$\newcommand{\bR}{\mathbb{R}}$ There are many sources explaining this old technique of determining the asymptotic behavior of certain integrals depending on a small parameter $\hbar$. However, in applications, the integrals do not quite fit the setup described in most books I have consulted and I thought it would be nice to present the general strategy. What follows is folklore, and even not the most general possible result, but I took great pain to highlight the key features one should look for when attempting to use the Laplace technique in a concrete case.
Consider an interval $(a,b)\subset \bR$ and a family of $C^2$-functions
$$\phi_\hbar: (a,b)\to \bR,\;\;\hbar>0, $$
where the interval $(a,b)$ could be finite, or infinite. We are interested in the behavior of the integral
$$I_\hbar:=\int_a^b e^{-\phi_\hbar(t)} dt $$
as $\hbar \searrow 0$ given that the functions $\phi_h$ satisfy certain conditions
$\mathbf{C}_1$ For any $\hbar>0$ the function $\phi_\hbar$ has a
unique critical point $\tau=\tau(\hbar)\in (a,b)$. Moreover, $\phi_\hbar''(\tau)>0$. In other words, $\tau$ is a
nondegenerate local minimum, and the uniqueness assumption implies that $\phi_\hbar$ achieves its
absolute minimum at $\tau$. We set $\newcommand{\si}{\sigma}$
$$ \si=\si(\hbar):= \frac{1}{\sqrt{\phi_\hbar''(\tau)}}. $$
$\mathbf{C}_2$ The numers $\tau(\hbar)$ and $\si(\hbar)$ satisfy the conditions
$$ \lim_{\hbar\to 0}\frac{\tau(\hbar)-a}{\si(\hbar)}=\lim_{\hbar\to 0}\frac{ b-\tau(\hbar)}{\si(\hbar)}=\infty. $$
$\mathbf{C}_3$
$$\lim_{\hbar\to 0}\bigl(\,\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)\,\bigr)=\frac{x^2}{2},\;\;\forall x\in\bR $$
$\mathbf{C}_4$ There exists $u:\bR\to \bR$ such that
$$\int_{\bR}e^{-u(x)} dx <\infty\;\;\mbox{and}\;\;\phi_\hbar(\tau+ \si x)-\phi_\hbar(\tau)\geq u(x),\;\;\forall \hbar,\;\;x\in J(\hbar). $$
Then, under the assumptions $\mathbf{C}_1,\dotsc,\mathbf{C}_4$ we have
$$ I_\hbar \sim\sqrt{2\pi} \si e^{-\phi_\hbar(\tau)}\;\;\mbox{as $\hbar\to 0$}. \label{A}\tag{A} $$
Proof of (\ref{A}). We make the change of variables $t=\tau+\si x$ in the integral $I_\hbar$ to conclude that
$$I_\hbar=\si e^{-\phi_\hbar(\tau)}\int_{J(\hbar)} e^{-\bigl(\;\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)\;\bigr)} dx, $$
where
$$ J(\hbar)= \Bigl[\frac{a-\tau}{\si},\frac{b-\tau}{\si}\Bigr]. $$
The condition $\mathbf{C}_2$ implies that the intervals $J(\hbar)$ expand to $\bR$ as $\hbar \to 0$ Set $\newcommand{\vfih}{\varphi_\hbar}$
$$\vfih(x): =\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau). $$
$\mathbf{C}_3$ implies that
$$\vfih(x)\to\frac{x^2}{2}\;\;\mbox{as $\hbar\to 0$} ,\;\;\forall x\in\bR. $$
We can now use $\mathbf{C}_4$ to invoke the dominated convergence theorem and conclude that
$$\lim_{\hbar\to 0} \int_{J(\hbar)} e^{-\vfih(x)} dx=\int_{\bR}e^{-\frac{x^2}{2}}=\sqrt{2\pi}. $$
This completes the proof of (\ref{A}).
Remark 1. (a)
Often in applications each of the functions $\phi_\hbar$ is convex. In such cases the bound $\mathbf{C}_4$ is a consequence of the bound
$$ \bigl|\;\phi'_\hbar(\tau\pm \si)\;\bigr|= O\Bigl(\frac{1}{\si}\Bigr) \;\;\mbox{as $\hbar\to 0$}\label{B}\tag{B}. $$
Indeed, $\vfih$ is convex and thus its graph its situated above either of the tangent lines at $x=\pm 1$. Thus
$$ \vfih(x) \geq \max\Bigl\{ \vfih'(1)(x-1)+\vfih(1),\;\;\vfih'(-1)(x+1) +\vfih(-1)\Bigr\}. $$
Observing that
$$\vfih(\pm 1)>\vfih(0)=0,\;\;0< \pm \vfih'(\pm 1)= \pm \si\phi_\hbar(\tau\pm \si)=O(1), $$
we deduce that in $\mathbf{C}_4$ we can choose $u(x)$ of the form $u(x)=C(|x|-1)$ for some positive constant $C$.
(b) Both conditions $\mathbf{C}_3$ and $\mathbf{C}_4$ would follow immediately if one can prove that there exists a $C^1$-function
$$\Psi:[0,\infty)\times \bR\to \bR,\;\;(\hbar,x)\mapsto \Psi(\hbar,x), $$
such that
$$\Psi(0,x)=\frac{x^2}{2},\;\;\Psi(\hbar,x)=\vfih(x),\;\;\forall x\in J(\hbar),\;\;\forall \hbar >0. $$
(c) The esence of the above results is that, under appropriate assumptions, we can replace $\phi_\hbar(t)$ with its $2$-nd order jet at $\tau$
$$ \phi_\hbar(t)\approx \phi_\hbar(\tau)+\frac{(t-\tau)^2}{2\si^2} $$
and deduce that
$$ I_\hbar\sim \int_a^b e^{-\phi_\hbar(\tau)-\frac{(t-\tau)^2}{2\si^2}} dt,\;\;\mbox{as $\hbar \to 0$} $$
Example 1. Let me illustrate how the above strategy works in the classical situation described in all the books on asymptotics of integrals. Consider a $C^2$ convex function $\phi:(-a, a)\to\bR$ with a unique minimum at $\tau =0$ and such that $\phi''(0)>0$. Set $\phi_\hbar(t)=\frac{1}{\hbar}\phi(t)$ so that
$$I_\hbar=\int_{-a}^ae^{-\frac{1}{\hbar}\phi(t)} dt. $$
In this case
$$\tau=0,\;\; \si =\sqrt{\frac{\hbar}{\phi''(0)}}. $$
The conditions $\mathbf{C}_1,\mathbf{C}_2$ are obviously satisfied. As for $\mathbf{C}_3$ we observe that in this case we have
$$\vfih(x)=\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)=\frac{1}{\hbar}\bigl(\,\phi(\si x)-\phi(0)\;\bigr) = \frac{1}{\hbar}\Bigl(\;\phi'(0)\si x+ \frac{\phi''(0)}{2}\si^2 x^2+ o(\si^2)\;\Bigr)= \frac{x^2}{2}+ o(1). $$
Condition (\ref{B}) is also satisfied since
$$\vfih(\pm 1)=\frac{1}{\sqrt{\hbar\phi''(0)}}\phi'\Bigl(\pm \sqrt{\frac{\hbar}{\phi''(0)}}\Bigr) \to \pm 1\;\;\mbox{as $\hbar\to 0$}. $$
Hence we conclude that
$$\int_{-a}^a e^{-\frac{1}{\hbar}\phi(t)} dt \sim\sqrt{\frac{2\pi\hbar}{\phi''(0)}}\;\;\mbox{as $\hbar\to 0$}. $$
Example 2. Consider the integral
$$ \Gamma(\lambda +1)=\int_0^\infty t^\lambda e^{-t} dt,\;\;\lambda \to \infty. $$
Observing that it has the form $I_\hbar$ where $a=0$, $b=\infty$, $\hbar=\frac{1}{\lambda}$ and
$$\phi_\hbar(t)= t-\lambda \log t. $$
In this case we have
$$\phi'_\hbar(t)=1-\frac{\lambda}{t},\;\; \phi_\hbar(t)=\frac{\lambda}{t^2}, \;\; \tau(\lambda)=\lambda,\;\;\si(\lambda)=\frac{1}{\sqrt{\lambda}}. $$
thus, the conditions ($\mathbf{C}_1$) and ($\mathbf{C}_2$) are satisfied. To verify ($\mathbf{C}_3$) observe that
$$\vfih(x)=\phi_\hbar(\tau+\si x)-\phi_\hbar(\tau)= (\lambda +\sqrt{\lambda}x)-\lambda\log(\lambda+\sqrt{\lambda}x)- \lambda -\lambda \log \lambda $$
$$=\sqrt{\lambda} x-\lambda\log\Bigl(1+\frac{x}{\sqrt{\lambda}}\Bigr). $$
The condition $\mathbf{C}_3$ now follows from the Taylor expansion of $\log(1+s)$ at $s=0$. To prove $\mathbf{C}_4$ we observe that in this case $\phi_\hbar$ is continuous, so it suffices to check (\ref{B}), i.e., $\vfih(\pm 1)=O(1)$. In this case we have
$$\vfih'(\pm1)=\sqrt{\lambda}-\lambda\log\Bigl(1\pm \frac{1}{\sqrt{\lambda}}\Bigr), $$
and (\ref{B}) follows by using the Taylor expansion of $\log(1+s)$ at $s=0$. In this case
$$e^{-\phi_\hbar(\tau)}=\lambda^\lambda e^{-\lambda}$$
and we deduce
$$ \Gamma(\lambda+1)\sim \sqrt{2\pi}\lambda^{\lambda-\frac{1}{2}}e^{-\lambda}\;\;\mbox{as $\lambda\to\infty$}. $$
Example 3. Suppose that $w:[0,\infty)\to\bR$ is a smooth function such that
$$w(t), \;w'(t),\;\;w''(t) >0,\;\;\forall t>T>1. $$
Then
$$\mu_\lambda=\int_0^\infty t^\lambda e^{-w(t)} dt <\infty,\;\;\forall \lambda >0$$
and I would like to investigate the behavior of $\mu_\lambda$ as $\lambda\to \infty$. The quantitites $\mu_k$, $k\in\mathbb{Z}_{\geq 0}$, are the moments of the measure $e^{-w(t)}dt$ on the positive semiaxis. Note that
$$\mu_\lambda=\int_0^T t^\lambda e^{-w(t)} dt+\int_T^\infty t^\lambda e^{-w(t)} dt. $$
Observe that
$$ \int_T^\infty t^\lambda e^{-w(t)} dt \geq T^\lambda\int_T^\infty e^{-w(t)} dt, $$
while
$$ T^{-\lambda} \int_0^T t^\lambda e^{-w(t)} dt= \int_0^T \left(\frac{t}{T}\right)^\lambda e^{-w(t)} dt \to 0\;\;\mbox{as $\lambda\to \infty$}. $$
Thus, as $\lambda \to \infty$ we have
$$\mu_\lambda\sim I_\lambda:=\int_T^\infty t^\lambda e^{-w(t)} dt. $$
Observe that
$$I_\lambda =\int_T^\infty e^{-\phi_\lambda(t)} dt,\;\;\phi_\lambda(t)=w(t)-\lambda\log t. $$
We will show that the Laplace method is applicable in this case if we assume that we have an asymptotic estimate $\newcommand{\bZ}{\mathbb{Z}}$
$$ tw'(t)\sim A t^\alpha(\log t)^p\;\;\mbox{as $t\to\infty$},\;\; A>0\;\;,\alpha>1,\;\;p\in {\bZ}_{\geq 0}\tag{$\ast$} \label{ast} $$
which is twice differentiable.
The critical points of $\phi_\lambda$ are solutions of the equality
$$\lambda=tw'(t). $$
Since $tw'(t)$ is increasing and $tw'(t)\to\infty$ as $t\to \infty$ we deduce that the above equation has a unique solution $\tau=\tau(\lambda)$ for $\lambda \gg 0$. The correspondence $\lambda\to \tau(\lambda)$ is smooth and $\tau(\lambda)\to \infty$ as $\lambda\to \infty$. This proves $\mathbf{C_1}$. In view of (\ref{ast}) we deduce that
$$\lambda(\tau)\sim At^\alpha(\log \tau)^p \;\;\mbox{as $\tau\to \infty$}. $$
Observe that
$$\phi_\lambda''(\tau)=w''(\tau)+\frac{\lambda}{\tau^2} = w''(\tau)+\frac{w'(\tau)}{\tau}=\frac{\tau w''(\tau)+w'(\tau)}{\tau}. $$
Hence
$$\si =\sqrt{\frac{\tau}{\tau w''(\tau)+w'(\tau)}}=\sqrt{\frac{\tau}{\lambda'(\tau)},}\;\;\frac{\tau}{\si}= \sqrt{\tau \lambda'(\tau)}=\sqrt{\tau^2w''(\tau) +\tau w'(\tau)}\to \infty. $$
This proves $\mathbf{C_2}$. Observe
$$ \phi_\lambda(\tau+\si x)-\phi_\lambda(\tau)= w(\tau+ \si x)-w(\tau) -\lambda\log \Bigl(1+\frac{\si}{\tau}x \Bigr) = w(\tau+ \si x)-w(\tau) -\tau w'(\tau)\log \Bigl(1+\frac{\si}{\tau}x \Bigr) $$
$$=\si^2\left( w''(\tau) +\frac{w'(\tau)}{\tau}\right)\frac{x^2}{2} +\frac{\si^3x^3}{3!}w^{(3)}(\theta) + O\left(\frac{\si^3w'(\tau) x^3}{\tau^2}\right),$$
for some $\theta=\theta(\tau,x)\in [\tau,\tau+\si x]$. Now observe that
$$ \frac{\si^3 w'(\tau)}{\tau^2}=\frac{\si}{\tau} \frac{\si^2w'(\tau)}{\tau}=\frac{\si}{\tau}\frac{w'(\tau)}{\tau w''(\tau)+ w'(\tau)}\leq \frac{\si}{\tau}\to 0,\;\;\mbox{as $\tau\to \infty$}. \tag{1}\label{2}$$
To verify $\mathbf{C}_3$ we need to prove that
$$\lim_{t\to \infty}\si^3w^{(3)}(\theta)=0.\tag{2} \label{1} $$
It is time to use (\ref{ast}). We have
$$w'(t)+ tw''(t) \sim A\alpha t^{\alpha-1}(\log t)^p+ Apt^{\alpha-1} (\log t)^{p-1}, $$
$$2w''(t) +tw^{(3)}(t)\sim A\alpha(\alpha-1)t^{\alpha-2}(\log t)^p+A\alpha pt^{\alpha-2}(\log t)^{p-1}+Ap(\alpha-1) t^{\alpha-2} (\log t)^{p-1}+ Ap(p-1)t^{\alpha-2}(\log t)^{p-2}. $$
Now observe that
$$ t^3w^{(3)}(t)\sim -2t^2 w''(t)+A\alpha(\alpha-1)t^{\alpha}(\log t)^p+Ap t^\alpha(2\alpha-1) (\log t)^{p-1}+ Ap(p-1)t^{\alpha}(\log t)^{p-2}, $$
and
$$ -2t^2w''(t)\sim 2tw'(t) -2A\alpha t^{\alpha}(\log t)^p-2 Apt^{\alpha} (\log t)^{p-1} = -2A(\alpha-1)t^\alpha(\log t)^p -2Apt^\alpha (\log t)^{p-1}.\tag{3}\label{3}$$
Hence
$$t^3w^{(3)} (t) \sim At^{\alpha}\Bigl(\;(\alpha-1)(\alpha-2)(\log t)^p + p(2\alpha-3)(\log t)^{p-1}+ p(p-1)(\log t)^{p-2}\;\Bigr).$$
Also
$$\frac{\si}{\tau}=\frac{1}{\sqrt{\tau\lambda'(\tau)}} \sim \frac{1}{\sqrt{A\alpha}}\tau^{-\frac{\alpha}{2}}(\log \tau )^{-\frac{p}{2}},\;\;\si \sim \frac{1}{\sqrt{A\alpha}}\tau^{1-\frac{\alpha}{2}}(\log \tau )^{-\frac{p}{2}}. \tag{4}\label{4}$$
We distinguish three cases.
Case 1. $\alpha \neq 2$ In this case $\si\to 0$ and $\newcommand{\ve}{\varepsilon}$
$$\theta= \tau(1 +c\frac{\si}{\tau}x) $$
$$w^{3}(\theta) =Ct^{\alpha-3} (\log t)^p\bigl( \;1+\ve(t)\;\bigr), $$
where $C=A(\alpha-1)(\alpha-2)\neq 0$ and $\ve(t)\to 0$ as $t\to \infty$. Then
$$ \si^3w^{(3)}(\theta) \sim\frac{C}{\sqrt{A\alpha}} t^{-\frac{\alpha}{2}} (\log t)^{-\frac{p}{2}} \to 0. $$
This proves (\ref{1}).
Case 2. $\alpha=2$ $p=1$ In this case (\ref{ast}) implies
$$\lim_{t\to\infty}w^{(3)}(t)=0,\;\;\si(t)=O(1)\;\;\mbox{as $t\to\infty$}. $$
which clearly implies (\ref{1}).
Case 3. $\alpha=2, p>0$ Proceed as in
Case 1.
Finally, we want to check $\mathbf{C}_4$. In our case $\phi_h$ is convex and we will verify (\ref{B}). We have
$$\vfih(x) = \frac{d}{dx}\Bigl( w(\tau+\si x) -w(\tau)-\tau w'(\tau)\log\Bigl(1+\frac{\si}{\tau}x\Bigr)\;\Bigr) $$
$$=\si w'(\tau+\si x)-\frac{\si w'(\tau)}{1+\frac{\si}{\tau}x}=\si w'(\tau)\left(1-\frac{1}{1+\frac{\si}{\tau}x}\right)+\frac{1}{2}\si^2x^2w''(\theta)=\frac{\si^2 w'(\tau)}{\tau}x^2+ \frac{1}{2}\si^2x^2w''(\theta)+ O\left( \frac{\si^3w'(\tau)}{\tau^2} x^2\right), $$
for some $\theta\in [\tau,\tau+\si]$. As in the proof of (\ref{2}) we deduce that
$$ \frac{\si^2w'(\tau)}{\tau}=O(1),\;\; \frac{\si^3w'(\tau)}{\tau^2}=o(1). $$
We only need to verify that
$$ \si^2w''(\theta)=O(1). $$
This follows from (\ref{3}) and (\ref{4}}).
