# How to find $\int_0^{\infty}\frac{dx}{(1+x^2)^4}$

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.

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Do you know anything about complex analysis? This type of definite integrals can typically be solved with that sort of techniques. The indefinite integral can be calculated as well, but that quite a painful exercise I'm afraid. – Myself Feb 14 '11 at 1:51

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.

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When integration $\cos^6 \theta$ or any trig integral, a beautiful resource is the Wikipedia list: en.wikipedia.org/wiki/… – Eric Naslund Feb 14 '11 at 2:02
@ Sivaram, that last bit where you actually evaluated the definite integral is an identity derived from Fourier Series, right? Do you know where I can find the formulae, I can't for the life of me find it Googling... – Uticensis Feb 14 '11 at 2:14
@Billare: The last integral can be found in the post here math.stackexchange.com/questions/20397/… as I have mentioned in my answer. – user17762 Feb 14 '11 at 2:18

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 $\int \cos^6 \theta \, d\theta$.)

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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: $$\color{#00f}{\large\int_0^{\infty}{\dd x \over \pars{1 + x^{2}}^{4}}} ={15\pi/16 \over 6}= \color{#00f}{\large{5 \over 32}\,\pi}$$