Reflection relation for gamma function

Statement

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

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

Facts used

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

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

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:

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

Then, we'll obtain:

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

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