Why do we use only upper half plane to do Residue Integration?

In the class of mathematics my professor showed me that "how to use the residue integration method?". And i doubt at almost the last step that professor did. he made a contour over the upper half plane of complex plan. and i didn't know that why did we must focus only upper half(that will cover poles on only upper plane and ignore another on lower plane)

My question is:

I doubt that why do we use only upper half of complex plane to do Residue Integration?

What's happened if i use lower plane or both lower and upper plane?

Solution 1:

Residue techniques often help one solve improper (real) integrals like: $\int_{-\infty}^\infty f(x)\,dx$.

In order to apply contour techniques one has to turn this real integral into a complex contour integral. This is done by taking a symmetric interval: $[-R,R]$ and adding in a upper semi-circle of radius $R$ (centered at the origin). Then as $R \to \infty$ one wants the semi-circle part to tend to zero so that in the limit one gets the integral over $(-\infty,\infty)$ (well, actually the principal value).

As your half circle gets larger and larger it captures more and more of the upper half plane, so in the limit you have essentially captured the entire upper half plane. Thus when you go to compute the contour integral using residues, you should only consider poles in the upper half plane.

One could just as easily use the lower half plane. Or if you where doing an integral of the form $\int_0^\infty f(x)\,dx$ you could use the first quadrant.

There are occasions where you might want to consider the entire plane, but none come to mind at the moment.

Solution 2:

This depends entirely on the function which you are integrating. Let's say that you are taking a Fourier transform:

$$\int_{-\infty}^{\infty} dx \, f(x) e^{i k x}$$

If you wish to use a contour integral to evaluate this integral, then your choice of whether you use the upper or lower half plane makes a huge difference. For $k \gt 0$, for example, the contour integral simply does not converge on a semicircle in the lower half plane as the radius of that semicircle goes to $\infty$. Thus, you would use the upper half plane. The opposite goes for $k \lt 0$.