Evaluating a complex contour

Question

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?

Answer 1

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

$I_{2n}=\frac{cos^{2n-1}xsinx}{2n}+\frac{2n-1}{2n}I_{2n-2}$

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 $\frac{(2n-1)(2n-3)...3.1}{2n(2n-2)...4.2} \int_{-\frac{\pi}{2}}^\frac{\pi}{2} cos^2y dy$

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

Answer 2

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.

Answer 3

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

Answer 4

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 $$\binom{2n}{n}\frac{\pi}{2^{2n}}=\lim_{R\to\infty}\oint_{C_R} f(z)dz=\int_{-\infty}^\infty\frac{1}{(1+x^2)^{n+1}}dx$$