Trying to show $\int_0^1 e^{-xt}sin(t) dt \sim \frac{1}{x^2}$

Question

I am using Laplace's Method and I am trying to show

$$I =\int_0^1 e^{-xt}sin(t) dt \sim \frac{1}{x^2}$$

$h(t) = -t$ has a maximum at $0$ and as it is a simple function there is no need to expand it.

We replace $I$ with

$$\int_0^{\delta} e^{-xt}sin(t) dt$$

where $\delta$ is so small that we can replace $sin(t)$ by the first term of it's taylor series, i.e. $1$.

We now have $$I =\int_0^1 e^{-xt}sin(t) dt \sim \int_0^{\delta} e^{-xt} dt = \frac{1 - e^{-\delta x}}{x}$$

And this goes to $\frac{1}{x}$ as $x \to \infty$

Why am I getting $\frac{1}{x}$ instead of $\frac{1}{x^2}$?

Answer 1

Just a simple mistake of the common but annoying "How could I have missed that?" type.

$$\sin t = t - \frac{t^3}{6} + O(t^5),$$

so you should look at the integral

$$\int_0^\delta te^{-xt}\,dt = -\frac{\delta e^{-\delta x}}{x} + \frac{1}{x}\int_0^\delta e^{-xt} = \frac{1-e^{-\delta x} - \delta xe^{-\delta x}}{x^2}\sim \frac{1}{x^2}.$$