0
$\begingroup$

Find the leading behavior of $\int_0^{\pi/2} e^{i s \cos(x)} dx$ as $s \to \infty$

I wanted to use Laplace's method, but then I saw the $i$ and now I don't know what to do. And furthermore the methods I was seeing seemed to depend in $\frac{d}{dx} \cos (x) \neq 0$ when $x = 0$ (that is where cosine takes its maximum, though in this context "maximum" doesn't make sense since $i \cos(x)$ is a complex number and thus does not have a "maximum"), so even then I seem to be stuck.

$\endgroup$
  • $\begingroup$ Are you familiar with the method of stationary phase or the saddle point method? $\endgroup$ – Antonio Vargas Aug 8 '17 at 17:42
  • $\begingroup$ hint: the contours of stationary phase are lines paralell to the imaginary axis $\endgroup$ – tired Aug 8 '17 at 22:03
2
$\begingroup$

The modulus of $$f(s)\stackrel{\text{def}}{=}\int_{0}^{\pi/2}e^{is\cos x}\,dx = \int_{0}^{\pi/2}e^{is\sin x}\,dx = \int_{0}^{1}e^{isx}\frac{dx}{\sqrt{1-x^2}} =\frac{\pi}{2}\left[J_0(s)+i H_0(s)\right]$$ decays like $\frac{C}{\sqrt{s}}$ for $s\to +\infty$. You may study how to apply Laplace method in the last section of these notes. An equivalent approach is to devise a differential equation fulfilled by $f(s)$ and to derive such bound from such a differential equation.

| cite | improve this answer | |
$\endgroup$

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.