# Help with Jordan's Lemma (Collorary)

1. Jan 22, 2015

### juan.

Hello everyone. I'm having some troubles with Jordan's Lemma and I can't find how to fix it, because everyone in internet and in the books I read do not prove this. Now, I'm following Wunsch (Complex Variables with Applications) but it only does this part for a easy problem.

For example, if I have an improper integral like this: (page 377 & 378 from Wunsch book)
$$\int_{-\infty}^{\infty} \frac{cos(3.x)} {(x-1)^2+1} dx$$
To solve it, I do this:
$$\int_{-\infty}^{\infty} \frac{exp(i.3.x)} {(x-1)^2+1} dx$$
Integrating that on the upper half circle, with $$R \rightarrow \infty$$ and using the Theorem of Residues:
$$\int_{-R}^{R} \frac{exp(i.3.x)} {(x-1)^2+1} dx+\int_{C_1}^{ } \frac{exp(i.3.z)} {(z-1)^2+1} dx=2.\pi.i .\sum\limits_{k}^{ } Res( \frac{exp(i.3.z)} {(z-1)^2+1}, z_k)$$

Now, I have to prove in a correct and properly way why this occurs:
$$\int_{C_1}^{ } \frac{exp(i.3.z)} {(z-1)^2+1} dx=0$$

I'm trying to use the ML inequality, like this:
$$| \int_{C_1}^{ } \frac{exp(i.3.z)} {(z-1)^2+1} dx | \leq M.L = M.\pi.R$$, because$$L=\pi.R$$ (it's the length of the semicircunference)
So, (I don't understand this step)
$$| \frac{exp(i.3.z)} {(z-1)^2+1} | \leq M$$

Now I have to put something like:
$$| \frac{exp(i.3.z)} {(z-1)^2+1} | \leq "something"$$
where I say that $$M = "something"$$
and $$"something" \rightarrow 0$$, so $$\int_{C_1}^{ } \frac{exp(i.3.z)} {(z-1)^2+1} dx=0$$

But I don't know how to do it in this case. I know I have to use, for example, $$z=R.exp(i.\theta)$$

For example, something I got from:
$$|exp(i.3.z)|$$
is this:
$$|exp(i.3.z)|=|exp(-3.y)|.|exp(i.3.x)|=|exp(-3.y)|$$
but I suppose that is not helpful for this problem

Can anyone help me? Thanks!!!

Last question: ML inequality is the same as Jordan's Lemma?

Last edited: Jan 22, 2015
2. Jan 22, 2015

### mathman

What is varying to make "something" -> 0?

As an aside the original integral is easily shown to exist. |integrand| < 1 for all x, while |integrand| < x^2 as |x| becomes infinite.

3. Jan 24, 2015

### Hawkeye18

Jordan's Lemma is not ML inequality, it is a bit stronger. But for your integral ML inequality is enough. Namely, in the upper half plane $|e^{i3z}|\le 1$. Therefore for $|z|=R>1$, $\operatorname{Im} z\ge 0$ one can estimate $$\left|\frac{e^{i3z}}{(z-1)^2+1}\right| \le \frac{1}{|(z-1)^2+1|}\le \frac1{(R-1)^2}$$. Then ML estimate gives that the integral over the semicircle of radius $R$ is estimated by $$\frac{\pi R}{(R-1)^2} \to 0$$ as $R\to\infty$.

This works for integrands of form $e^{iaz} g(z)$, $a> 0$ such that $\lim_{R\to\infty}RM(R) =0$, where $M(R)$ is the maximum of $|g(z)|$ on the semicircle of radius $R$.

Jordan's lemma asserts that the integral over the semicircle tends to $0$ even if we only have $\lim_{R\to\infty}M(R) =0$ and the ML estimate does not help. You can look it up in the Wikipedia