How to solve $\int_0^{\infty}\frac{x^n}{(x^2+1)^n}\,\mathrm dx$ for $n\ge 2$?

Question

I am trying to find some closed form answer for the integral $$\int_0^{\infty}\frac{x^n}{(x^2+1)^n}\,\mathrm dx,\quad n\ge 2$$ I am not sure if a closed form exists and I have been trying this integral for hours.

Any tips or hints would be appreciated.

Answer 1

Sub $x=\tan{t}$ and get

$$\int_0^{\pi/2} dt \, \sin^n{t} \, \cos^{n-2}{t} $$

which one may recognize as a Beta function. Further to this, sub $y=\sin{t}$ and get

$$\int_0^1 dy \, y^n \, (1-y^2)^{\frac{n-3}{2}} = \frac12 \int_0^1 du \, u^{\frac{n-1}{2}} (1-u)^{\frac{n-3}{2}}$$

which is

$$\frac{\Gamma \left ( \frac{n+1}{2} \right ) \Gamma \left ( \frac{n-1}{2} \right )}{2 \Gamma(n)} $$

Answer 2

Let $x=\tan t,u=\sin t$. Then \begin{eqnarray} \int_0^{\infty}\frac{x^n}{(x^2+1)^n}dx&=&\int_0^{\pi/2}\sin^nt\cos^{n-2}tdt\\ &=&\int_0^{1}u^n(1-u^2)^{\frac{n-3}2}du\\ &=&\frac12\int_0^{1}u^{\frac{n-1}{2}}(1-u)^{\frac{n-3}2}du \end{eqnarray} Then using $$ B(p,q)=\int_0^1u^{p-1}(1-u)^{q-1}du $$ you can easily get the answer.

Answer 3

Both other answers have made indirect use of Wallis' integrals to arrive at the integral form of

Euler's beta function. However, this can be done directly, by substituting $t=\dfrac1{x^2+1}~.~$ In fact,

all integrals of the form $~\displaystyle\int_0^\infty\frac{x^{k-1}}{\Big(x^p+a^p\Big)^n}~dx~$ can be evaluated by a similar approach, letting

first $~x=au,~$ and then $~t=\dfrac1{u^p+1}~,~$ finally yielding $~\dfrac{a^{k-np}}p~B\bigg(\dfrac kp~,~n-\dfrac kp\bigg),~$ which, for

integer values of n, can be substantially simplified by recursively employing $\Gamma(x+1)=x~\Gamma(x),~$

in conjunction with Euler's reflection formula for the $\Gamma$ function.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\infty} {x^{n} \over \pars{x^{2} + 1}^{n}} \,\dd x\,\right\vert_{\,\Re\pars{n}\ >\ 1}} \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\int_{0}^{\infty} x^{\color{red}{n/2 + 1/2} - 1}\,\,\,\,\pars{1 + x}^{-n}\,\,\dd x \end{align} Note that $\ds{\pars{1 + x}^{-n} = \sum_{k = 0}^{\infty}{-n \choose k}x^{k} = \sum_{k = 0}^{\infty}{n + k - 1 \choose k}\pars{-1}^{k}x^{k} = \sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{n + k} \over \Gamma\pars{n}}{\pars{-x}^{k} \over k!}}$.

Then, with Ramanujan's Master Theorem, \begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\infty} {x^{n} \over \pars{x^{2} + 1}^{n}} \,\dd x\,\right\vert_{\,\Re\pars{n}\ >\ 1}} = {1 \over 2}\,\Gamma\pars{\color{red}{{n \over 2} + {1 \over 2}}} {\Gamma\pars{n - \bracks{\color{red}{n/2 + 1/2}}} \over \Gamma\pars{n}} \\[5mm] = &\ \bbx{\Gamma\pars{n/2 + 1/2}\Gamma\pars{n/2 - 1/2} \over 2\Gamma\pars{n}}\\ & \end{align}