7
$\begingroup$

It is a standard trick for evaluating difficult integrals along the real line to consider a closed-contour and "blow-up" the complex part till it vanishes, leaving us with the residues picked up along the way. This is usually done by bounding the magnitude of the integral from above with something that tends to $0$. My question is, is there an intuitive reason for why we should expect this?

My thought was that, if we project the complex plane onto a Riemann sphere, the path which appears to enlarge is actually shrunk to the point at $\infty$. I drew a (not very pretty) picture to illustrate what I mean:

enter image description here

In blue is a semi-circular contour in the upper-half plane, centered at the origin, and in red is the stereographic projection of this contour onto a Riemann sphere. It is clear that as the radius of the blue semi-circle $\to\infty$, the red semi-circle that corresponds to the complex portion of the contour is shrunk to the north pole of the sphere.

Is this a good way of thinking about it and is this a good enough reason for the integral along that arc to vanish (as long as it does not encounter poles on the way)?

$\endgroup$
3
$\begingroup$

I don't think this is a useful way to think about it. If $z$ is the standard coordinate on the complex plane, then we can define a coordinate $w = 1/z$ which includes $\infty$ on the Riemann sphere. Then we have $$ f(z)\, dz = -\frac{1}{w^2}f\left(\frac{1}{w}\right)dw ~. $$ So it really depends on $f$ (obviously!).

The only 'heuristic explanation' I can think of for why the trick works so often is that if the integral converges on the real line, then the integrand must be dying away fairly quickly at $\pm\infty$. Therefore it is 'fairly likely' to also die away quickly over a whole half-circle of angles.

$\endgroup$
5
$\begingroup$

I think there is what scientists call "selection bias" here.

Typically the kind of integrals you're evaluating with contour integration are improper integrals which converge. For instance, integrals like $$\int_0^\infty \frac{P(x)}{Q(x)},$$

where $\deg {Q} > \deg{P}+1$ and $Q$ has nonzero constant term.

If you turn such an integral into a complex function $P(z)/Q(z)$, then of course it will go to zero over (say) a large semicircular path, because it's bounded by something like

$$\frac {C\cdot R^{\deg{P}}} {D\cdot R^{\deg{Q}}}\cdot \pi R,$$

which will go to zero because the hypothesis $\deg {Q} > \deg{P}+1$ guarantees it!

$\endgroup$
2
$\begingroup$

I think that is an interesting way of looking at complex numbers and visualizing the contour. That said, I'm not entirely sure how useful it is in general. You might recall that we use a semicircular contour in only some special cases (for example, those integrals to which Jordan's lemma applies). Many other integrals which may be evaluated using contour integration techniques will not converge along a semicircular contour and other contours are therefore used. (For example, see this problem).

That said, all of these cases do involve evaluating the integral at the north pole on the Riemann sphere. I am just not sure what additional insight that brings.

$\endgroup$
0
$\begingroup$

This trick does not work with all complex contour integrals, but in particular it works for integrals of functions of the form $\frac{P(z)}{Q(z)}$ where $\deg Q \ge \deg P + 2$.

The reasoning for this is simple:

$$\left| \int_\Gamma \frac{P(z)}{Q(z)}\ dz \right| \le ML$$

where $M = \max_\Gamma \left|\frac{P(z)}{Q(z)}\right|$ is the maximum value of the function along the curve, and $L$ is the length of the curve.

As the radius of the path $R \to \infty$, we know that $\frac{P(z)}{Q(z)}$ goes like $\frac{1}{R^2}$, so $ML \sim \frac{1}{R} \to 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.