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
  • $\begingroup$ This post (math.stackexchange.com/questions/2274972/…) is related to the question at hand. The draft of the article posted on arXiv (arxiv.org/abs/1809.03933, math.CO) has now actually been accepted for publication in a special issue of the journal Axioms. The modified draft of the article contains some more useful related integral transforms for anyone who finds this answer interesting or of use in some application. $\endgroup$ – mds May 18 at 17:35

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