Calculate $\int_{0}^{\infty}\frac{1}{(x^2 + t^2)^n}dx$ for $t > 0, n \ge 1$ The problem is as follows:
Show that for all $t > 0, n \ge 1$,
$$
\int_{0}^{\infty}\frac{1}{(x^2 + t^2)^n}dx = {2n-2 \choose n -1}\frac{\pi}{(2t)^{2n -1}}
$$
What I have so far:
Let $f(t) = \int_{0}^{\infty}\frac{1}{(x^2 + t^2)^n}dx$
$$
\frac{df}{dt} = \int_{0}^{\infty}\frac{\partial}{\partial t}\frac{1}{(x^2+t^2)^n}dx
$$
After simplification
$$
\frac{df}{dt} = \int_{0}^{\infty}-2nt(x^2+t^2)^{-n-1}dx
$$
However, I am not sure how to integrate this and WolframAlpha outputs a complicated solution, so any help would be greatly appreciated.
 A: Let $$ I_{n} = \int_{0}^{\infty} \frac{dx}{(x^2 + t^2)^n} $$ then by using integration by parts this becomes
\begin{align}
I_{n} &= \left[\frac{x}{(x^2 + t^2)^{n+1}}\right]_{0}^{\infty} + 2 \, n \, \int_{0}^{\infty} \frac{x^2 \, dx}{(x^2 + t^2)^{n+1}} \\
&= 2 \, n \, \int_{0}^{\infty} \frac{(x^2 + t^2) - t^2}{(x^2 + t^2)^{n+1}} \, dx \\
I_{n} &= 2 \, n \, \left(I_{n} - t^2 \, I_{n+1}\right).
\end{align}
This gives
$$I_{n+1} = \frac{2 n -1}{2 \, n \, t^2} \, I_{n}. $$
When $n=1$ the integral $I_{1}$ gives the result
$$ I_{1} = \int_{0}^{\infty} \frac{dx}{x^2 + t^2} = \frac{1}{t} \, \left[ \tan^{-1}\left(\frac{x}{t}\right) \right]_{0}^{\infty} = \frac{\pi}{2 \, t}.$$
From here the values of $I_{n}$ can be obtained.
A method using a generating function goes as follows.
\begin{align}
\sum_{n=1}^{\infty} I_{n} \, y^n &= \sum_{n=1}^{\infty} \, \int_{0}^{\infty} \left(\frac{y}{x^2 + t^2}\right)^n \, dx \\
&= \int_{0}^{\infty} \frac{y \, dx}{x^2 + t^2 - y} \\
&= \frac{\pi}{2} \, \frac{y}{\sqrt{t^2 - y}} \\
&= \frac{\pi}{2} \, \sum_{k=0}^{\infty} \binom{-1/2}{k} \, \frac{(-1)^{k}}{t^{2k+1}} \, y^{k+1} = \frac{\pi}{2} \, \sum_{n=1}^{\infty} \binom{-1/2}{n-1} \, \frac{(-1)^{n-1}}{t^{2n-1}} \, y^n.
\end{align}
This leads to
$$ I_{n} = \frac{\pi}{2} \, \binom{-1/2}{n-1} \, \frac{(-1)^{n-1}}{t^{2n-1}} = \frac{\pi \, n \, C_{n-1}}{(2 t)^{2n-1}}, $$
where $C_{n}$ is the $n^{th}$ Catalan number.
