Finding asymptotic expansion using Laplace's method

Consider the integral $\displaystyle I$ defined by

$\displaystyle \displaystyle I = \int^1_0 e^{-xt} t \cos t dt$.

Use Laplace's method to obtain the first term of the asymptotic expansion of the integral as $\displaystyle x \to \infty$.

How can I start? Can someone point me to somewhere with a similar example?

Thank you.

Re: Finding asymptotic expansion using Laplace's method

Quote:

Originally Posted by

**math2011** Consider the integral $\displaystyle I$ defined by

$\displaystyle \displaystyle I = \int^1_0 e^{-xt} t \cos t dt$.

Use Laplace's method to obtain the first term of the asymptotic expansion of the integral as $\displaystyle x \to \infty$.

How can I start? Can someone point me to somewhere with a similar example?

Thank you.

I don't see that this is Laplace's method but:

For large $\displaystyle x$ the function $\displaystyle e^{-xt}$ is small everywhere except close to $\displaystyle t=0$, so we expand the other term in the integrand about $\displaystyle t=0$:

$\displaystyle \int_0^1 e^{-xt}t \cos(t)\; dt = \int_0^1 e^{-xt}t\ [1-t^2/2+...]\; dt $

Truncating after the first term gives:

$\displaystyle \int_0^1 e^{-xt}t \cos(t)\; dt \sim \int_0^1 e^{-xt}t\; dt=x^{-2}[1-(x+1)e^{-x}] $

and you can probably replace the square brackets by 1.

CB

Re: Finding asymptotic expansion using Laplace's method

Thanks! I can see that the approximated solution is similar to the solution. But this question is definitely a Laplace method question. How can I apply the Laplace method to the question?

Re: Finding asymptotic expansion using Laplace's method

Quote:

Originally Posted by

**math2011** Thanks! I can see that the approximated solution is similar to the solution. But this question is definitely a Laplace method question. How can I apply the Laplace method to the question?

Well that was the version of Laplace's method from the first part of this document, except the maximum is of $\displaystyle \phi(x)$ is at an end point of the interval so not quadratic, and so this is covered by the middle paragraph of the further notes section.

CB

Re: Finding asymptotic expansion using Laplace's method

According to the tutorial 2.8 referred to by the middle paragraph of the further notes section, the higher boundary $\displaystyle 1$ should be taken to $\displaystyle \infty$. This is what I got after doing that.

$\displaystyle \int^1_0 e^{-xt} t \cos(t) dt =& \int^1_0 e^{-xt} t \left(1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \cdots \right) dt \\ \sim & \int^\infty_0 e^{-xt} t dt \\ =& -\frac{1}{x} \left[ e^{-xt} t \right]^\infty_0 - \int^\infty_0 \left(-\frac{1}{x}\right) e^{-xt} dt \\ =& -\frac{1}{x} \left[ e^{-xt} t \right]^\infty_0 - \frac{1}{x^2} \left[ e^{-xt} \right]^\infty_0 \\ =& \frac{1}{x^2}$

Thanks for helping me getting into this topic.