Closed form for the integral $\int_0^\infty t^s/(1+t^2)$

Question

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.

Answer 1

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

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

Answer 2

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 $x \to \infty$.

$$\int_{0}^{\infty} \frac{x^{m-1}}{(1+x)^{m-1}} \frac{1}{(1+x)^{n-1}} \frac{1}{(1+x)^2} dx$$

$$=\int_{0}^{\infty} \frac{x^{m-1}}{(1+x)^{m+n}} dx$$