# Are residues useful for proper integrals?

1. Mar 8, 2014

### jinawee

Calculating residues are useful when we are trying to solve some improper integral, because the Cauchy principal value will be the sum of residues inside the path taken (if the integral along the complex path tends towards 0).

When we have a proper integral of trigonometric functions, this is useful too.

But in general, are residues useful calculating proper integral?

It would be nice if I could apply the power of residues to evaluate an integral involving rational functions.

2. Mar 8, 2014

### SteamKing

Staff Emeritus
Although residues were developed primarily for evaluating complex integrals, they can be used to evaluate the integrals of real functions as well. Any decent text on complex integration should include a section on this topic.

This paper has some examples:

http://people.math.gatech.edu/~cain/winter99/supplement.pdf

3. Mar 9, 2014

### jinawee

I'm aware of that use, but I'm asking for definite integrals.

In general,

$$\int^{R}_{-R}f(x)dx+\int_{C_R} f(z)dz=2\pi i\sum \mathrm{Res}$$

For example,

$$\int^{10}_{-10}\frac{x^2}{x^6+1}dx=\frac{\pi}{3}-\int_{C_R} \frac{z^2}{z^6+1}dz$$

Where CR is a circular contour from 10 to -10. The problem is that the contour integral is not zero, so we would have something like this:

$$\int^{10}_{-10}\frac{x^2}{x^6+1}dx=\frac{\pi}{3}-\int_{C_R} \frac{x^2}{x^6+1}dz=\frac{\pi}{3}-i \int^{\pi}_{0} \frac{10^3 e^{i3\theta}}{10^6 e^{i6\theta}+1} d \theta$$

This method seems to complicate things. But is there any case where it's useful?

4. Mar 9, 2014

### SteamKing

Staff Emeritus
It's not clear why you want to use residues for this particular problem. It can be handled relatively simply by using substitution.

5. Mar 9, 2014

### jinawee

Sorry, I didn't explain myself correctly. It was just an example.

I just want to know if it could be useful for other definite integral that doesn't involve trigonometric functions (eliptic, hyperbolic...).

6. Mar 9, 2014

### lurflurf

wikipedia http://en.wikipedia.org/wiki/Methods_of_contour_integration gives

$$\int_0^3 \frac{x^{\frac{3}{4}} (3-x)^{\frac{1}{4}}}{5-x}\,dx = \frac{\pi}{2\sqrt{2}} \left(17 - 40^{\frac{3}{4}} \right)$$

Periodic and improper examples are more common because closing the contour trades one integral for another. In order for the trade to be an improvement we need an integral to be easily found, zero, or expressible in terms of another. This happens often for periodic and improper examples, less so in other cases. Notice the above example takes advantage of another opportunity, branch cuts.