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!
 A: Integrate
$$\left( \frac t{(1+t^2)^{n-1}}\right)’
=\frac{3-2n}{(1+t^2)^{n-1}}+ \frac{2n-2}{(1+t^2)^{n}}
$$
to establish $I_n= \int_0^\infty \frac{1}{(1+t^2)^{n}}dt=\frac{2n-3}{2n-2}I_{n-1}
$.
Then\begin{align}
\int_0 ^\infty\frac{dx}{(x^2+b^2)^4}
=\frac{1}{b^7}\int_0^{\infty}\frac{dt}{(t^2+1)^4}
=\frac1{b^7}\frac56\frac34\frac12\int_0^\infty \frac{dt}{1+t^2}= \frac{5\pi}{32b^7}
\end{align}
A: Since you are not looking for a solution but rather a hint in the right direction, I will outline the solution process for you and leave the rest for you to fill in.
Let us first take at the denominator and let us solve
$$(x^2 + b^2)^4 = 0.$$
It is easy to show that $x = \pm i b$, each with multiplicity 1. Next, consider a contour in the complex plane consisting of (a) an upper half disk of radius $R$ (denoted $C_a$) and (b) the real segment $[-R, R]$ (denoted $C_b$). Denote the union of these two contours as $C$.
The idea here is to let $R \rightarrow \infty$ and use Cauchy's theorem. Observe that
$$\int_C \frac{1}{(z^2 + b^2)^4}dz = \int_{C_a}\frac{1}{(z^2 + b^2)^4}dz + \int_{C_b}\frac{1}{(z^2 + b^2)^4}dz\\ \implies \int_C \frac{1}{(z^2 + b^2)^4}dz = \int_{C_a}\frac{1}{(z^2 + b^2)^4}dz + \int_{-\infty}^\infty\frac{1}{(x^2 + b^2)^4}dx\\
\implies \int_{0}^\infty\frac{1}{(x^2 + b^2)^4}dx =  \frac{1}{2}\int_C \frac{1}{(z^2 + b^2)^4}dz - \frac{1}{2}\int_{C_a}\frac{1}{(z^2 + b^2)^4}dz.$$
Here the last implication follows because your integrand is an even function. Let us consider the last integral first. Using the ML inequality it is easy to show that this integral tends to 0 as $R \rightarrow \infty$, I leave it to you to show this.
Hence our integral reduces to evaluating
$$\frac{1}{2}\int_C \frac{1}{(z^2 + b^2)^4}dz$$
which we will evalaute using the residue theorem. Here only the pole $z = ib$ is within our contour. Furthermore, note that we can write
$$\int_C \frac{1}{(z^2 + b^2)^4}dz = \int_C \frac{1}{(z + ib)^4(z - ib)^4}dz$$
so it is a pole of order 4.
By Cauchy's Theorem you know that
$$\int_C \frac{1}{(z^2 + b^2)^4}dz = 2\pi i \cdot \text{Res}(ib).$$
So what remains is for you to calculate the residue at this particular pole. Can you take it from here?
A: This is the more bone-headed approach, but it works.
Letting $x=bt,$ you get the integral $$\frac{1}{b^7}\int_0^{\infty}\frac1{(t^2+1)^4}\,dt\tag 1$$ Using Wolfram Alpha, I get the partial fractions:
$$\frac1{(t^2+1)^4}=\frac{5i}{32}\left(\frac{1}{t+i}-\frac1{t-i}\right)-\frac5{32}\left(\frac{1}{(t+i)^2}+\frac1{(t-i)^2}\right)-\frac i8\left(\frac1{(t+i)^3}-\frac{1}{(t-i)^3}\right)+\frac1{16}\left(\frac{1}{(t+i)^4}+\frac1{(t-i)^4}\right)$$
The pairs cancel out in the integral from $0$ to $\infty,$ except for the first terms.
And $$\frac{5i}{32}\left(\frac{1}{t+i}-\frac{1}{t-i}\right)=\frac{5}{16}\frac{1}{t^2+1}$$
So you get tbe result:
$$\frac{5}{16b^7}\int_{0}^{\infty}\frac{dt}{t^2+1}=\frac{5\pi}{32b^7}$$

The classic substitution is $t=\tan\theta$ so $dt=\sec^2\theta\,d\theta.$ and $1+t^2=\sec^2\theta.$ Then the integral becomes:
$$\frac{1}{b^7}\int_{0}^{\pi/2}\cos^6\theta\,d\theta$$
But using $\cos(\theta)=(e^{i\theta}+e^{-i\theta})/{2},$ you get: $$\cos^6(\theta)=\frac{1}{32}\left(\cos(6\theta)+6\cos(4\theta)+15\cos(2\theta)+10\right)$$
The $\cos(2k\theta)$ integrals are zero.
This shows more generally why:
$$\int_{0}^{\infty}\frac{dt}{(1+t^2)^n}=\int_0^{\pi/2}\cos^{2n-2}\theta\,d\theta=\frac{\binom{2n-2}{n-1}\pi}{2^{2n-1}}$$
A: $$I=\int_0^\infty \frac{dx}{(x^2+b^2)^4}$$
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:$$
I=\frac{I_4}{b^7}=\frac{5!!}{6!!}\frac{\pi}{2}\frac{1}{b^7}=\frac{5\pi}{32b^7}$$
A: 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.
A: There are posted plenty of nice solutions. Just to add one more - evaluation via beta-function.
For $n\in N$
$$I=\int_0^\infty \frac{dt}{(t^2+b^2)^n}=\frac{1}{b^{2n-1}}\int_0^\infty \frac{dx}{(x^2+1)^n}$$
Making the substitution $t=\frac{1}{1+x^2}$
$$I=\frac{1}{2\,b^{2n-1}}\int_0^1t^{n-\frac{3}{2}}(1-t)^{-\frac{1}{2}}dt=\frac{1}{2\,b^{2n-1}}\frac{\Gamma\Big(n-\frac{1}{2}\Big)\Gamma\Big(\frac{1}{2}\Big)}{\Gamma(n)}$$
$$=\frac{\pi}{2^nb^{2n-1}}\frac{(2n-3)!!}{(n-1)!}=\frac{\pi}{2\,b^{2n-1}}\frac{(2n-3)!!}{(2n-2)!!}$$
