How to find the formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}$, where $n\in N$?

Question

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.

Answer 1

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

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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 $(2)$

Answer 2

Denote $I_n = \int_0^\infty \frac1{(1+x^2)^n}dx$, and note that $$\left( \frac x{(1+x^2)^{n-1}} \right)’ = \frac {2(n-1)}{(1+x^2)^{n}}- \frac {2n-3}{(1+x^2)^{n-1}} $$ Integrate both sides to get the recursion $I_n= \frac{2n-3}{2(n-1)}I_{n-1} $, along with $I_1=\frac\pi2$. Thus $$I_n = \frac{(2n-3)!(n-1)\pi }{[2^{n-1} (n-1)!]^2}$$

Answer 3

Along the line of @robjohn's solution, consider the change of variables $u=\frac{1}{1+x^2}$. Then $$du=-2 u^2\Big(\frac{1}{u}-1\Big)^{1/2} dx=-\frac12 u^{3/2}(1-u)^{-1/2}dx$$ and so $$\int^\infty_0\frac{dx}{(1+x^2)^n}=\frac12\int^1_0u^{n-\tfrac12-1}(1-u)^{\tfrac12-1}\,du=\frac12B\big(n-\tfrac12,\frac12\big)=\frac12\frac{\Gamma(n-\tfrac12)\Gamma(\tfrac12)}{\Gamma(n)}$$

The identity of the OP follows from properties of the Gamma function.

Answer 4

If you use complex analysis, then there is a pole at $x = i$

If we take the contour of the semi-circle in the upper half-plane, then when $n\ge 1$ the integral along the semi-circle vanishes. That means that we can focus on the residual at $i.$

$\int_0^\infty \frac {1}{(x^2 + 1)^n} \ dx = \frac {\pi i}{(n-1)!} \frac {d^{n-1}}{dx^{n-1}} \frac {1}{(x+i)^n}$ evaluated at $i$

$\frac {\pi (2n-2)!}{((n-1)!)^2(2)^{2n-1}}$

Since it seem like you are interested:

$\int_0^\infty \frac {1}{x^n + 1}\ dx = \frac {\pi \csc \frac {\pi}{n}}{n}$

And, I am thinking that $\int_0^\infty \frac {1}{(x^m + 1)^n}\ dx$ will give something along the lines of $\frac {\pi \csc \frac {\pi}{m}}{m^n} {2n-2 \choose n-1}.$ But, I haven't worked it out.

Answer 5

Thanks to Mr Nejimban, I am now going to evaluate the integral by converting it into a Wallis integral by the substitution $x=\tan \theta$, which yields

$$I_{n}(1)= \int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}} =\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \theta d \theta}{\sec ^{2 n} \theta} =\int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d \theta $$

By the Wallis Formula for cosine, $$ \int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d\theta =\frac{\pi}{2} \prod_{k=1}^{n-1} \frac{2 k-1}{2 k}, $$ Simplifying yields $$ \begin{aligned} \int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta &=\frac{\pi}{2} \prod_{k=1}^{n-1}\left(\frac{2 k-1}{2 k} \cdot \frac{2 k}{2 k}\right) \\ &=\frac{\pi}{2}\cdot \frac{(2 n-2) !}{2^{2 n-2}[(n-1) !]^{2}} \\ &=\frac{\pi}{2^{2 n-1}}\left(\begin{array}{c} 2 n-2 \\ n-1 \end{array}\right) \end{aligned} $$

$$ \int_{0}^{\infty} \frac{d x}{\left(x^{2}+a\right)^{n}} \stackrel{x \rightarrow \frac{x}{\sqrt{a}}}{=} \frac{1}{a^{n-\frac{1}{2}}} I_{n}(1)=\frac{\pi}{a^{\frac{2n-1}{2} } \cdot 2^{2 n-1}}\left(\begin{array}{c} 2 n-2 \\ n-1 \end{array}\right) $$ In particular, $$ \boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}=\frac{\pi}{2^{2 n-1}}\left(\begin{array}{c} 2 n-2 \\ n-1 \end{array}\right)} $$

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\dd x \over \pars{x^{2} + 1}^{n}} \right\vert_{\Re(n)\ >\ 1/2}} \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\int_{0}^{\infty}{x^{\color{#f00}{1/2} - 1} \over \pars{1 + x}^{n}}\,\dd x \end{align}

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Since $\ds{\pars{1 + x}^{-n} = \sum_{k = 0}^{\infty}{-n \choose k}x^{k} = \sum_{k = 0}^{\infty}\bracks{{n + k - 1\choose k}\pars{-1}^{k}}x^{k} = \sum_{k = 0}^{\infty}\color{#f00}{\Gamma\pars{n + k} \over \Gamma\pars{n}}{\pars{-x}^{k} \over k!};}$ \begin{align} & \overbrace{\bbox[5px,#ffd]{\left.\int_{0}^{\infty}{\dd x \over \pars{x^{2} + 1}^{n}} \right\vert_{\Re(n)\ >\ 1/2}} = {1 \over 2}\,\Gamma\pars{1 \over 2}{\Gamma\pars{n - 1/2}\over \Gamma\pars{n}}} ^{\ds{Ramanujan's\ Master\ Theorem}} \\[5mm] = & {\pi \over 2}{\pars{n - 3/2}! \over \pars{n - 1}!\pars{-1/2}!} = \bbox[5px,#ffd]{{\pi \over 2}{n - 3/2 \choose n - 1}} \end{align}