How would you compute for the definite integral of
$$\int_0^{\infty}\frac{dx}{(1+x^2)^4}$$
I know that integral of $\displaystyle \frac1{(1+x^2)}$ equals $\tan^{-1}x$. I tried using integration by parts without much luck. My teacher pointed me to special functions by which I found out about the hypergeometric distribution. Although I don't know how to apply it to this problem.
Anybody know how to use special functions or how to go about this problem?
The method given below works in general for any integral of the form $\displaystyle \int_{0}^{\infty} \frac{dx}{(1+x^2)^n}$.
Plug in $x = \tan(\theta)$. The integral becomes
$$\displaystyle \int_{0}^{\infty} \frac{dx}{(1+x^2)^4} = \int_{0}^{\pi/2} \frac{\sec^2(\theta) d \theta}{(1+\tan^2(\theta))^4} = \int_{0}^{\pi/2} \cos^{6}(\theta) d\theta = \frac{5}{6}\frac{3}{4}\frac{1}{2}\frac{\pi}{2} = \frac{5}{32}\pi$$ where the last integral has been done in a previous post here. The integration over there has been done for $\sin^n(\theta)$ but the same method works for $\cos^n(\theta)$.
If you are familiar with complex analysis, you could solve it using complex analysis by extending the function into the complex domain, choose a semicircular contour with the diameter along the real axis. Look for the poles inside the contour, (there are four poles at $z=+i$) and compute the residues. Let the radius of the semicircle tend to infinity and evaluate the integral using Cauchy residue theorem.
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} {\cal F}\pars{\mu}&\equiv \int_0^{\infty}{\dd x \over \mu + x^{2}} =\mu^{-1/2}\int_0^{\infty}{\dd x \over 1 + x^{2}} = \half\,\pi\mu^{-1/2} \end{align} \begin{align} {\cal F}^{'''}\pars{\mu}&\equiv -3!\int_0^{\infty}{\dd x \over \pars{\mu + x^{2}}^{4}} =\half\,\pi\,\totald[3]{\mu^{-1/2}}{\mu} =\half\,\pi\,\pars{-\,\half}\pars{-\,{3 \over 2}}\pars{-\,{5 \over 2}}\mu^{-7/2} \end{align} Set $\mu = 1$ in both members: $$ \color{#00f}{\large\int_0^{\infty}{\dd x \over \pars{1 + x^{2}}^{4}}} ={15\pi/16 \over 6}= \color{#00f}{\large{5 \over 32}\,\pi} $$
Integration by parts does in fact help. Let $I_n = \int (1+x^2)^{-n}dx$. Then multiply by 1 and integrate by parts: $$I_n = x (1+x^2)^{-n} - \int x \cdot (-n)2x (1+x^2)^{-n-1}dx = \dots = x (1+x^2)^{-n} + 2n(I_n - I_{n+1}).$$ (In the "..." step, write $x^2=(x^2+1)-1$ in the numerator and then divide.)
From this you can solve for $I_{n+1}$ in terms of $I_n$, and since you know $I_1 = \arctan x + C$, you can recursively compute $I_2$, $I_3$, $I_4$, etc.
(This is pretty much the same method that Sivaram pointed you to for finding $\int \cos^6 \theta \, d\theta$.)
Ten years later (with Residue Calculus).
If $\gamma^R=\gamma_1^R\cup \gamma_2^R$ is the contour $$ \gamma_R^1(t)=t, \,\,\,t\in [-R,R], \quad \gamma_R^2(t)=R\mathrm{e}^{it}, \,\,\,t\in [0,\pi], $$ then $$ \int_{\gamma_R}\frac{dz}{(1+z^2)^n}=2\pi i\, \mathrm{Res}\bigg(\frac{1}{(1+z^2)^n},z=i\bigg). $$ But $$ \bigg|\int_{\gamma_R^2}\frac{dz}{(1+z^2)^n}\,\bigg|= \bigg|\int_0^\pi\frac{iR\mathrm{e}^{it}dt}{(1+\big(R\mathrm{e}^{it})\big)^n} \bigg|\le \int_0^\pi \frac{Rdt}{(R^2-1)^n}=\frac{\pi R}{R^2-1)^n}\to 0, $$ as $R\to\infty$, while $$ \lim_{R\to\infty}\int_{\gamma_R^1}\frac{dz}{(1+z^2)^n}=\int_{-\infty}^\infty\frac{dx}{(1+x^2)^n}. $$ Hence $\int_{-\infty}^\infty\frac{dx}{(1+x^2)^n}=2\pi i\, \mathrm{Res}\bigg(\frac{1}{(1+z^2)^n},z=i\bigg).$
We use the following fact.
Fact. If $f$ has a pole of order $m$ at $z=a$, then $$ \mathrm{Res}(f,a)=\frac{1}{(m-1)!} \left((z-a)^mf(z)\right)^{(m-1)}_{z=a} $$
This is straightforward since $$ f(z)=a_{-m}(z-a)^{-m}+\cdots+a_{-1}(z-a)^{-1}+a_0+a_{1}(z-a)^{1}+\cdots $$ and hence $$ (z-a)^mf(z)=a_m+a_{-m+1}(z-a)^{1}\cdots+a_{-1}(z-a)^{m-1}+a_0(z-a)^{m}+\cdots $$ and thus $$ \big((z-a)^mf(z)\big)^{(m-1)}=(m-1)!a_{-1}+m!a_0(z-a)^1+\cdots $$ and finally $\,\,\frac{1}{(m-1)!} \big((z-a)^mf(z)\big)^{(m-1)}_{z=a}=a_{-1}$.
In our case $f(z)=\frac{1}{(1+z^2)^n}$ has a pole of order $n$, and hence $$ \mathrm{Res}\bigg(\frac{1}{(1+z^2)^n},z=i\bigg)= \frac{1}{(n-1)!} \left( (z-i)^n \frac{1}{(1+z^2)^n} \right)^{(n-1)}_{z=i}=\frac{1}{(n-1)!} \left(\frac{1}{(z+i)^n} \right)^{(n-1)}_{z=i} \\ =\frac{1}{(n-1)!}\cdot (-1)^{n-1}\cdot n\cdot(n+1)\cdots(2n-2)\cdot \left(\frac{1}{(z+i)^{2n-1}}\right)_{z=i}\\ =(-1)^{n-1}\binom{2n-2}{n-1}\cdot\frac{1}{(2i)^{2n-1}}=\frac{-i}{2^{2n-1}}\binom{2n-2}{n-1} $$ Finally $$ \int_{-\infty}^\infty\frac{dx}{(1+x^2)^n}=2\pi i\, \mathrm{Res}\bigg(\frac{1}{(1+z^2)^n},z=i\bigg)=2\pi i\,\frac{-i}{2^{2n-1}}\binom{2n-2}{n-1}=\frac{\pi}{2^{2n-2}}\binom{2n-2}{n-1}. $$
