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?
 A: $$J(b)=\int_0^\infty \frac{1}{x^2+b}dx=\frac{\pi}{2\sqrt{b}}$$
$$I=\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{db^{n-1}}J(b)|_{b=a^2}$$
$$\frac{d^{n-1}}{db^{n-1}}J(b)=\frac{\pi}{2}(-1)^{n-1}\frac{(2n-3)!!}{2^{n-1}}\frac{1}{b^{\frac{2n-1}{2}}}$$
$$I=\frac{\pi}{2}\frac{1}{(n-1)!}\frac{(2n-3)!!}{2^{n-1}}\frac{1}{a^{2n-1}}$$
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:
$$I=\frac{\pi}{2}\binom{2n-2}{n-1}\frac{1}{4^{n-1}}\frac{1}{a^{2n-1}},~~~n=1,2,3...$$
A: 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}
A: Maple says: if $a,n$ are real numbers, $a>0, n>\frac12$, then
$$
\int_0^\infty\frac{dx}{(x^2+a^2)^n} = 
\frac{\sqrt{\pi}\,\Gamma(n-\frac12)}{2 \,a^{2n-1}\,\Gamma(n)}
$$
If $n$ is a positive integer, this is, indeed, equal to
$$
I_n = (-1)^{n+1} \binom{-1/2}{n-1} \cdot \frac{\pi}{2 \, a^{2n-1}}
\tag1$$

Note: We can summarize $(1)$ as
$$
\sum_{n=1}^\infty I_n = {\frac {-\pi a}{2\sqrt {{a}^{2}+1}\left( a+\sqrt {{a}^{2}+1
} \right)}}  \quad\text{for all } a>1
$$
A: $$I_n = \int \frac{dx}{(x^2+a^2)^{n}} =\frac 1{a^{2n-1}}\int \frac{dx}{(t^2+1)^{n}}$$
Using the Gaussian hypergeometric function
$$I_n=\frac t{a^{2n-1}}  \,\, _2F_1\left(\frac{1}{2},n;\frac{3}{2};-t^2\right)$$
$$J_n = \int_0^\infty \frac{dx}{(x^2+a^2)^{n}}=\frac {\sqrt \pi}{2\,a^{2n-1}}\,\frac{\Gamma \left(n-\frac{1}{2}\right)}{\Gamma (n)}$$ which is identical to your conjecture.
A: 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
$$\boxed{\int_{0}^{\frac{\pi}{2}} \frac{a \sec ^{2} \theta d \theta}{\left(a^{2} \tan ^{2} \theta+a^{2}\right)^{n}}= \frac{\pi}{(2a)^{2 n-1}}\left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}$$
A: $$
I=\int_0^{\infty}\frac{dx}{(x^2+a^2)^n}=\frac{1}{2}\int_{-\infty}^{\infty}\frac{dx}{(x^2+a^2)^n}
$$
The function $\displaystyle f(z)=\frac{1}{(z^2+a^2)^n}$ admits $\pm ia$ as the pole of order $n$ and
\begin{eqnarray}
\text{Res}(f,ia)&=&\lim_{z\to ia}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}\left[(z-ia)^nf(z)\right]\cr
&=&\lim_{z\to ia}\frac{(-1)^{n-1}n(n+1)\cdots(2n-2)}{(n-1)!(z+ia)^{2n-1}}\cr
&=&\frac{(-1)^{n-1}n(n+1)\cdots(2n-2)}{(n-1)!(2ia)^{2n-1}}\cr
&=&\frac{n{2n \choose n}}{2i(2n-1)(2a)^{2n-1}}
\end{eqnarray}
Given $R>|a|$, let's consider the integral
$$
J_R=\int_{\Gamma_R}f(z)dz,
$$
where $\Gamma_R$ consists of the segment $[-R,R]$ and the upper half circle of radius $R$ centered at the origin ( oriented counterclockwise).
On one hand, we have thanks to the residue theorem
$$\tag{1}
J_R=2\pi i \text{Res}(f, ia)=\frac{\pi n{2n \choose n}}{(2n-1)(2a)^{2n-1}}.
$$
On the other hand, we have
$$
J_R=\underbrace{\int_{-R}^Rf(x)dx}_{J_R^1}+\underbrace{\int_0^{\pi}\frac{iRe^{it}dt}{(R^2e^{2it}+a^2)^n}}_{J_R^2}.
$$
Since
$$
|J_R^2|\le \int_0^{\pi}\frac{Rdt}{(R^2-a^2)^n}=\frac{\pi R}{(R^2-a^2)^n},
$$
we deduce that $J_R^2\to 0$ as $R \to \infty$. It follows that
$$
J_R=\lim_{R\to\infty}J_R^1=\int_{-\infty}^{\infty}f(x)dx.
$$
Hence
$$
I=\int_0^{\infty}\frac{dx}{(x^2+a^2)^n}=\frac{1}{2}J_R=\frac{\pi n{2n \choose n}}{2(2n-1)(2a)^{2n-1}}.
$$
