Find the value of $\int_0^{\infty} \frac{x^{a-1}\,\mathrm{d}x}{1+x^{2}},\,$ where $\,0<a<2$

Question

Using the Residue Theorem calculate $$ \int_0^{\infty} \frac{x^{a-1}\,\mathrm{d}x}{1+x^{2}},\,\,\,\, \text{where}\,\,\,0<a<2. $$ My solution comes out to be
$$ \frac{\pi}{2}[i^{a-1}-(-i)^{a-1}]$$

How to proceed from here since the solution of a rational expression must be independent of $i$

Answer 1

I imagine you have integrated $\,\,f(z)=\dfrac{z^{a-1}}{z^2+1},\,$ along the contour $\,\gamma=\gamma_1\cup\gamma_2\cup\gamma_3\cup\gamma_4,\,$ where $$ \gamma_1=[-R,-\varepsilon],\quad\gamma_2=\{\varepsilon\mathrm{e}^{i(\pi-t)}: t\in[0,\pi]\},\quad \gamma_3=[\varepsilon,R],\quad\gamma_4=\{R\mathrm{e}^{it}: t\in[0,\pi]\}. $$ In such case $$ z^{a-1}=\mathrm{e}^{(a-1)\log z}, $$ where $\log z$ is a branch of logarithm defined in $$ \Omega=\mathrm{C}\setminus \{it: t\in (-\infty,0]\}, $$ as follows. If $z=r\mathrm{e}^{i\vartheta}$, then $\log z=\log r+i\vartheta$, with $$ \vartheta\in (-\pi/2,3\pi,2). $$ So, in such case, Residue Theorem provides $$ \int_\gamma f(z)\,dz=2\pi i\,\mathrm{Res}\left(\frac{z^{a-1}}{1+z^2},i\right)=2\pi i\cdot\frac{i^{a-1}}{2i}=\pi\mathrm{e}^{i(a-1)\pi/2}=-i\pi \mathrm{e}^{ia\pi/2}. $$ Clearly $$ \lim_{R\to\infty,\,\varepsilon\to 0}\int_{\gamma_1}f(z)\,dz= \int_{-\infty}^0\frac{z^{a-1}}{z^2+1}=(-1)^{a-1}\int_0^\infty\frac{x^{a-1}\,dx}{1+x^2}=-\mathrm{e}^{ia\pi}\int_0^\infty\frac{x^{a-1}\,dx}{1+x^2}, $$ since $(-1)^{a-1}=\mathrm{e}^{i(a-1)\pi}=-\mathrm{e}^{ia\pi},\,$ and $$ \lim_{R\to\infty,\,\varepsilon\to 0}\int_{\gamma_3}f(z)\,dz= \int_0^{\infty}\frac{z^{a-1}}{z^2+1}=\int_0^\infty\frac{x^{a-1}\,dx}{1+x^2}. $$ Meanwhile, $\lim_{\varepsilon\to 0}\int_{\gamma_2}f(z)\,dz=0$ and $\lim_{R\to\infty}\int_{\gamma_4}f(z)\,dz=0$. Hence, altogether $$ (1-\mathrm{e}^{ia\pi})\int_{0}^\infty\frac{x^{a-1}\,dx}{1+x^2}= \lim_{\varepsilon\to 0,\,\,R\to\infty}\int_\gamma f(z)\,dz= 2\pi i\,\mathrm{Res}\left(\frac{z^{a-1}}{1+z^2},i\right) =-i\pi \mathrm{e}^{ia\pi/2}. $$ Hence $$ \int_{0}^\infty\frac{x^{a-1}\,dx}{1+x^2}=\frac{i\pi \mathrm{e}^{ia\pi/2}}{\mathrm{e}^{ia\pi}-1}=\frac{i\pi }{\mathrm{e}^{ia\pi/2}-\mathrm{e}^{-ia\pi/2}}=\frac{\pi}{2\sin (a\pi/2)}. $$

Answer 2

In this answer, it is shown that for $m\gt0$ and $-1<n<m-1$ $$ \frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x $$ Plug in $n=a-1$ and $m=2$ to get that for $0\lt a\lt2$ $$ \int_0^\infty\frac{x^{a-1}}{1+x^2}\,\mathrm{d}x=\frac\pi2\csc\left(\frac{\pi a}2\right) $$