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.
2026-04-06 19:35:32.1775504132
Integral formula for $\frac{1}{\Gamma(z)}$
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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 $\dfrac{\pi}{\Gamma(z)}$.