<s>[INST] Conjectured closed form of $\int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x$ Consider
$$ I = \int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x $$
where $a > 0$ is a constant. We can evaluate this using the Leibnitz theorem for specific values of $n$. Mathematica can solve it for specific values of $n$ as well but not the generalized form. Here, I conjecture a generalization -
$$ I = (-1)^{n+1} \binom{-1/2}{n-1} \cdot \frac{\pi}{2 \, a^{2n-1}} $$
This can be further simplified by
$$ \binom{-1/2}{n} = \left(\frac{-1}{4}\right)^n \binom{2n}{n} $$
Is this conjecture true?
 [/INST] Let's first try to find a recursive relation of $I_n$. In fact using integration by part,
\begin{align}
I_{n} &= \int_0^\infty \frac{1}{(x^2 + a^2)^n}\mathrm dx\\
&= \int_0^\infty \frac{x^2 + a^2}{(x^2 + a^2)^{n+1}}\mathrm dx\\
&= \frac{1}{2n} I_{n} + a^2 I_{n+1}. 
\end{align}
So $$I_{n+1} = \frac{1}{a^2} \frac{2n-1}{2n} I_n = \frac{1}{a^2}\frac{2n(2n-1)}{4n^2} I_n$$
Use induction to prove that:
$$I_n = \frac{1}{a^{2(n-1)}} \frac{(2(n-1))!}{4^{n-1} ((n-1)!)^2} I_1$$

Another way to compute $I_n$ for general case (even with $n$ is non-integer) is
\begin{align}
I_n &= \int_{0}^{\infty} \frac{1}{(x^2 + a^2)^n}\mathrm dx\\
&=_{t := \frac{x}{a}} \frac1{a^{2n-1}}\int_0^{\infty}\frac{1}{(1+t^2)^n}\mathrm dt\\
&=_{u := \frac1{1+t^2}} \frac1{a^{2n-1}}\int_1^0 u^{n} (\frac{1}{2}(-\frac1{u^2}(\frac{1-u}{u})^{-\frac12})) \mathrm d u\\
&= -\frac1{2a^{2n-1}}\int_0^1 u^{n-\frac32}(1-u)^{-\frac12}\mathrm d u\\
&= -\frac{1}{2a^{2n-1}} \beta (n-\frac12, \frac12)
\end{align}
</s>