TITLE: Integral formula for $\frac{1}{\Gamma(z)}$
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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 $\dfrac{\pi}{\Gamma(z)}$.