Euler reflection formula proof using contour integration

$$\int_{0}^{\infty}\frac{x^{\alpha}}{x+1}\,dx=- \pi \csc(\pi \alpha)$$

$$\textit{proof}$$

Consider the following function

$$f(z) = \frac{z^{\alpha}}{1+z} = \frac{e^{\alpha \log(z)}}{1+z}$$

As we know the function \( \log(z) \) is multi-valued defined as

$$\log(z) = \ln|z|+i\theta +2k\pi i$$

This maps the complex plain more than once in order to restrict it and make use of the integration path we set

$$\log(z) = \ln|z|+i\theta \,,\, \theta \in[0,2\pi)$$

This function is single-valued and on the integration path we have

$$e^{\alpha \log(z)} =x^{\alpha} e^{i\theta \alpha}= \begin{cases}
x^{\alpha} & \theta \to 0\\
x^{\alpha} e^{2 \pi \alpha i} & \theta \to 2\pi
\end{cases}$$

Integrating around the following contour

By the residue theorem we have

$$\oint f(z) \,dz = 2\pi i\, \mathrm{Res}(f,-1)$$

Which becomes

$$\int_{C_{\rho}}f(z)\,dz + \int_{C_{\epsilon}}f(z)\,dz+\int_{\epsilon}^{\rho}f(z)dz-\int_{\epsilon}^{\rho}f(z)dz =2\pi i\, \mathrm{Res}(f,-1) $$

By the chosen branch of the algorithm we have

$$\int_{C_{\rho}}f(z)\,dz + \int_{C_{\epsilon}}f(z)\,dz+\int_{\epsilon}^{\rho}\frac{x^{\alpha}}{x+1}\,dx-e^{2\pi i \alpha}\int_{\epsilon}^{\rho}\frac{x^{\alpha}}{x+1}\,dx =2\pi i\, \mathrm{Res}(f,-1) $$

Note for the circle \( |z| = \rho \) we have

$$\left| \oint_{|z|=\rho}f(z) \,dz \right| \leq 2\pi \rho\, \mathrm{max}\left| \frac{ z^{\alpha}}{1+z}\right|\leq 2\pi \rho \frac{|z|^{\alpha}}{||z|-1|} =2\pi \frac{{\rho}^{\alpha +1}}{|\rho -1|}$$

Note the triangle inequality

$$|\omega +1 | \geq ||\omega |-|1||$$

and the ML inequality

$$\left| \int_{\Gamma}f(z)\,dz\right| \leq M \,l(\Gamma)\,\,, \, M = \mathrm{max}(|f|)$$

Note that if \( -1 <\alpha < 0  \)  we get

$$\left| \lim_{\rho \to \infty}\oint_{|z|=\rho}f(z) \,dz \right| \leq 2\pi \frac{{\rho}^{\alpha +1}}{|\rho -1|} \to 0$$

Similarly we have

$$\left| \lim_{\epsilon\to 0}\oint_{|z|=\epsilon}f(z) \,dz \right| \leq 2\pi \frac{{\epsilon}^{\alpha +1}}{|\epsilon -1|} \to 0$$

We deduce that as \( \epsilon \to 0 \) and \( \rho \to \infty \)

$$(1-e^{2\pi i \alpha})\int_{0}^{\infty}\frac{x^{\alpha}}{x+1}\,dx =2\pi i\, \mathrm{Res}(f,-1) \,\,\, , \, -1 < \alpha < 0$$

Notice that

$$\mathrm{Res}(f,-1) = \lim_{z \to -1} e^{\alpha \log(z)} = e^{\alpha (\ln|-1|+ \pi i )} = e^{\alpha \pi i}$$

Hence we have

$$\int_{0}^{\infty}\frac{x^{\alpha}}{x+1}\,dx =2\pi i\,\frac{ e^{\alpha \pi i}}{1-e^{2\alpha \pi i}} = \frac{2\pi i}{e^{-\pi i \alpha }-e^{\pi i \alpha }} =- \pi \csc(\pi \alpha)$$

Note we can deduce the Euler reflection formula

$$\Gamma(\alpha)\Gamma(1-\alpha) = \pi \csc(\pi \alpha)$$

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