- #1
Leo_rr
- 1
- 0
I'm trying to use the saddle point method to solve the following integral:
Z = (1/sqrt{2 pi t}) ∫_{1}^{infinity} ds (1/sqrt{2 pi s}) exp{ p [-s ln(s/t) +s] } cos(2 pi L~ p ~ s), as p → infinity
Mod edit to make integral more readable:
$$Z = \frac{1}{\sqrt{2\pi t}} \int_1^{\infty} \frac 1 {\sqrt{2 \pi s}} e^{p(-s \ln(s/t) + s)} \cos(2 \pi L p s)ds$$
$$\lim_{p \to \infty}Z = ? $$
Is this the correct integral?
where L and p are Integers. If we remove the fast oscillatory cosine, the Sadde Point method is straightforward and the solution is:
Z = e^{pt}
Introducing a fast oscillatory function in the integral, one would expect the value of Z to be much smaller, but the direct (or maybe naive) application of the method results in the exact same result, and that's absurd! Have anybody faced a similar problem?
Z = (1/sqrt{2 pi t}) ∫_{1}^{infinity} ds (1/sqrt{2 pi s}) exp{ p [-s ln(s/t) +s] } cos(2 pi L~ p ~ s), as p → infinity
Mod edit to make integral more readable:
$$Z = \frac{1}{\sqrt{2\pi t}} \int_1^{\infty} \frac 1 {\sqrt{2 \pi s}} e^{p(-s \ln(s/t) + s)} \cos(2 \pi L p s)ds$$
$$\lim_{p \to \infty}Z = ? $$
Is this the correct integral?
where L and p are Integers. If we remove the fast oscillatory cosine, the Sadde Point method is straightforward and the solution is:
Z = e^{pt}
Introducing a fast oscillatory function in the integral, one would expect the value of Z to be much smaller, but the direct (or maybe naive) application of the method results in the exact same result, and that's absurd! Have anybody faced a similar problem?
Last edited by a moderator:
