Evaluate $\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^z}$ where $\Re(z)>\frac{1}{2}$

Question

I am dealing with the following complex integration:

$$\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^z} \quad\text{ where }\quad\Re(z)>\frac{1}{2}$$

But I do not know how exactly to deal with the $z$.

Answer 1

Using Schwinger parametrization one can write $$\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^z}dx=\frac{1}{\Gamma(z)}\int_{-\infty}^{\infty}\int_0^\infty t^{z-1}e^{-t(1+x^2)}dtdx$$ Then switching the order of integration we have $$\frac{1}{\Gamma(z)}\int_0^{\infty}t^{z-1}e^{-t}\int_{-\infty}^\infty e^{-tx^2}dxdt$$ The second integral is a Gaussian integral and can be found by using the substitution $s=\sqrt{t}x\rightarrow ds=\sqrt{t}dx$ which gives $$\int_{-\infty}^\infty e^{-tx^2}dx=\frac1{\sqrt{t}}\int_{-\infty}^\infty e^{-s^2}ds=\sqrt{\frac{\pi}t}$$ So the integral becomes $$\frac{1}{\Gamma(z)}\int_0^{\infty}t^{z-1}e^{-t}\sqrt{\frac{\pi}t}dt=\frac{\sqrt{\pi}}{\Gamma(z)}\int_0^{\infty}t^{z-1.5}e^{-t}dt$$ and then the integral left is the Gamma function - $\Gamma(z-0.5)$ - because $$\int_0^\infty t^{z-1}e^{-t}dt=\Gamma(z)$$ Giving the final answer of $$\boxed{\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^z}dx=\sqrt{\pi}\frac{\Gamma(z-0.5)}{\Gamma(z)}}$$

Answer 2

Let $f(z)$ be given by

$$f(z) =\int_0^\infty \frac1{(1+x^2)^z}\,dx$$

Since the integrand is an even function of $x$ we have after enforcing the substitution $x\mapsto \sqrt x$

$$\begin{align} f(z)&=\int_0^\infty \frac{x^{-1/2}}{(1+x)^z}\,dx\\\\ &=B(1/2,z-1/2)\\\\ &=\sqrt \pi \,\frac{\Gamma(z-1/2)}{\Gamma(z)} \end{align}$$