laplace method $\sim \frac12 \sqrt{\frac{\pi}x}$ 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?
 A: 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.
A: 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
$$ I=e^x \frac{1}{2}\sqrt{2 \pi \sigma^2}=\frac{e^x}{2}\sqrt{\frac{\pi}{x}}$$
