What is $\int_0^\infty\frac{x^{z-1}}{x+1} dx$?

Question

I was looking around the internet for a simple(r) proof of the Gamma Reflection Formula. I found this: Detailed explanation of the Γ reflection formula understandable by an AP Calculus student, and did not understand the last integration: $$\displaystyle\int\limits_0^\infty\frac{v^{z-1}}{v+1} dv$$

Can anyone help me? Explanations understandable by an AP Calculus student would be great!

Answer 1

Splitting the integral as $\int_0^\infty \frac{v^{z-1}}{1+v}\,dv=\int_0^1\frac{v^{z-1}}{1+v}\,dv+\int_1^\infty \frac{v^{z-1}}{1+v}\,dv$, and enforcing the substation $v\to 1/v$ in the integral that extends from $1$ to $\infty$, expanding $\frac1{1+v}$ as $\sum_{n=0}^\infty (-1)^nv^n$, and interchanging the order of the series and the integral, we can write

$$\begin{align} \int_0^\infty \frac{v^{z-1}}{1+v}\,dv&=\int_0^1\frac{v^{z-1}}{1+v}\,dv+\int_1^\infty \frac{v^{z-1}}{1+v}\,dv\\\\ &=\int_0^1\frac{v^{z-1}+v^{-z}}{1+v}\,dv\\\\ &=\sum_{n=0}^\infty (-1)^n \int_0^1 (v^{n+z-1}+v^{n-z})\,dv\\\\ &=\sum_{n=0}^\infty (-1)^n\left(\frac{1}{n+z}+\frac{1}{n+1-z}\right) \tag 1\\\\ &=\frac{\pi}{\sin(\pi z)} \end{align}$$

where I showed in the appendix of THIS ANSWER using real analysis methods only that $(1)$ is the partial fraction expansion of $\pi \csc(\pi z)$.

Answer 2

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With $\ds{\pars{~\Re\pars{z - 1} > - 1\ \mbox{and}\ \Re\pars{z - 1} < 0~} \implies \bbx{\ds{0 < \Re\pars{z} < 1}}}$:

\begin{align} \int_{0}^{\infty}{v^{z - 1} \over v + 1}\,\dd v & \,\,\,\stackrel{t\ =\ 1/\pars{v + 1}}{=}\,\,\, \int_{1}^{0}t\,\pars{{1 \over t} - 1}^{z - 1}\pars{-\,{1 \over t^{2}}}\dd t = \int_{0}^{1}t^{-z}\,\pars{1 - t}^{z - 1}\,\dd t \\[5mm] & = {\Gamma\pars{-z + 1}\Gamma\pars{z} \over \Gamma\pars{1}} = \bbx{\ds{\pi \over \sin\pars{\pi z}}} \end{align}