# Integral

• Apr 28th 2009, 11:15 AM
Moo
Integral
Hi

So this is something that came from a problem we did in probability. I hope it's a correct result, since I modified it a bit.
So I know the probabilitistic way to prove it. If you can find, then do it (but you'd have to know where you're going).
Otherwise, I don't know if there is a calculus approach lol, hence this thread.

Find $\displaystyle \int_{\mathbb{R}} \frac{1}{(1+u^2)^k} ~du$
for any positive integer k.

Spoiler:
$\displaystyle \sqrt{\pi} \cdot \frac{\Gamma\left(k-\frac 12\right)}{\Gamma(k)}$
• Apr 28th 2009, 01:07 PM
Opalg
One method would be to adapt the reduction formula in this thread (modifying it so as to work for the interval $\displaystyle (-\infty,\infty)$ instead of [0,1]).
• Apr 28th 2009, 02:10 PM
NonCommAlg
another way: letteing $\displaystyle t=\sin^2 \theta$ in the beta function formula we get $\displaystyle B(x,y)=2 \int_0^{\frac{\pi}{2}} (\sin \theta)^{2x-1} (\cos \theta)^{2y-1} \ d \theta, \ \ x > 0, \ y> 0.$

now suppose $\displaystyle \frac{1}{2} < k \in \mathbb{R}$ and let $\displaystyle u=\tan \theta.$ then: $\displaystyle I_k=\int_{\mathbb{R}} \frac{du}{(1 + u^2)^k}=2 \int_0^{\frac{\pi}{2}} \cos^{2k-2} \theta \ d \theta=B \left(\frac{1}{2}, k -\frac{1}{2} \right).$ therefore:

$\displaystyle I_k=\frac{\Gamma (\frac{1}{2}) \Gamma (k - \frac{1}{2})}{\Gamma(k)}= \frac {\sqrt{\pi}\Gamma (k - \frac{1}{2})}{\Gamma(k)}.$
• Apr 29th 2009, 09:52 AM
TheEmptySet
Quote:

Originally Posted by Moo
Hi

So this is something that came from a problem we did in probability. I hope it's a correct result, since I modified it a bit.
So I know the probabilitistic way to prove it. If you can find, then do it (but you'd have to know where you're going).
Otherwise, I don't know if there is a calculus approach lol, hence this thread.

Find $\displaystyle \int_{\mathbb{R}} \frac{1}{(1+u^2)^k} ~du$
for any positive integer k.

Spoiler:
$\displaystyle \sqrt{\pi} \cdot \frac{\Gamma\left(k-\frac 12\right)}{\Gamma(k)}$

Here is another way (I'm in complex this term so here is goes)

Consider the closed arc in the complex plane. Let R be real and R > 1

$\displaystyle \gamma_1=-R+2R(t), 0 \le t \le 1$ and $\displaystyle \gamma_2=Re^{it}, 0 \le t \le \pi$

This forms a simple closed curve in the complex plane.

By the cauchy integral formula

$\displaystyle \oint \frac{1}{(z^2+1)^k}dz=\oint \frac{\frac{1}{(z+i)^k}}{(z-i)^k}dz=\frac{2\pi i}{(k-1)!}\frac{d^k}{dz^k}\left( \frac{1}{(z+i)^k}\right)_{z=i}=\frac{\pi}{2^{2k-2}}\left( \frac{(2k-2)!}{((k-1)!)^2}\right)$

Now if we consider

$\displaystyle \int_{\gamma_1} \frac{1}{(z^2+1)^k}dz +\int_{\gamma_2} \frac{1}{(z^2+1)^k}dz$

$\displaystyle \int_{-R}^{R} \frac{1}{(x^2+1)^k}dx +\int_{\gamma_2} \frac{1}{(z^2+1)^k}dz$

Note that on the circular arc that $\displaystyle \left| \frac{1}{(z^2+1)^k}\right| \le \frac{1}{(R^2-1)^k}$

Using the ML estimate

$\displaystyle \left| \int_{\gamma_2} \frac{1}{(z^2+1)^k}dz\right| \le \frac{\pi R}{(R^2-1)^k}$

Now if we take the limit as $\displaystyle R \to \infty$

We end up with

$\displaystyle \int_{-\infty}^{\infty}\frac{1}{(1+x^2)^k}=\frac{\pi}{2^{ 2k-2}}\left( \frac{(2k-2)!}{((k-1)!)^2}\right)$