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?

Letting $x=a\tan \theta$ transform the integral into a Beta function
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{a \sec ^{2} \theta d \theta}{\left(a^{2} \tan ^{2} \theta+a^{2}\right)^{n}}
&=\int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d \theta =\frac{1}{2 a^{2 n-1}} B\left(\frac{1}{2}, n-\frac{1}{2}\right)
\end{aligned}
$$
Using the property of Beta function, we have
$$
\begin{aligned}
B\left(\frac{1}{2}, n-\frac{1}{2}\right) &= \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(n-\frac{1}{2}\right)}{\Gamma(n)} \\
&= \frac{1}{(n-1)!} \Gamma\left(\frac{1}{2}\right)\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\left(n-\frac{5}{2}\right) \cdots\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)\\&= \pi\frac{(2 n-3)(2 n-5) \cdots 1}{(n-1) ! 2^{n-1}} \times \frac{(2 n-2)(2 n-4) \cdots 2}{(2 n-2)(2 n-4) \cdots 2}\\&= \pi\frac{(2 n-2) !}{(n-1) ! 2^{2 n-2}(n-1) !}\\&= \frac{\pi}{2^{2 n-2}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)
\end{aligned}
$$
Now we can conclude that
$$\boxed{\int_{0}^{\frac{\pi}{2}} \frac{a \sec ^{2} \theta d \theta}{\left(a^{2} \tan ^{2} \theta+a^{2}\right)^{n}}= \frac{\pi}{(2a)^{2 n-1}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}$$
