How can the residue theorem be used to calculate contour integrals?

[HJ Farnsworth]

Greetings,

I am a bit confused on an aspect of contour integrals. Wikipedia gives the following example of an integral where the residue theorem can be used (it also finds the integral in a couple of other ways, but I am mainly interested in doing it using residues):

We want to calculate

\int_{- \infty}^\infty \frac{1}{(x^2+1)^2}\,dx.

To this end, we take the integral in the complex plane, and instead consider

\oint_{- \infty}^\infty \frac{1}{(z^2+1)^2}\,dz.

The singularities of f(z)=\frac{1}{(z^2+1)^2} are at i and -i. The Laurent series about i has a residue of -i/4. In the upper half of the complex plane, we can choose our contour to be the semicircle with endpoints at opposite values along the real axis enclosing the pole at i, which is a closed loop, so that the residue theorem gives that the value of the closed integral is \pi/2.

Up to there, I am perfectly fine. However, how does this tell us the value of the original (the dx) integral? For that integral to be \pi/2, the integral over the arc part of the semicircle would have to be 0. How do we know that this is the case? Furthermore, since the original integral will be less than \pi/2 if the integration limits aren't \pm \infty, but since we can still enclose the same pole with finite integration limits, the arc part of the integral must be something other than 0 except in the limit where the integration limits become \pm \infty.

So, I guess what I am saying is that I will sometimes see examples where the residue theorem is used in which something is assumed about part of the integral that doesn't seem obvious to me, but that is crucial toward getting the correct result for the original, real integral, as in the example above. Is there something that I am missing?

Thanks for helping in advance.

-HJ Farnsworth

 

[R136a1]

Let $\alpha_r$ be the arc with radius $r$. You need to prove

\lim_{r\rightarrow +\infty} \int_{\alpha_r} \frac{1}{(x^2 + 1)^2}dx = 0

Or you could prove the same thing with absolute values. Note that

\frac{|(x^2 + 1)^2|}{|x^3|}\rightarrow +\infty

So there exists a $C$ such that for $x$ with large enough modulus holds that

|(x^2 + 1)^2| \geq C |x^3|. Thus for $r$ large enough:

\left|\int_{\alpha_r} \frac{1}{(x^2 + 1)^2}dx \right| \leq \int_{\alpha_r} \frac{1}{|(x^2 + 1)^2|}dx \leq \frac{1}{C} \int_{\alpha_r} \frac{1}{|x^3|}dx

the latter integral is easy to calculate and you see that it goes to $0$.

 

[HJ Farnsworth]

Thanks for replying, R136a1.

That's a cool argument, and definitely does the trick. So basically, when using the residue theorem, I will still usually need to do some outside-the-box thinking to focus in on the part of the integrals I want to evaluate, like your derivation above. That's a relief, I thought I might be missing something more fundamental built into the theorem itself.

I'm guessing, though, that for the most part there's a few tricks like that that come in handy so frequently that people just use them without proving them when they use them, since it is expected knowledge if you are using contour integrals (just as I'm not going to bother proving integration by parts in ordinary calculus whenever I use it - I just expect people to know what I'm doing). Does this guess match people's experience?

 

[jackmell]

Thanks for replying, R136a1.
I'm guessing, though, that for the most part there's a few tricks like that that come in handy so frequently that people just use them without proving them when they use them, since it is expected knowledge if you are using contour integrals (just as I'm not going to bother proving integration by parts in ordinary calculus whenever I use it - I just expect people to know what I'm doing). Does this guess match people's experience?

No. It's not a trick. Your job is to carefully and meticulously analyze every leg of the contour. For example, there are some problems in which the integral will not go to zero over such a path.

 

[andrien]

you should take a look at jordan's[/PLAIN] lemma