Help with Jordan's Lemma (Collorary)

[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, becauseL=\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 problemCan anyone help me? Thanks!

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

 

[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.

 

[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