Closed form for $\int_0^\infty\frac{1}{(1+x^2)^s}\,dx$ when $s\in (0.5,\infty)\setminus\mathbb{N}$ I know that the improper integral 
$$
\int_0^\infty\frac{1}{(1+x^2)^s}\,dx
$$
is convergent for $s>0.5$ and divergent otherwise. Furthermore, it has a closed form for $s \in \mathbb{N}$ (this can be obtained using the Residue Theorem). My question is to know whether it has a closed form for any non-integer $s >0.5$.
 A: It depends what you mean by closed form. Changing variables to $u=x^2$, we have that the integral is equal to
$$ I= \frac{1}{2}\int_0^{\infty} \frac{u^{-1/2}}{(1+u)^s} \, du, $$
where by definition
$$ \Gamma(s) = \int_0^{\infty} t^{s-1} e^{-t} \, dt $$
Now, using
$$ \frac{1}{(1+u)^s} = \frac{1}{\Gamma(s)} \int_0^{\infty} \alpha^{s-1} e^{(1+u)\alpha} \, d\alpha, $$
and changing the order of integration gives
$$ I = \frac{1}{2\Gamma(s)}\int_0^{\infty} \alpha^{s-1}e^{-\alpha} \left( \int_0^{\infty} u^{-1/2} e^{-u\alpha} du \right) \, d\alpha = \frac{1}{2\Gamma(s)}\int_0^{\infty} \alpha^{s-1/2-1}\Gamma(1/2) e^{-\alpha} d\alpha = \frac{\Gamma(1/2)\Gamma(s-1/2)}{2\Gamma(s)}. $$
Now, to answer your question, $\Gamma(1/2)=\sqrt{\pi}$, basically because you can substitute to turn this Gamma integral into the Gaussian integral, $e^{-x^2}$. Similarly, since $ \Gamma(s+1) = s\Gamma(s) $, if $s$ is an integer or a half-integer, you can reduce final form of the value of the integral to a product of factorials (possibly also including a $\pi$). For other values of $s$, there are not known closed forms in terms of more elementary functions for the Gamma-function. As far as I am aware, anyway.
A: By substituting $u=\frac{1}{1+x^2}$ we have, through the Euler's Beta function:
$$\forall s>\frac{1}{2},\qquad \int_{0}^{+\infty}\frac{dx}{(1+x^2)^s}= \frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(s-\frac{1}{2}\right)}{\Gamma(s)}.$$
Since the Fourier transform of $\frac{1}{1+x^2}$ is $e^{-|t|}$, for integer values of $s$ we can read the above line in terms of a convolution of Laplace distributions.
