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

Solution 1:

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.

Solution 2:

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

Solution 3:

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