evaluation of oscillatory integrals

by Sangoku
Tags: evaluation, integrals, oscillatory
Sangoku is offline
Aug26-07, 03:34 AM
P: 21
HOw can you evaluate (for big 'u') the exponential integral

[tex] \int_{0}^{1}dx f(x)e^{iux} [/tex] u-->oo

if you use Numerical methods, since exp(iux) oscillates rapidly you cannot apply any of them, also since we have inside exponential (iux) you can not apply 'Stationary phase method' unless you make the change.

[tex] \int f(g(t)exp(iug(t)) [/tex]

and then apply stationary phase finding [tex] g'(t)=0 [/tex] to evaluate the integral but i am not pretty sure.
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AiRAVATA is offline
Aug26-07, 10:47 AM
P: 173
You should check out Olver's book Asymptotics and Special Functions. I'm pretty sure you'll find something there.
jostpuur is offline
Aug27-07, 12:11 AM
P: 1,983
If f is continuous, then taking f out of the integral with the intermediate value theorem, and calculating the integral of the exponential expression, should enable you to prove that this approaches zero when u approches infinity.

jostpuur is offline
Aug28-07, 01:30 AM
P: 1,983

evaluation of oscillatory integrals

And remember that in

\int\limits_0^1 f(x) e^{iux} dx = f(\xi_u) \int\limits_0^1 e^{iux} dx

the [itex]\xi_u[/itex] depends on u. I think suprisingly many of my mistakes have been related to first using some intermediate value theorem, and then ignoring on what the xi depends.
Haelfix is offline
Aug28-07, 07:44 AM
Sci Advisor
P: 1,663
You can try a physics trick called "wick rotation" by making a change of variables to the real plane (to get rid of the annoying 'i' in the top integral), you just have to check to see if the analytic continuation makes sense there.

But i'd first go with the method of steepest descent, it seems tailor made for this type of integral after a suitable variable change f(x) --> h(Z) and ix -- > g(Z). I don't see why its not defined in this case. So long as h(Z) and g(Z) are analytic functions in some contour of the complex plane (check this) and u --> infinity, you're set.

Of course if f(x) is some pathological beast, you're probably out of luck.
jostpuur is offline
Aug28-07, 08:20 AM
P: 1,983
My two previous posts were example of how first making a mistake, and then fixing it, can lead into unnecessarily complicated way. This is what I should have said

\big| \int\limits_0^1 f(x) e^{iux} dx\big| \leq \big(\underset{x\in [0,1]}{\textrm{sup}} |f(x)|\big) \big| \int\limits_0^1 e^{iux} dx\big| \underset{u\to\infty}{\to} 0

If the supremum doesn't exist, then you'll have to know something else, and use it somehow.
Canerg is offline
Aug25-11, 06:57 AM
P: 12
Please this problem is similar to your problem. But i could'nt solve please help me
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jostpuur is offline
Aug25-11, 03:27 PM
P: 1,983
Holy crap! I've been writing nonsense here in 2007.

If you want to use the intermediate value theorem

\int\limits_0^1 f(x)\phi(x)dx = f(\xi) \int\limits_0^1 \phi(x)dx

then [itex]\phi[/itex] should be real and non-negative.

Also, if you set [itex]f(x)=e^{-i\pi x}[/itex] and [itex]u=2\pi[/itex], then

\int\limits_0^1 f(x) e^{iux}dx = \frac{2i}{\pi}


\big(\underset{x\in [0,1]}{\textrm{sup}}\; |f(x)|\big) \int\limits_0^1 e^{iux}dx = 0

So my inequality wasn't fine.

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