# Thread: Residue Theorem/Complex Methods Question

1. ## Residue Theorem/Complex Methods Question

Hi all, can't really figure this one out, any help will be much appreciated.

Find using complex methods lim as r --> infinity for the integral, from -r to r, x/(1 + x + x^2).

Also, why is the limit necessary?

2. Originally Posted by h2osprey
Hi all, can't really figure this one out, any help will be much appreciated.

Find using complex methods lim as r --> infinity for the integral, from -r to r, x/(1 + x + x^2).
I am not exactly sure what you are asking. I will write what I think you are asking. You want to find $\displaystyle \int_{-\infty}^{\infty} \frac{x}{x^2+x+1} dx$. Immediately there is a problem. This integral diverges. So let us use a different example, say $\displaystyle \int_{-\infty}^{\infty}\frac{dx}{x^2+x+1}$. To find this integral we define $\displaystyle f(z) = (x^2+x+1)^{-1}$. And find the residues in the upper half-plane. Then we define a contour integral over the semicircle from $\displaystyle -R$ to $\displaystyle R$ (where $\displaystyle R$ is suffiiciently large) in the upper-half plane. We compute this integral in two ways. The first way is by using the Residue theorem and the other using the direct definition of contour integration. Then we take the limit as $\displaystyle R\to \infty$ and argue that the semi-circle integral vanishes, and we are just left an integral from $\displaystyle (-\infty,\infty)$ which is what we want.

3. Thanks for the prompt reply. I am aware that the integral diverges, I was thinking that that's the reason why lim r-->infinity is used instead of the taking the integral directly from （infinity， -infinity）

However I'm afraid I don't see how this applies to the question. I understand your example and am able to solve it, but how does it apply to my question? I am unable to show that the semi-circle integral vanishes.

Thanks again!

4. Originally Posted by h2osprey
Thanks for the prompt reply. I am aware that the integral diverges, I was thinking that that's the reason why lim r-->infinity is used instead of the taking the integral directly from （infinity， -infinity）
No you never take a integral from $\displaystyle (-\infty,\infty)$ directy. You first take it from $\displaystyle (-R,R)$ and then take limit.

However I'm afraid I don't see how this applies to the question. I understand your example and am able to solve it, but how does it apply to my question? I am unable to show that the semi-circle integral vanishes.
Say we have $\displaystyle f(z) = (z^2+z+1)^{-1}$. The semi-circle integral is,
$\displaystyle \int_0^{\pi} \frac{Rie^{i\theta}}{R^2e^{2i\theta}+Re^{i\theta}+ 1}d\theta$
Now use estimation techniques,
$\displaystyle \left| \int_0^{\pi} \frac{Rie^{i\theta}}{R^2e^{2i\theta}+Re^{i\theta}+ 1}d\theta \right| \leq$$\displaystyle \int_0^{\pi}\left| \frac{Rie^{i\theta}}{R^2e^{2i\theta}+Re^{i\theta}+ 1}\right| d\theta \leq \int_0^{\pi} \frac{R}{R^2 - R - 1}d\theta = \frac{\pi R}{R^2-R-1}$
But now note,
$\displaystyle \lim_{R\to \infty} \frac{\pi R}{R^2 - R - 1} = 0$
Because the degree of numerator is less than degree of denominator.
The problem is when you have $\displaystyle f(z) = z(z^2+z+1)^{-1}$ thne by trying to use the same estimation trick you run into a problem. Because then you would end up with,
$\displaystyle \lim_{R\to \infty} \frac{\pi R^2}{R^2 - R - 1} = \pi$
.
Thus, the semi-circle integral does not vanish. But that still is not the major probem. The serious problem is that the integral that you are trying to find diverges. So there is not point in doing this computation anyway.

5. I think I've solved it - the semi-circle integral doesn't vanish, but can be determined taking a limit to infinity. I tried two different approaches and got different answers when assuming the integral vanishes, and so kind of realised that it can't vanish, and guessed from the two values what the value of the integral should be. And when I worked out that integral I got that value - I guess that's the reason why the follow-up to the question asks why it is necessary to take a limit.

Thanks anyway!