# Thread: help for asymptotic expansion

1. ## help for asymptotic expansion

Hi im new to this forum and was wondering if someone could help me with the following problem

use a suitable series expansion to find an asymptotic expansion as x approaches infinity for

the integral from 0 to infinity of [ exp(-x(t^2))sin(t) ]

i tried expanding the sine part and integrating term by term. i got stuck doing integrating term by term. i dont know how to do integral im not sure if i should use a different expansion?

2. Originally Posted by jules
Hi im new to this forum and was wondering if someone could help me with the following problem

use a suitable series expansion to find an asymptotic expansion as x approaches infinity for

the integral from 0 to infinity of [ exp(-x(t^2))sin(t) ]

i tried expanding the sine part and integrating term by term. i got stuck doing integrating term by term. i dont know how to do integral im not sure if i should use a different expansion?

We want an expansion of

$
I(x) = \int_0^{\infty} \exp(-x t^2)\sin(t) dt
$

for large positive $x$.

First put $u=\sqrt{x}t$, then the integral becomes:

$
I(x) = \frac{1}{\sqrt{x}}\int_0^{\infty} \exp(- u^2)\sin(u/\sqrt{x}) du
$

Now as $x$ is large positive and $\exp(-u^2)$ decays rapidly for large $u$ we can expand the $\sin(u/\sqrt{x})$ to get:

$
I(x) = \frac{1}{\sqrt{x}}\int_0^{\infty} \exp(- u^2)\left[ \sum_0^{\infty}\frac{(u/\sqrt{x})^{2n+1}(-1)^{n}}{(2n+1)!} \right] du
$

Hopefully you can take it from here.

RonL

3. Hi
Thanks that helped. Though im still not clear on how to do that integral??
Thanks

4. Originally Posted by jules
Hi
Thanks that helped. Though im still not clear on how to do that integral??
Thanks
Standard integral:

$
\int_0^{\infty} \exp(-u^2) ~u^k du = \frac{1}{2} \Gamma\left(\frac{2k+1}{2}\right)
$

In the case of the power you have the $\Gamma$ will simplify to a factorial.

RonL