Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2+1)^3} dx$

Question

Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2+1)^3} dx$ I wasnt exactly sure how to approach this. I saw some similar examples that used Cauchy's theorem.

Answer 1

$$\int_{-\infty}^{+\infty}\frac{dx}{(x^2+1)^3} = 2\pi i\cdot\operatorname{Res}\left(\frac{1}{(z^2+1)^3},z=i\right)=\pi i\left.\frac{d^2}{dz^2}\frac{(z-i)^3}{(z^2+1)^3}\right|_{z=i}=\color{red}{\frac{3\pi}{8}}. $$

Answer 2

Another way is a trig substitution $x = \tan u$ which reduces the integrand to $$ \int \frac{dx}{(x^2+1)^3} = \int \frac{\sec u \tan u}{\sec^6 u} du = \int \sin u \cos^4 u du = \frac{\cos^5 u}{5} $$ and the rest is arithmetic

Answer 3

$x=\tan{u}\rightarrow \frac{dx}{1+x^2}=du$

So the problem reduces to

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^4{u}\space du$

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \int_{-\infty}^{\infty}{\dd x \over \pars{x^{2} + 1}^{3}}} =\left.\totald[2]{}{\mu}\int_{0}^{\infty}{\dd x \over x^{2} + \mu} \right\vert_{\, \mu\ =\ 1} =\left.\totald[2]{}{\mu}\pars{\mu^{-1/2}\int_{0}^{\infty}{\dd x \over x^{2} + 1}} \right\vert_{\, \mu\ =\ 1} \\[5mm]&={\pi \over 2}\,\left.\totald[2]{\mu^{-1/2}}{\mu}\right\vert_{\, \mu\ =\ 1} ={\pi \over 2}\,\pars{{3 \over 4}\,\mu^{-5/2}}_{\, \mu\ =\ 1} =\color{#66f}{\large{3\pi \over 8}} \end{align}