I want to find a following integral
$$\int_0^\infty \frac{t^s}{1+t^2} \,\frac{dt}{t}$$
where $s \in \mathbb{C}$, $\Re(s) \in (0, 2)$ and want to find a closed form for it. I think it should be $\frac{1}{2} \Gamma(s/2)\Gamma(1-s/2)$ but I'm not completely certain and, if that's true, I'm not sure how to go about showing it.
Hint
Use the general formula
$$\int^\infty_0 \frac {t^{x-1}}{(1+t)^{x+y }}dt =\frac {\Gamma (x)\Gamma (y)}{\Gamma (x+y)}$$
After applying the substitution $t^2 \to t$.
You can also simplify using the Euler reflection formula
$$\Gamma (x)\Gamma (1-x) =\pi \csc \pi x$$
As a proof to Zaid's formula,
Define,
$$B(m,n)=\int_{0}^{1} u^{m-1}(1-u)^{n-1} du$$
Now let $u=\frac{x}{1+x}$. Then $du=\frac{1}{(1+x)^2} dx$. Note $x=-\frac{u}{u-1}$. Now as $u \to 1^-$ then $x \to \infty$.
$$\int_{0}^{\infty} \frac{x^{m-1}}{(1+x)^{m-1}} \frac{1}{(1+x)^{n-1}} \frac{1}{(1+x)^2} dx$$
$$=\int_{0}^{\infty} \frac{x^{m-1}}{(1+x)^{m+n}} dx$$
Let $x=\tan \theta$, then $$ \begin{aligned} I & =\int_0^{\frac{\pi}{2}} \tan ^s \theta d \theta \\ & =\int_0^{\frac{\pi}{2}} \sin ^{2\left(\frac{s+1}{2}\right)-1} \theta \cos ^{2\left(\frac{-s+1}{2}\right)-1} \theta d \theta \\ & =\frac{1}{2} B\left(\frac{s+1}{2}, \frac{-s+1}{2}\right) \\ & =\frac{\pi}{2} \csc \left(\frac{\pi(1-s)}{2}\right)\cdots (*) \\ & =\frac{\pi}{2} \sec \left(\frac{s}{2}\right) \end{aligned} $$ where $(*)$ using the reflection property of Beta Function.
