<s>[INST] 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!
 [/INST] $$I=\int_0^\infty \frac{dx}{(x^2+b^2)^4}$$
First I’ll start by ridding the integrand of $b$
$$I=\int_0^\infty \frac{dx}{(x^2+b^2)^4}=\int_0^\infty\frac{dx}{b^8((\frac{x}{b})^2+1)}$$
Substitute $u=\frac{x}{b}$
$$I=\frac{1}{b^7}\int_0^\infty\frac{dx}{(x^2+1)^4}$$
Now we will start building a reduction formula for the indefinite integral $\int \frac{dx}{(x^2+1)^n}$
$$J_{n}=\int \frac{dx}{(x^2+1)^n}=\int \frac{x^2+1-x^2}{(x^2+1)^n}dx=\int\frac{dx}{(x^2+1)^{n-1}}-\int\frac{x\cdot x}{(x^2+1)^n}dx$$
We can recognize that the first integral is $J_{n-1}$, and we will use integration by parts on the second, where $u=x$ and $dv=\frac{x}{(x^2+1)^n}$
$$J_{n}=J_{n-1}-\left[\frac{-x}{(2n-2)(x^2+1)^{n-1}}+\frac{1}{2n-2}\int \frac{dx}{(x^2+1)^{n-1}} \right]$$
Simplify
$$J_{n}=J_{n-1}-\frac{1}{2n-2}J_{n-1}+\frac{x}{(2n-2)(x^2+1)^{n-1}}=\frac{2n-3}{2n-2}J_{n-1}+\frac{x}{(2n-2)(x^2+1)^{n-1}}$$
Now we’ll introduce a new sequence, $I_{n}$, and find a recursive formula using our information on $J_{n}$
$$I_{n}=\int_0^\infty\frac{dx}{(x^2+1)^n}=\frac{2n-3}{2n-2}I_{n-1}+\left. \frac{x}{(2n-2)(x^2+1)^{n-1}}\right |_0^\infty=\frac{2n-3}{2n-2}I_{n-1}$$
Therefore
$$\frac{I_{n}}{I_{n-1}}=\frac{2n-3}{2n-2}$$
$$\frac{I_{n}}{I_{n-1}}\frac{I_{n-1}}{I_{n-2}} \frac{I_{n-2}}{I_{n-3}}\cdots\frac{I_{3}}{I_{2}}\frac{I_{2}}{I_{1}}=\frac{2n-3}{2n-2}\frac{2n-5}{2n-4}\frac{2n-7}{2n-6}\cdots\frac{3}{4}\frac{1}{2}$$
On the left side we cancel almost everything. The right side can be rewritten using double factorials.
$$\frac{I_{n}}{I_1}=\frac{(2n-3)!!}{(2n-2)!!}$$
$$I_1=\int_0^\infty \frac{dx}{x^2+1}=\frac{\pi}{2}$$
Therefore
$$I_{n}=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}$$
We can calculate the original integral like this:$$
I=\frac{I_4}{b^7}=\frac{5!!}{6!!}\frac{\pi}{2}\frac{1}{b^7}=\frac{5\pi}{32b^7}$$
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