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

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.

• You would have a better statement for more general, real values of $n$ by restricting to $n \gt 1$. – Ron Gordon Oct 23 '15 at 16:44

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

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

• You are right. I made a mistake. – uniquesolution Oct 23 '15 at 16:56
• I just wrote $(2+x^2)$ instead of $(1+x^2)$. Nothing deep. – uniquesolution Oct 23 '15 at 16:57

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.

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.


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}