Past Paper Question:
Use Laplace’s method to find the leading-order asymptotic behaviour of the integral
$$\begin{align*} & I\left( x\right) =\int _{0}^{\infty }\dfrac {1} {\left( 1+t\right) }e^{x\left( 1-t^{2}\right) }dt\\ & x\rightarrow \infty \end{align*}$$
Wasn't given many example in lecture notes on how to actually apply the Laplace method; we studied the general case mainly. Trying to learn from examples.
My Attempt:
This is an integral of the form $$I\left( x\right) =\int _{a}^{b}f\left( t\right) e^{xg\left( t\right) }dt$$
In this case $f\left(t\right) = \dfrac {1} {1+t}$ and $g\left( t\right) =\left( 1-t^{2}\right)$. Thus the maximum is at $t=0$, so for our purposes this is the "Case $3$ Laplace method: The maximum is at $t=c$ where $a<c<b$ with $f\left( c\right) \neq 0.$"Step $1$: Split the integral into a local and non-local part and estimate the non-local contribution.
Step $2$: Use Taylor expansion around $c$ to obtain $I\left( x;\varepsilon \right)$, then rescale the integration variable to remove $x$ from the exponential.
Step $3$: Replace both limits of the integration of $I\left( x;\varepsilon \right)$ by $\pm\infty$ and calculate the associated error.
At the moment this fails the definition of case $3$ we apply the following amendment to the question, "Step $0$:"
$$I\left( x\right) =\dfrac{1}{2}\int _{-\infty}^{\infty }\dfrac {1} {\left( 1+t\right) }e^{x\left( 1-t^{2}\right) }dt\\$$
Best Answer
We'll consider the integral
$$ J(x) = \int_0^\infty \frac{1}{1+t} e^{-xt^2}\,dt, $$
which is in the form
$$ \int_0^\infty f(t) e^{xg(t)}\,dt $$
with
$$ f(t) = \frac{1}{1+t} \qquad \text{and} \qquad g(t) = -t^2. $$
The basic idea of the Laplace method is that the main contribution to the integral $J(x)$ comes from the points $t$ where the integrand $f(t)e^{xg(t)}$ is largest.
As $x$ grows larger, the behavior of the factor $f(t)$ matters less and less; where $g(t)$ is large $f(t)e^{xg(t)}$ will be large, and where $g(t)$ is small $f(t)e^{xg(t)}$ will be small. This means that we're really interested in the maxima of $g(t)$ -- these will be the points $t$ where the integrand $f(t)e^{xg(t)}$ is largest when $x$ is large.
For this problem the function $g(t) = -t^2$ has a single maximum at $t=0$. By the heuristics aboove, the large-$x$ behavior of $J(x)$ should be very similar to the large-$x$ behavior of an integral over a small neighborhood of this point:
$$ J(x) \approx \int_0^\delta \frac{1}{1+t} e^{-xt^2}\,dt. $$
Our goal is to justify this and to further simplify the integrand to get a simplified expression for the approximation.
The first thing we want to do is show that the integral over the region $[\delta,\infty)$ is very small. Here we can get the estimate
$$ \begin{align} 0 \leq \int_\delta^\infty \frac{1}{1+t} e^{-xt^2}\,dt &< \frac{1}{1+\delta} \int_\delta^\infty e^{-xt^2}\,dt \\ &= \frac{1}{1+\delta} e^{-\delta^2x} \int_\delta^\infty e^{x (\delta^2-t^2)}\,dt \\ &\leq \frac{1}{1+\delta} e^{-\delta^2x} \int_\delta^\infty e^{\delta^2-t^2}\,dt \tag{1} \end{align} $$
for all $x \geq 1$. Thus
$$ J(x) = \int_0^\delta \frac{1}{1+t}e^{-xt^2}\,dt + O\!\left(e^{-\delta^2x}\right) \tag{2} $$
for all $x \geq 1$.
Our next goal is to study this remaining integral and to show that, to leading order, it is larger than the error term $O(e^{-\delta^2x})$. To do this we will use the fact that
$$ \lim_{t \to 0} \frac{1}{1+t} = 1. $$
Let
$$ h(t) = \frac{1}{1+t} - 1 = \frac{-t}{1+t}, $$
so that
$$ \int_0^\delta \frac{1}{1+t} e^{-xt^2}\,dt = \int_0^\delta e^{-xt^2}\,dt + \int_0^\delta h(t) e^{-xt^2}\,dt. \tag{3} $$
We expect the second integral to be smaller than the first since $h(t) \approx 0$ for small $\delta$. Indeed,
$$ \begin{align} \int_0^\delta e^{-xt^2}\,dt &= \int_0^\infty e^{-xt^2}\,dt - \int_\delta^\infty e^{-xt^2}\,dt \\ &= \frac{1}{2} \sqrt{\frac{\pi}{x}} + O\!\left(e^{-\delta^2 x}\right), \tag{4} \end{align} $$
where we used an argument very similar to $(1)$ to obtain the $O(\cdots)$ estimate, and
$$ \begin{align} \left| \int_0^\delta h(t) e^{-xt^2}\,dt \right| &\leq \int_0^\delta |h(t)| e^{-xt^2}\,dt \\ &< \int_0^\delta te^{-xt^2}\,dt \\ &< \int_0^\infty te^{-xt^2}\,dt \\ &= \frac{1}{2x}. \tag{5} \end{align} $$
Combining $(4)$ and $(5)$, $(3)$ becomes
$$ \int_0^\delta \frac{1}{1+t} e^{-xt^2}\,dt = \frac{1}{2} \sqrt{\frac{\pi}{x}} + O\!\left(\frac{1}{x}\right), $$
and combining this with $(2)$ yields
$$ \int_0^\infty \frac{1}{1+t} e^{-xt^2}\,dt = \frac{1}{2} \sqrt{\frac{\pi}{x}} + O\!\left(\frac{1}{x}\right) $$
for all $x > 1$.