Let $c>0$. How to prove that for any complex number $z$, $$\frac{1}{\Gamma(z)}=\frac{1}{2\pi}\int_{-\infty}^\infty (c+it)^{-z}e^{c+it}\,dt?$$ where $\Gamma(z)$ is the Gamma function.


In two steps:

  • Consider the right side as a complex contour integral. The integrand has one branch point $t=ic$ in the upper half-plane. Introduce the branch cut $B=[ic,i\infty)$ and deform the contour of integration so that it runs counterclockwise from $i\infty$ to $i\infty$ around $B$. Combined with the definition of the gamma function, this will give you something proportional to $\Gamma(1-z)\sin\pi z $.

  • Apply Euler's reflection formula to replace $\Gamma(1-z)\sin\pi z $ by $\dfrac{\pi}{\Gamma(z)}$.

  • $\begingroup$ The answer neglects a convergence condition and misses the analytic continuation step. In step 1, when $\Re z\ge 1$ the integral from $ic$ diverges; when $\Re z\le 0$ the integral on the semicircle diverges. The branch cut procedure works only for $0\le \Re z<1$. We have to show the original integral on the real axis on the right hand side of the first equation is entire in $z$. Having established the equality for $0\le \Re z<1$, we can then analytic continue the identity to the entire complex plane. $\endgroup$ – Hans Jul 30 '17 at 9:46

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