Solving a Very Tricky Integral

[mvillagra]

Hi, can someone give me a hand with this "little" integral please.

[tex]\int (\cos{k})^{t-s+1}(b-a\sin^2k)^{s/2}e^{-ik(n-1)}dk[/tex]

where
t is the time, which is discrete
s is between 0 and t
k has domain [-pi, pi]
n is a natural number
a, b are complex constants

actually this integral is the binomial expansion of this other integral

[tex]\int (\cos{k}+\sqrt{b-a\sin^2k})^t\cos k \quad e^{-ik(n-1)}dk[/tex]

there are some other constants, which I omitted. Also, I am omitting the coefficients and summation of the binomial expansion in the first integral.

I'm looking for a closed form, and I tried using asymptotic approximations but it doesn't work because in general to solve it using asymptotic techniques you need to write it in this form:

[tex]f(k)e^{\phi(k,t)}[/tex]

and I can't do that. Maybe there are some other asymptotic techniques that I don't know, maybe you can also give me a hand on this.

Thanks in advance!

 

[flatmaster]

That looks painfull. Sorry, but I have no ideas.

 

[mvillagra]

Maybe Laplaces's method could work. It requires an integral with the form:

[tex]I_n=\int_a^b \phi(k)[f(x)]^tdk[/tex]

Take a maximum point [tex]\xi[/tex] of the function [tex]f[/tex], then the integral is:

[tex]I_n\sim \phi(\xi)[f(\xi)]^{t+1/2}\left[\frac{-2\pi}{tf{''}(\xi)}\right]^{1/2}\textrm{ as }t\to\infty[/tex]

I think that this method only requires [tex]f[/tex] and [tex]\phi[/tex] to be analytic.

 

[jeff1evesque]

I am not fully sure right now how to incorporate it, but it seems as though you have "[URL identity[/URL] in your original equation. Perhaps you can rewrite [tex]e^{-ik(n-1)}[/tex] in terms of complex numbers (cosine and sine). You would have [tex]e^{-ik(n-1)} = e^{i(1k- kn)} = \frac{e^{ik}}{e^{kn}} = \frac{cos(1k)+isin(1k)}{cos(kn)+isin(kn)}.[/tex] If that seems like a good idea, maybe someone else could go into this integration in more detail (using euler's identity).

If that helps at all, that would be great. Otherwise, forgive me, and you can always check answers to various integrations online- Wolfram offers a website that does nearly all integrations.

I did out your"binomial expansion" equation and the results can be found:
http://integrals.wolfram.com/index....sin(k))^2k)^(t)cos(x)e^(-ix(n-1)&random=false

Goodluck,JL

 

[mvillagra]

I am not fully sure right now how to incorporate it, but it seems as though you have "[URL identity[/URL] in your original equation. Perhaps you can rewrite [tex]e^{-ik(n-1)}[/tex] in terms of complex numbers (cosine and sine). You would have [tex]e^{-ik(n-1)} = e^{i(1k- kn)} = \frac{e^{ik}}{e^{kn}} = \frac{cos(1k)+isin(1k)}{cos(kn)+isin(kn)}.[/tex] If that seems like a good idea, maybe someone else could go into this integration in more detail (using euler's identity).

I did out your"binomial expansion" equation and the results can be found:
http://integrals.wolfram.com/index....sin(k))^2k)^(t)cos(x)e^(-ix(n-1)&random=false

I tried to decompose it but the integral stays ugly

Also, I've checked the link, but the input was incorrect, there was one parenthesis missing and solved a different integral. I fixed it, and the answer I got was "Mathematica could not find a formula for your integral. Most likely this means that no formula exists"

Thank you very much for the reply