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

About YiMin

This is just a nerd PhD student of Math@UT Austin.