Solve $\int_{0}^\infty \frac{1}{(x^2 +b^2)^4}\ dx$ I ran into this integral in the context of Quantum Mechanics, and I don't really know how to tackle it. Here, $b$ is simply a real constant, which we can assume is positive.
$$\int\limits_0 ^\infty\frac{1}{(x^2+b^2)^4}\ dx$$
It doesn't look like I can use "traditional" methods to solve it, so I was thinking to maybe try to transform it to complex integral somehow and apply Cauchy's theorem or something, but I'm unsure if that would even work.
Any nudge in the right direction would be appreciated!

Here's a short proof for a generalization of the problem. From this answer you know that the integral $I_{\color{blue}{n}} = \int_0^\infty\frac{\mathrm{d}x}{(x^2+1)^{\color{blue}{n}}}$ follows the recursion $I_{n+1}=\frac{2n-1}{2n}I_n$. Solving the recursion gives:
$$
I_{n} = \frac{2n-3}{2n-2}I_{n-1} =  \frac{(2n-3)(2n-5)}{(2n-2)(2n-4)}I_{n-2}= \dots = \frac{(2n-3)(2n-5)\dots(5)(3)(1)}{(2n-2)(2n-4)\dots(6)(4)(2)}I_1
$$
And since $I_1 = \int_{0}^{\infty} \frac{\mathrm{d}x}{(1+x^2)^1} = \arctan(x)\Big\vert_{0}^{\infty} = \frac{\pi}{2}$, recalling the definition of the double factorial you get the closed form
$$
I_n =\int_0^\infty\frac{1}{(u^2+1)^{n}}\mathrm{d}u = \frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}\qquad \text{for} \quad \qquad n \in \mathbb{N}
$$
But since $ \int_0^\infty\frac{1}{(u^2+1)^{n}}\mathrm{d}u = \int_0^\infty\frac{1}{\left(\frac{(ub)^2}{b^2}+\frac{b^2}{b^2}\right)^{n}}\mathrm{d}u = b^{2n}\int_0^\infty\frac{1}{\left((ub)^2+b^2\right)^{n}}\mathrm{d}u  $, taking the substitution $x = ub$ gives
$$
\boxed{\int_{0}^{\infty} \frac{1}{(x^2 +b^2)^n}\mathrm{d}x=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2|b|^{2n-1}} \qquad \text{for} \quad b\neq  0, \ n \in \mathbb{N}}
$$
which for your particular case gives $\int_{0}^{\infty} \frac{1}{(x^2 +b^2)^4} = \frac{5!!}{6!!}\frac{\pi}{2b^{2(4)-1}} = \frac{5\pi}{32b^{7}} $ as expected.
