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Can anyone please advise me how to integrate expressions of the form $\text{sinc}\,(x) / (1-x)_n$ along the real axis?

Using a CAS, one could suggest that $$ n! \int_{-\infty}^\infty \frac{\sin \pi x}{\pi x (1-x)_n}dx=2^n,\ n=1,2,\ldots $$ But I don't know how to prove this.

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  • $\begingroup$ A few manipulations give the equivalent integral $$\int_{-\infty}^\infty \frac{\Gamma(1+n)}{\Gamma(1+t)\Gamma(1-t+n)}\mathrm dt=\int_{-\infty}^\infty \binom{n}{t}\mathrm dt$$ $\endgroup$ – J. M. is a poor mathematician Apr 10 '16 at 10:50
  • $\begingroup$ Thank you for the interesting observation, but how about integrals of binomial coefficients? Could you please suggest some source on how to integrate them? $\endgroup$ – user2835965 Apr 10 '16 at 12:19
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Using contour integration one can show that \begin{align} &\int_{-\infty}^\infty \frac{\sin \pi x}{x \cdot (1-x)\ldots (n-x)}dx\\ &=\text{Im}\left\{\pi i \left(\underset{z=0}{\text{res}}-\sum_{k=1}^n \underset{z=k}{\text{res}}\right)\frac{e^{\pi i z}}{z \cdot (1-z)\ldots (n-z)}\right\}\\ &=\pi\left(\frac{1}{n!}-\sum_{k=1}^n\frac{(-1)^k}{k\cdot (1-k)\ldots (-1)\cdot 1\ldots (n-k)}\right)\\ &=\pi\left(\frac{1}{n!}+\frac{1}{n!}\sum_{k=1}^n\binom{n}{k}\right)=\frac{2^n\pi}{n!},\quad n=1,2,3\ldots \end{align}

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