# Thread: Asymptotic evaluation of an integral..

1. ## Asymptotic evaluation of an integral..

Let be the integral:

Int(-oo,oo)dx f(x)exp(iux) I would like to know if there is any method to

evaluate it for u-->oo

-Stationary phase method:=can't be applied since y=x has no "stationary points" for any real x.

- by the way if we had f(x)=g(x)+ih(x) could the integral above be evaluated splitting it into the real and complex part?..thanks.

2. Originally Posted by lokofer
Let be the integral:

Int(-oo,oo)dx f(x)exp(iux) I would like to know if there is any method to

evaluate it for u-->oo

If |f(x)|^2 is integrable over (-inf,inf) then your integral goes to 0 almost
everywere as u goes to infinity, in fact it's absolute value squared is itself
integrable from -inf to inf.

(This is a fourier transform).

RonL

3. Originally Posted by lokofer
Let be the integral:

Int(-oo,oo)dx f(x)exp(iux) I would like to know if there is any method to

evaluate it for u-->oo

-Stationary phase method:=can't be applied since y=x has no "stationary points" for any real x.

- by the way if we had f(x)=g(x)+ih(x) could the integral above be evaluated splitting it into the real and complex part?..thanks.
Int(-oo,oo)dx f(x)exp(iux)=Int(-inf,inf) [g(x)+ih(x)][cos(ux)+isin(ux)] dx

................................ ..=Int(-inf,inf) {[g(x)cos(ux)-h(x)sin(x)]+i[g(x)sin(ux)+h(x)cos(ux)]}dx

....................................=Int(-inf,inf)[g(x)cos(ux)-h(x)sin(x)]dx +
.........................................i.Int(-inf,inf)[g(x)sin(ux)+h(x)cos(ux)]dx

RonL