Let $p\in \mathbb{R}$, I would like to investigate the asymptotic behavior of the following integral:
$$\int_0^1 e^{x(tp-\frac{1}{2}t^2)}dt$$
as $x\to\infty$. In particular, I would like to know how it depends on $p$ and $x$?.
If $p\in (0,1)$, then the maximum of $t\mapsto at-\frac{1}{2}t^2$ over $[0,1]$ is achieved at $p\in (0,1)$. Hence applying Laplace's method (e.g., here),
$$
\int_0^1 e^{x(tp-\frac{1}{2}t^2)}dt\sim \sqrt{\frac{2\pi}{x}}e^{x\frac{p^2}{2}}
$$
as $x\to \infty$.
In this case, is it possible to quantify how the error term
$$\left|\int_0^1 e^{x(tp-\frac{1}{2}t^2)}dt-\sqrt{\frac{2\pi}{x}}e^{x\frac{p^2}{2}} \right|$$
depends on $p$?
Alternatively, if $p\not \in (0,1)$, then the maximum of $t\mapsto at-\frac{1}{2}t^2$ over $[0,1]$ is obtained at boundary, and the first order derivative at the maximizer is not zero. In this case, I am not sure how to apply the classical result.
Could someone help to explain how to study the asymptotic behavior of the integral for different $p$?
Best Answer
I shall mainly address your first question. Your integral can be re-written in the form $$ \exp \left( {\frac{1}{2}p^2 x} \right)\int_0^1 {\exp \left( { - \frac{x}{2}(t - p)^2 } \right)dt} = \sqrt {\frac{{2\pi }}{x}} \exp \left( {\frac{1}{2}p^2 x} \right)(1 + R(x,p)), $$ where $$ R(x,p) = - \frac{1}{2}\left( {\operatorname{erfc}\left( {p\sqrt {\frac{x}{2}} } \right) + \operatorname{erfc}\left( {(1 - p)\sqrt {\frac{x}{2}} } \right)} \right), $$ with $\operatorname{erfc}$ being the complementary error function. By $\S7.12(\text{i})$, we have that $$ \left| {R(x,p)} \right| \le \frac{1}{{\sqrt {2\pi x} }}\left( {\frac{1}{p}\exp \left( { - \frac{1}{2}p^2 x} \right) + \frac{1}{{1 - p}}\exp \left( { - \frac{1}{2}(1 - p)^2 x} \right)} \right) $$ provided $x>0$ and $0<p<1$. If $p\to 0$ or $p\to 1$, the saddle coalesces with an endpoint. In that case, the asymptotics is given by the exact form of $R(x,p)$. See $\S2.3(\text{v})$ for more details. Outside of $(0,1)$, one may use $\operatorname{erfc}(-w)=2-\operatorname{erfc}(w)$.