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I need to show the following result:

$$ \int_{-\infty}^\infty \frac{1}{(1+x^2)^{n+1}}dx\, = \frac{1\cdot 3\cdot\ldots\cdot(2n-1)}{2\cdot 4\cdot\ldots\cdot(2n)}\pi $$

With n=1,2,3,...

This function has a pole at i and -i. I've tried a semicircle in the upperhalf of the plain, but the residue then goes to infinity. I've also tried a rectangle in the upperhalf that stays beneath i, but all 3 sides that do not include the integral we're looking for go to zero because of the R in the denominator.

Anyone with tips?

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What do you mean by "the residue then goes to infinity"? –  mrf Aug 7 '12 at 9:00

4 Answers 4

Integrating over a large semicircle in the upper halfplane works well.

Write $$\frac{1}{(1+z^2)^{n+1}} = \dfrac{\dfrac{1}{(z+i)^{n+1}}}{\quad(z-i)^{n+1}\quad}.$$

Hence, the residue at $z=i$ is the $1/n!$ times the $n$th derivative of $1/(z+i)^{n+1}$ evaluated at $z=i$, i.e. $$ \begin{split} \operatorname{Res}\limits_{z=i} \frac{1}{(1+z^2)^{n+1}} &= \frac{1}{n!}\frac{d^n}{dz^n}\left( \frac{1}{(z+i)^{n+1}} \right)\Bigg|_{z=i} \\ &= \frac{1}{n!}(-1)^n(n+1)(n+2)\cdots(2n)\frac{1}{(2i)^{2n+1}} \\ &=\frac{(n+1)(n+2)\cdots(2n)}{n!\cdot2^{2n}\cdot 2i} \\ &= \frac{n!\cdot(n+1)\cdots(2n)}{(n!)^2\cdot 2^{2n}\cdot 2i} \\ &= \frac{(2n)!}{(2\cdot 4\cdot 6\cdots(2n))^2\cdot 2i} \\ &= \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n)\cdot 2i} \\ \end{split} $$ I think you can do the last step yourself.

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Thanks, that seems conclusive! The last steps I can do myself indeed. I also understand all of the rewriting parts, but how did you get from the rewritten integrand to the first way of writing the residue? Can't seem to reproduce that step. (So this: the residue at z=i is the 1/n! times the nth derivative of 1/(z+i)n+1 evaluated at z=i) –  boudewijn Aug 7 '12 at 10:29
Rewriting of the integrand shows that it has a pole of order $n$ at $z=i$. There are several ways to compute the residue of higher order poles, but this is usually the easiest way. See your textbook or en.wikipedia.org/wiki/… –  mrf Aug 7 '12 at 11:57

Substitute $x^2=\dfrac{u}{1-u}$ and $\mathrm{d}x=\dfrac{\mathrm{d}u}{2\,u^{1/2}(1-u)^{3/2}}$ $$ \begin{align} \int_{-\infty}^\infty\frac1{(1+x^2)^{n+1}}\mathrm{d}x &=2\int_0^\infty\frac1{(1+x^2)^{n+1}}\mathrm{d}x\\ &=\int_0^1(1-u)^{n-1/2}u^{-1/2}\mathrm{d}u\\ &=\mathrm{B}(n+1/2,1/2)\\ &=\frac{\Gamma(n+1/2)\Gamma(1/2)}{\Gamma(n+1)}\\ &=\frac{(n-\frac12)(n-\frac32)(n-\frac52)\dots\frac12\Gamma(\frac12)}{n(n-1)(n-2)\dots1}\Gamma(\tfrac12)\\ &=\frac{(2n-1)(2n-3)(2n-5)\dots1}{2n(2n-2)(2n-4)\dots2}\pi \end{align} $$

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if x=tany, $dx=sec^2ydy$ so y varies from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$

$\int 1 / (1+x^{2})^{n+1}dx = \int \frac{sec^2ydy}{(sec^2y)^{n+1}} dy= \int cos^{2n}y dy = I_{2n}(say)$

Here is how to derive the reduction formula

So, $I_m=\frac{cos^{m-1}xsinx}{m}+\frac{m-1}{m}I_{m-2}$


Now if take definite integral of $I_{2n}$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$,

$\int_{-\frac{\pi}{2}}^\frac{\pi}{2} cos^{2n}y dy =[\frac{cos^{2n-1}xsinx}{2n}]_{-\frac{\pi}{2}}^\frac{\pi}{2}+\frac{2n-1}{2n}\int_{-\frac{\pi}{2}}^\frac{\pi}{2} cos^{2n-2}y dy=\frac{2n-1}{2n}\int_{-\frac{\pi}{2}}^\frac{\pi}{2} cos^{2n-2}y dy$ (as the first integral is 0)

$I_2= \int cos^2y dy=\int \frac{1+cos2y}{2} dy=\frac{y}{2}+\frac{sin2y}{2}+C$ where C is the indeterminate constant of indefinite integral.

The definite integral of $I_{2}$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ will be $\pi$

So, the definite integral of $I_{2n}$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ will be $\frac{(2n-1)(2n-3)...3.1}{2n(2n-2)...4.2} \int_{-\frac{\pi}{2}}^\frac{\pi}{2} cos^2y dy$

$=\frac{(2n-1)(2n-3)...3.1}{2n(2n-2)...4.2} \pi$

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Taking the semicircular contour $$C_R:=\left(\gamma_R:=\{z\in\Bbb C\;:\;|z|=R\,\,,\,Im(z)\geq 0\}\right)\cup [-R,R]\,\,,\,R>1$$
the domain within this path contains one pole of the function, and:

$$f(z)=\frac{1}{(1+z^2)^{n+1}}\Longrightarrow Rez_{z=i}(f)=\frac{1}{n!}\lim_{z\to i}\frac{d^n}{dz^n}\left[(z-i)^{n+1}f(z)\right]=$$ $$=\frac{1}{n!}\lim_{z\to i}\frac{d^n}{dz^n}\left[\frac{1}{(z+i)^{n+1}}\right]=\frac{1}{n!}\lim_{z\to i}\frac{(-1)^n(n+1)(n+2)\cdot\ldots\cdot(2n)}{(z+i)^{2n+1}}=$$ $$=\frac{1}{n!}\frac{(-1)^n(n+1)(n+2)\cdot\ldots\cdot(2n)}{(2i)^{2n+1}}=\frac{1}{n!}\frac{(-1)^n(n+1)(n+2)\cdot\ldots\cdot(2n)}{2^{2n+1}(-1)^ni}=$$ $$=\binom{2n}{n}\frac{1}{2^{2n+1}i}$$

We thus get by Cauchy's Integral Theorem that $$\oint_{C_R} f(z)dz=\binom{2n}{n}\frac{\pi}{2^{2n}}\,\,\,\,\,(**)$$

But on $\,\gamma_R\,$ we have $\,z=Re^{it}\,$ , so $$\left|\int_{\gamma_R}f(z)dz\right|\leq\max_{z\in\gamma_R}\left|\frac{1}{(1+R^2e^{2it})^{n+1}}\right|\pi R\leq\frac{\pi R}{(1-R^2)^{n+1}}\xrightarrow [R\to\infty]{} 0$$

So taking the limit when $\,R\to\infty\,$ in (**) we get $$\binom{2n}{n}\frac{\pi}{2^{2n}}=\lim_{R\to\infty}\oint_{C_R} f(z)dz=\int_{-\infty}^\infty\frac{1}{(1+x^2)^{n+1}}dx$$

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