By the generalization in my post,we are going to evaluate the integral $$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}},$$
where $n\in N.$
First of all, let us define the integral $$I_n(a)=\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} \textrm{ for any positive real number }a.$$
Again, we start with $$I_1(a)=\int_{0}^{\infty} \frac{d x}{x^{2}+a}= \left[\frac{1}{\sqrt{a}} \tan ^{-1}\left(\frac{x}{\sqrt{a}}\right)\right]_{0}^{\infty} = \frac{\pi}{2 }a^{-\frac{1}{2} } $$
Then differentiating $I_1(a)$ w.r.t. $a$ by $n-1$ times yields
$$
\int_{0}^{\infty} \frac{(-1)^{n-1}(n-1) !}{\left(x^{2}+a\right)^{n}} d x=\frac{\pi}{2} \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) \cdots\left(-\frac{2 n-3}{2}\right) a^{-\frac{2 n-1}{2}}
$$
Rearranging and simplifying gives $$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} =\frac{\pi a^{-\frac{2 n-1}{2}}}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)}
$$
Putting $a=1$ gives the formula of our integral
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}} =\frac{\pi}{2^{n}(n-1) !} \prod_{k=1}^{n-1}(2 k-1)= \frac{\pi}{2^{2 n-1}} \left(\begin{array}{c}
2 n-2 \\
n-1
\end{array}\right)}$$
For verification, let’s try $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{10}} &= \frac{\pi}{2^{19}}\left(\begin{array}{c}
18 \\
9
\end{array}\right) =\frac{12155 \pi}{131072} ,
\end{aligned}
$$
which is checked by Wolframalpha .
Are there any other methods to find the formula? Alternate methods are warmly welcome.
Join me if you are interested in creating more formula for those integrals in the form $$
\int_{c}^{d} \frac{f(x)}{\left(x^{m}+1\right)^{n}} d x.
$$
where $m$ and $n$ are natural numbers.
Best Answer
Beta Function
One approach is to use the Beta Function: $$ \begin{align} \int_0^\infty\frac{\mathrm{d}x}{(1+x^2)^n} &=\frac12\int_0^\infty\frac{x^{-1/2}\,\mathrm{d}x}{(1+x)^n}\tag{1a}\\[6pt] &=\tfrac12\operatorname{B}\left(\tfrac12,n-\tfrac12\right)\tag{1b}\\ &=\tfrac12\frac{\sqrt\pi\cdot\sqrt\pi\frac{(2n-2)!}{2^{2n-2}(n-1)!}}{(n-1)!}\tag{1c}\\[3pt] &=\frac\pi{2^{2n-1}}\binom{2n-2}{n-1}\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: substitute $x\mapsto x^{1/2}$
$\text{(1b)}$: Beta Function
$\text{(1c)}$: $\Gamma(1/2)=\sqrt\pi$, $\Gamma(n-1/2)=\sqrt\pi\frac{(2n-2)!}{2^{2n-2}(n-1)!}$, $\Gamma(n)=(n-1)!$
$\text{(1d)}$: simplify
Contour Integration
Another approach is to use contour integration. First, we compute the residue of $\frac1{\left(z^2+1\right)^n}$ at $z=i$: $$\newcommand{\Res}{\operatorname*{Res}} \begin{align} \Res_{z=i}\frac1{\left(z^2+1\right)^n} &=\Res_{z=i}\frac1{(z-i)^n}\frac1{(z+i)^n}\tag{2a}\\ &=\left[u^{-1}\right]\frac1{u^n}\frac1{(u+2i)^n}\tag{2b}\\[3pt] &=\left[u^{-1}\right]\frac1{(2i)^n}\frac1{u^n}\frac1{\left(1+\frac{u}{2i}\right)^n}\tag{2c}\\ &=\frac1{(2i)^n}\binom{-n}{n-1}\left(\frac1{2i}\right)^{n-1}\tag{2d}\\ &=\frac{-i}{2^{2n-1}}\binom{2n-2}{n-1}\tag{2e} \end{align} $$ Explanation:
$\text{(2a)}$: $\frac1{z^2+1}=\frac1{z-i}\frac1{z+i}$
$\text{(2b)}$: substitute $z\mapsto u+i$; the residue at $u=0$ is the coefficient of $u^{-1}$
$\text{(2c)}$: pull out a factor of $(2i)^{-n}$ to prepare for using the Binomial Theorem
$\text{(2d)}$: $\left[u^{-1}\right]u^{-n}(1+\frac{u}{2i})^{-n}=\left[u^{n-1}\right]\left(1+\frac{u}{2i}\right)^{-n}$
$\phantom{\text{(2d):}}$ which can be computed using the Binomial Theorem
$\text{(2e)}$: simplify using negative binomial coefficients
Now it is fairly easy to compute $$ \begin{align} \int_0^\infty\frac{\mathrm{d}x}{(x^2+1)^n} &=\frac12\int_\gamma\frac{\mathrm{d}z}{(z^2+1)^n}\tag{3a}\\[3pt] &=\pi i\Res_{z=i}\frac1{\left(z^2+1\right)}\tag{3b}\\ &=\frac\pi{2^{2n-1}}\binom{2n-2}{n-1}\tag{3c} \end{align} $$ Explanation:
$\text{(3a)}$: use the contour $\gamma=\lim\limits_{R\to\infty}[-R,R]\cup Re^{i[0,\pi]}$
$\phantom{\text{(3a):}}$ which circles the singularity at $i$ once counter-clockwise
$\phantom{\text{(3a):}}$ the integral along the arc vanishes as $R\to\infty$
$\text{(3b)}$: Residue Theorem
$\text{(3c)}$: apply $(2)$