laplace method $\sim \frac12 \sqrt{\frac{\pi}x}$

Question

Use Laplace's method to show that $$I(x)=\int \limits_0^{\infty}\frac{e^{x(2t-t^2)}}{1+t^2}dt \, \, \sim \, \, \frac12 \sqrt{\frac{\pi}x}$$ as $x \rightarrow \infty$.

So we make the top limit into $w$ with $w \rightarrow \infty$.

Let $f(t)=1/(1+t^2)$, $g(t)=2t-t^2$ and $g'(t)=2-2t$

So $$\int \limits_0^w \frac{f(t)}{xg'(t)} \frac d{dt}(e^{xg(t)})dt = \bigg[\frac{f(t)}{xg'(t)} e^{xg(t)} \bigg]_0^w - \int \limits_0^w \frac d{dt}\bigg(\frac{f(t)}{g'(t)}\bigg) e^{xg(t)}dt$$

Since $g'(t) \neq 0$ for $t \in [0,w]$ and either $f(0)$ or $f(w)$ is not $0$, then $$I(x) \sim \frac{f(w)}{xg'(w)} e^{xg(w)} -\frac{f(0)}{xg'(0)} e^{xg(0)} =A-B$$

$B=1/2x$ which is $0$ as $x \rightarrow \infty$ but how do we evaluate $A$?

Am I even on the right tracks for this?

Answer 1

This approach will not work because $g'(t)$ has a simple zero at $t=1$, which means that

$$ \frac{d}{dt} \frac{f(t)}{g'(t)} $$

will have a pole of order $2$ at $t=1$, and thus the integral

$$ \int_0^\infty \left(\frac {d}{dt}\frac{f(t)}{g'(t)}\right) e^{xg(t)}\,dt $$

will not exist (even in a principal value sense).

The "right" approach is to note that for large $x$ the main contribution to the integral comes from a neighborhood of $t=1$, and so

$$ \begin{align} \int_0^\infty \frac{e^{x(2t-t^2)}}{1+t^2}\,dt &\sim \left. \frac{1}{1+t^2} \right|_{t=1} \int_0^\infty e^{x(2t-t^2)}\,dt \\ &\sim \left. \frac{1}{1+t^2} \right|_{t=1} \int_{-\infty}^{\infty} e^{x(2t-t^2)}\,dt \\ &= \frac{e^x}{2} \sqrt{\frac{\pi}{x}}. \end{align} $$

This is basically the usual structure of a Laplace method argument.

Answer 2

Completing squares, $2t -t^2 = -(t-1)^2+1$ Hence

$$\frac{e^{x(2t-t^2)}}{1+t^2}= e^x \frac{e^{-x(t-1)^2}}{1+t^2}$$

The numerator corresponds to a gaussian centered a $t=1$ with variance $\sigma^2=1/(2x)$. Hence as $x\to \infty$ the numerator will be approximately constant $1+t^2\approx 2$ and the integral gives

$$ I=e^x \frac{1}{2}\sqrt{2 \pi \sigma^2}=\frac{e^x}{2}\sqrt{\frac{\pi}{x}}$$