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?

$$J(b)=\int_0^\infty \frac{1}{x^2+b}dx=\frac{\pi}{2\sqrt{b}}$$
$$I=\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{db^{n-1}}J(b)|_{b=a^2}$$
$$\frac{d^{n-1}}{db^{n-1}}J(b)=\frac{\pi}{2}(-1)^{n-1}\frac{(2n-3)!!}{2^{n-1}}\frac{1}{b^{\frac{2n-1}{2}}}$$
$$I=\frac{\pi}{2}\frac{1}{(n-1)!}\frac{(2n-3)!!}{2^{n-1}}\frac{1}{a^{2n-1}}$$
where
$$(2n-3)!!=\frac{(2n-2)!}{2^{n-1}(n-1)!}$$
$$I=\frac{\pi}{2}\frac{1}{(n-1)!}\frac{(2n-2)!}{(n-1)!~4^{n-1}}\frac{1}{a^{2n-1}}$$
Final result:
$$I=\frac{\pi}{2}\binom{2n-2}{n-1}\frac{1}{4^{n-1}}\frac{1}{a^{2n-1}},~~~n=1,2,3...$$
