Using Laplace’s method to find the leading-order of an integral with two variables in the exponent 
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 $0$:"
  $$I\left( x\right) =\dfrac{1}{2}\int _{-\infty}^{\infty }\dfrac {1} {\left( 1+t\right) }e^{x\left( 1-t^{2}\right) }dt\\$$

 A: 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 $x > 1$.
