1
$\begingroup$

I need a clarification on the utility of Jordan's lemma. I think I have understood the theorem and its implications. It basically implies that if you have a function like $g(z) =f(z)e^{iz}$ it suffices for $f(z)$ to tend to zero at infinity in order to have a negligible integral of $g(z)$ along a sufficiently big semicircle in the upper half plane.

However, It confuses me that Wikipedia provides, as an example, the evaluation of this integral: $\int_{-\infty}^\infty \frac{\cos x}{1+x^2}\,dx$

Then it uses the fact that $f(z)=\frac{e^{iz}}{1+z^2}$ satisfies Jordan's lemma. But it seems to me that this particular function satisfies the Estimation lemma in the upper part of the Gauss graph, a stronger condition, and Jordan's lemma is not needed

1 Answers 1

1

That is not a great example, for as you mention the Estimation Lemma suffices there. A better example would be for the evaluation of $\int^{\infty}_{-\infty} \frac{\sin x}{x} dx.$

For that problem the usual estimate leads to the circular arc contributing $\mathcal{O}(1)$ so we truly need Jordan's lemma.