Reflection relation for gamma function

Statement

Let $\Gamma$ denote the gamma function. Then, for any $z$ that is not an integer, we have:

$\Gamma (z)\Gamma (1-z)={\frac {\pi }{\sin(\pi z)}}$

Facts used

• Recurrence relation for gamma function: This shows that it suffices to prove the result for $\operatorname {Re} (z)\in (0,1)$
• The following identity, for $0<\operatorname {Re} (a)<1$:

$\int _{0}^{1}{\frac {v^{a-1}}{1+v}}={\frac {\pi }{\sin \pi z}}$

Proof

Proof of the integration identity

Pick a branch of the logarithm that is slit along the positive real axis, and thus rewrite:

$v^{a-1}=e^{(a-1)\log v}$

Where the branch of logarithm is chosen so that the approach to the positive real axis from the upper half-plane side is the usual logarithm.

Next, apply the keyhole contour integration method to compute this integral. The inner and outer circle integrals approach zero, and if we define:

$I::=\int _{0}^{1}{\frac {v^{a-1}}{1+v}}$

Then, we'll obtain:

$I(e^{2\pi i(a-1)}-1)=2\pi i\operatorname {res} ({\frac {e^{(a-1)\log v}}{1+v}};-1)$

The residue at $-1$ is simple $e^{\pi i(a-1)}$ and simplifying the expression, we obtain the required expression for $I$.