Jordan’s Lemma.

August 9, 2011

[This post is for those of you who are already comfy with doing some basic contour integrals in complex analysis.]

 

So you’re sitting around, evaluating contour integrals, and everything is fine.  Then something weird comes up.  You’re asked to evaluate an integral that looks like

\displaystyle \int_{-\infty}^{\infty} e^{aix}g(x)\, dx

for g(x) is continuous.  Eek.  Don’t panic though, because Camille Jordan’s gonna help you out.

image 

The following lemma is often called Jordan’s Lemma, and it’s used quite a bit when you have things that look like the above integral.  In particular, Fourier stuff looks like this quite a bit.  There is a more general case, but I’ll just state the one we’ll use for this complex stuff.

 

Lemma (Jordan’s):  Let f is analytic at every point in the closed upper-half plane (that is, if z = x + yi, the closed upper half plane is all points such that y \geq 0) which is exterior to some half-circle of radius R_{0} > 0.  Let C_{R} denote z = Re^{i\theta}\,\, (0\leq \theta \leq \pi) for R > R_{0}.  If for each z on C_{R} we have some constant M(R) (depending on R) such that |f(z)| \leq M(R) and, furthermore, we have M(R) \rightarrow 0 as R\rightarrow \infty, then for a > 0 we have

\displaystyle \lim_{R\rightarrow\infty}\int_{C_{R}}f(z)e^{iaz}\, dz = 0

 

[By the way, this lemma was lovingly paraphrased from the one in Brown’s Complex Analysis book, which I highly recommend.] 

A lemma like this looks somewhat complicated, and the conclusion looks like it was made specifically to prove something about a very specific type of thing.  Maybe that’s why they called it a lemma instead of a theorem.  Either way, in a nutshell, the theorem says that if we have an analytic function with some poles bounded by some half circle in the upper-half plane, then if we can bound the f(z) part of our integral by some constant depending on R that goes to zero, we get the e^{aiz} part for free.  Usually, this lemma is used to show that the C_{R} part of the integral goes to 0 so that we can equate our "real-axis integral" to a contour one and  use Cauchy’s integral formula. 

Before we prove this, let’s just show an example of how it works.

 

Example!

Let’s take \displaystyle \int_{0}^{\infty}\frac{\cos(ax)}{x^{2} + 1} \, dx for a > 0

Alright, this will be the real part of \displaystyle \int_{0}^{\infty}\frac{e^{iax}}{x^{2} + 1} \, dx (if you don’t see this, write out e^{iax} in terms of sine and cosine and expand) so it suffices to find this integral and take the real part. 

Taking f(z) = \frac{1}{z^{2} + 1} and the contour C_{R} as above and the real line segment [-R,R], we could easily evaluate this using Cauchy if only the C_{R} part went to 0.  BUT, we have the lemma above!  Note that

\displaystyle |f(z)| = \frac{1}{|z^{2} + 1|} \leq \frac{1}{|R^{2} - 1|}

is valid on C_{R} (and we get the R^{2} - 1 by applying that "other" triangle inequality |x + y| \geq ||x| - |y||).  Note also that the right-hand-side goes to 0 as R\rightarrow\infty, which is enough to tell us, by Jordan’s lemma, that the integral of our function along C_{R} is zero.  Nice.  Now we have that the "real-axis part", which is equal to our original integral if we divide it by 2 (since the original one is only from 0 to \infty), is just equal the integral around the semi-circle.  By Cauchy, this means:

\displaystyle \int_{0}^{\infty}\frac{\cos(ax)}{x^{2} + 1} \, dx = REAL[\frac{1}{2} 2\pi i Res_{i}] = REAL[\pi i Res_{i}]

where the big REAL stands for the real part and Res_{i} is the residue at i.  Again, the \frac{1}{2} is there because the original integral is only from 0 to \infty.  Now the problem reduces to finding the residue at i.  Luckily, this is not so bad so use your own method of finding the residue! 

It turns out that the residue is \frac{e^{-a}}{2i}.  Thus, we finally have,

\displaystyle \int_{0}^{\infty}\frac{\cos(ax)}{x^{2} + 1} \, dx = REAL[\frac{\pi}{2} e^{-a}] = \frac{\pi}{2}e^{-a}.

 

Awesome.

 

The Proof of the Lemma.

The proof is surprisingly simple.  We need one little inequality (not surprisingly called Jordan’s Inequality) before we begin, though.  And this will be a picture proof. 

 

Lemma (Jordan’s Inequality): \displaystyle \int_{0}^{\pi} e^{-R\sin(\theta)}\, d\theta < \frac{\pi}{R}, where R > 0

 

Kind of complicated looking, but it turns out to be really nice to prove.  Okay.  I’ll just give you a hint, then you can prove the rest.  Note that \sin(\theta) \geq \frac{2\theta}{\pi} if \theta \in [0,\frac{\pi}{2}]

imageBam.  I read some class notes on an analytic proof of this fact which stated that one needed to use a strange (and quite complicated!) inequality to prove this; when I first went over this, I just Taylor expanded sine a bit.  Perhaps I’m missing something, but if you don’t trust my picture-proof or Taylor-proof, then there are other proofs available for you to go through!

Now write down that integral in the lemma from 0 to \frac{\pi}{2}, and replace -R\sin(\theta) with R\frac{2\theta}{\pi}.  Now integrate.  Since \sin is symmetric about \frac{\pi}{2}, just double the evaluation to get the required inequality.

 

Once we have this, Jordan’s lemma is easy.

 

Proof of Jordan’s Lemma.  Okay, so we have a function and all that other stuff holds that we needed in the statement of the lemma.  Now,

\displaystyle \int_{C_{R}}f(z)e^{aiz}\, dz = \int_{0}^{\pi}f(Re^{i\theta})e^{iaRe^{i\theta}}Re^{i\theta}\, d\theta

by just using the parameterization we have.  Luckily we can reduce a lot of this.  Okay, first, we know |f(z)| \leq M(R) by the statement, and we have that

\displaystyle |e^{iaRe^{i\theta}}| = |e^{iaR(\cos(\theta) + i\sin(\theta))}| \leq |e^{iaR( i\sin(\theta))}|

\displaystyle \leq |e^{-aR\sin(\theta)}| = e^{-aR\sin(\theta)}

note the last part is true since e is always positive, and, at this point, you might also see where that inequality is going to come into play.  Replacing these estimates into our original integral, we get

\left|\displaystyle \int_{0}^{\pi}f(Re^{i\theta})e^{iaRe^{i\theta}}Re^{i\theta}\, d\theta\right| \leq M(R)R\int_{0}^{\pi}e^{-aR\sin(\theta)}\, d\theta \leq \frac{M(R)\pi}{a}

But since \frac{\pi}{a} is just some positive constant, we have that since M(R)\rightarrow 0, the entire integral must go to 0 as R\rightarrow\infty.  Nice!  This is exactly what we wanted to prove.  \Box

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7 Responses to “Jordan’s Lemma.”

  1. zariski said

    Can I have a question?

    Your example can be solved by the residue theorem.
    What kind of problem is better using the Jordan lemma than using the residue theorem?
    Thanks in advance. :)

  2. Anonymous said

    Don’t answer that question and don’t write the proofs of any more famous results. PREPARE FOR THE QUAL!

    • James said

      Ha, I actually don’t know a good answer for that question. And this came up in studying for the qual!

      • Anonymous said

        The proof of Jordan’s Lemma is not coming up on the qual! You don’t need to write that up. Stop wasting time.

  3. Anonymous said

    Prove that the indefinite integral of an L^1 function is absolutely continuous. This is a good question for the analysis qual.

  4. Anonymous said

    To prove that sin(\theta)>=(2/pi)*(\theta) in that interval, just use the concavity of the function. It’s second derivative is negative in that interval.

  5. Thanks for the explanations! I’m an Electrical Engineering student and I found it very helpful :)

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