Integral

**Moo** · April 28th 2009, 11:15 AM

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

Answer should be :

Spoiler:

**Opalg** · April 28th 2009, 01:07 PM

One method would be to adapt the reduction formula in this thread (modifying it so as to work for the interval $(-\infty,\infty)$ instead of [0,1]).

**NonCommAlg** · April 28th 2009, 02:10 PM

another way: letteing $t=\sin^2 \theta$ in the beta function formula we get $B(x,y)=2 \int_0^{\frac{\pi}{2}} (\sin \theta)^{2x-1} (\cos \theta)^{2y-1} \ d \theta, \ \ x > 0, \ y> 0.$

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

$I_k=\frac{\Gamma (\frac{1}{2}) \Gamma (k - \frac{1}{2})}{\Gamma(k)}= \frac {\sqrt{\pi}\Gamma (k - \frac{1}{2})}{\Gamma(k)}.$

**TheEmptySet** · April 29th 2009, 09:52 AM

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

Answer should be :

Spoiler:

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

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

This forms a simple closed curve in the complex plane.

By the cauchy integral formula

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

Now if we consider

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

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

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

Using the ML estimate

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

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

We end up with

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

Thread: Integral

LinkBack

Thread Tools

Display

Integral

Similar Math Help Forum Discussions

Use Cauchy's integral theorem for derivatives to evaluate the contour integral.

Graph the region of integration and evaluate the double integral (integral from 0 to

Explaining result: comparing line integral and two-form integral

write [integral of] dx/SQRT(sinx) in terms of an elliptic integral

Evaluating a double integral to find its signle integral.

Search Tags