Help: apply the residue theorem

**Mentia** · December 7th 2008, 08:27 AM

First I must apologize as this thread is a copy of an earlier thread that I posted in the wrong place.

I am having trouble evaluating two improper integrals by applying complex analysis. The two integrals are

(1)
$ \int _{-\infty }^{\infty }\frac{x}{x^{2} -2x+2} dx$

and

(2)
$ \int _{0}^{\infty }\frac{x^{2} }{\left(x^{2} -2x+2\right)^{2} } dx$

The reason I feel (1) is non-trivial is because the degree of the denominator is not 2 or more greater than that of the numerator, so the integration over a semi-circle in the complex plane doesn't really apply. However, using numerical analysis it seems that the Cauchy principle value of (1) is $\pi$ , and if you blindly integrate over the semi-circle ignoring the denominator/numerator problem you will get $\pi$ from the residue theorem. I assume this is just coincidence, however is there someone out there might be able to shed a bit more light? I know that (1) is divergent, but why does it seem to converge to $\pi$ ?

The problem I am having with (2) is similar. Although there is no problem with the degrees of the denominator and numerator, the limits of integration are not correct to use the semi-circle. Further, it would be okay to use the semi-circle if the function was symmetric but it's not. I looked for a transform that would make it symmetric but I failed. I made a function that is symmetric and has the same values in the positive reals as (2):

(3)
$\int _{-\infty}^{\infty }\frac{x^{2} }{\left(x^{2} -2\left|x\right|+2\right)^{2} } dx$

I had hoped to integrate this over the semi-circle in the complex plane and divide by two at the end since it's symmetric. Unfortunately the abs(x) seems to cause problems. It seems to me that the function is analytic and should work and it's easy to find the poles, but what order are they? Can anyone tell me why (3) doesn't work, or provide an alternate way of evaluating the integral using the residue theorem or something similar?

**ThePerfectHacker** · December 7th 2008, 08:42 AM

Originally Posted by Mentia

$ \int _{-\infty }^{\infty }\frac{x}{x^{2} -2x+2} dx$

This does not converge.
What you can do the following write,
$\int \limits_{-\infty}^{\infty} \frac{x}{x^2 - 2x + 2} dx = \int \limits_{-\infty}^{\infty} \frac{x}{(x-1)^2 + 1} dx$
Let $t=x-1$ then we get,
$\int \limits_{-\infty}^{\infty} \frac{t+1}{t^2 + 1} dt = \int \limits_{-\infty}^{\infty} \frac{t}{t^2+1}dt + \int \limits_{-\infty}^{\infty} \frac{dt}{t^2+1}$
The second integral converges to $\pi$ .
The first integral does not converge unless you look at Cauchy Principal Value, which is zero for the function is odd.

**Mentia** · December 7th 2008, 08:51 AM

Thanks very much for your prompt reply! The problem (2) seems clear now with your transformation.

**Mentia** · December 7th 2008, 10:34 AM

Okay I'm still experiencing problems...

The problem crops up when I transform (2) using ThePerfectHacker's suggestion.

Its all fine and good until I split up the integral:

$\int_{0}^{\infty} \frac{x^{2}}{\left(\left(x-1\right)^{2}+1\right)^{2}} \,dx$ = $\int_{-\left(1\right)}^{\infty} \frac{\left(t+1\right)^{2}}{\left(\left(t\right)^{ 2}+1\right)^{2}} \,dt$

= $\int_{-\left(1\right)}^{0} \frac{\left(t+1\right)^{2}}{\left(t^{2}+1\right)^{ 2}} \,dt+ \int_{0}^{\infty} \frac{\left(t+1\right)^{2}}{\left(t^{2}+1\right)^{ 2}} \,dt$

= $\int_{-\left(1\right)}^{0} \frac{\left(t+1\right)^{2}}{\left(t^{2}+1\right)^{ 2}} \,dt+ \int_{-\left(\infty\right)}^{\infty} \frac{1}{2} \left( \frac{t^{2}}{\left(t^{2}+1\right)^{2}} + \frac{1}{\left(t^{2}+1\right)^{2}} \right)\,dt+ \int_{0}^{\infty} \frac{2 t}{\left(t^{2}+1\right)^{2}} \,dt$

Now all of these can be done without much problem except the last part:

$\int_{0}^{\infty} \frac{2 t}{\left(t^{2}+1\right)^{2}} \,dt$

This is an odd integrand and so I cant just go from -inf to +inf and divide by 2 like the others. Anyone know how to use the residue thm to evaluate this last integral?

**ThePerfectHacker** · December 7th 2008, 11:52 AM

Originally Posted by Mentia

$ \int _{0}^{\infty }\frac{x^{2} }{\left(x^{2} -2x+2\right)^{2} } dx$

Let $t=x-1$ to get,
$\int_0^{\infty} \frac{(t+1)^2}{(t^2+1)^2}dt = \int_0^{\infty} \frac{t^2}{(t^2+1)^2} dt + \int_0^{\infty} \frac{2t}{(t^2+1)^2}dt + \int_0^{\infty} \frac{dt}{(t^2+1)^2}$
Combine the first and third integral,
$\int_0^{\infty} \frac{dt}{t^2+1} + \int_0^{\infty} \frac{2t}{(t^2+1)^2}dt$
You can use residues to compute this but it is not necessary.
We get, by finding primitives, that this is,
$\tan^{-1} t \bigg|_0^{\infty} - \frac{1}{t^2+1}\bigg|_0^{\infty} = \frac{\pi}{2} + 1$

**Mentia** · December 7th 2008, 12:29 PM

Thanks again for your response, ThePerfectHacker.

It seems you forgot to change the bounds from 0->Infinity to -1->Infinity when you made the transformation. This changes the value slightly.

The main reason I am agonizing over this is because we are required to use residue theorem to solve the problem. How would you use residues to find:

$ \int_{0}^{\infty} \frac{2 t}{\left(t^{2}+1\right)^{2}} \,dt $

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