# Contour Integral

On Sophex, I was confused with the semi-circle method, then I found this method with a full circle…

Proof of Euler reflection Formula: $\Gamma(x)\Gamma(1-x)=\displaystyle \frac{\pi}{\sin\pi x}$

Scheme: Use Contour Integral along $\mathcal{C}$, $\mathcal{C}$ consists of two circles with radii of $R$ and $\epsilon$.

$\mathcal{C}_2$ and $\mathcal{C}_4$ are divided by the negative axis.

$\displaystyle\int_{\mathcal{C}}\frac{z^{x-1}}{1-z}\mathrm{d}z=-2\pi i$

$\displaystyle\int_{\mathcal{C}_1}\frac{z^{x-1}}{1-z}\mathrm{d}z=\int_{-\pi}^{\pi}\frac{i R^x e^{i x\theta}}{1-Re^{i\theta}}\mathrm{d}\theta$

$\displaystyle\int_{\mathcal{C}_3}\frac{z^{x-1}}{1-z}\mathrm{d}z=\int_{\pi}^{-\pi}\frac{i\epsilon^x e^{i x\theta}}{1-\epsilon e^{i\theta}}\mathrm{d}\theta$

$\displaystyle\int_{\mathcal{C}_2}\frac{z^{x-1}}{1-z}\mathrm{d}z=\int_{-R}^{-\epsilon}\frac{z^{x-1}}{1-z}\mathrm{d}z=\int_{R}^{\epsilon}\frac{t^{x-1}e^{ix\pi}}{1+t}\mathrm{d}t$

Here we replace $z=t e^{i\pi}$

$\displaystyle\int_{\mathcal{C}_4}\frac{z^{x-1}}{1-z}\mathrm{d}z=\int_{\epsilon}^R\frac{t^{x-1}e^{-i x\pi}}{1+t}\mathrm{d}z$

Here we replace $z=t e^{-i\pi}$, because of the principle value.

Then $R\rightarrow \infty$, $\epsilon\rightarrow 0^{+}$, it is done.

$\Box$