Laplace's method

Lower bound: Let ${\displaystyle \varepsilon >0}$. Since ${\displaystyle f''}$ is continuous there exists ${\displaystyle \delta >0}$ such that if ${\displaystyle |x_{0}-c|<\delta }$ then ${\displaystyle f''(c)\geq f''(x_{0})-\varepsilon .}$ By Taylor's Theorem, for any ${\displaystyle x\in (x_{0}-\delta ,x_{0}+\delta ),}$

${\displaystyle f(x)\geq f(x_{0})+{\frac {1}{2}}(f''(x_{0})-\varepsilon )(x-x_{0})^{2}.}$

Then we have the following lower bound:

${\displaystyle {\begin{aligned}\int _{a}^{b}e^{nf(x)}\,dx&\geq \int _{x_{0}-\delta }^{x_{0}+\delta }e^{nf(x)}\,dx\\&\geq e^{nf(x_{0})}\int _{x_{0}-\delta }^{x_{0}+\delta }e^{{\frac {n}{2}}(f''(x_{0})-\varepsilon )(x-x_{0})^{2}}\,dx\\&=e^{nf(x_{0})}{\sqrt {\frac {1}{n(-f''(x_{0})+\varepsilon )}}}\int _{-\delta {\sqrt {n(-f''(x_{0})+\varepsilon )}}}^{\delta {\sqrt {n(-f''(x_{0})+\varepsilon )}}}e^{-{\frac {1}{2}}y^{2}}\,dy\end{aligned}}}$

where the last equality was obtained by a change of variables

${\displaystyle y={\sqrt {n(-f''(x_{0})+\varepsilon )}}(x-x_{0}).}$

Remember ${\displaystyle f''(x_{0})<0}$ so we can take the square root of its negation.

If we divide both sides of the above inequality by

${\displaystyle e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}}$

and take the limit we get:

${\displaystyle \lim _{n\to \infty }{\frac {\int _{a}^{b}e^{nf(x)}\,dx}{e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}}}\geq \lim _{n\to \infty }{\frac {1}{\sqrt {2\pi }}}\int _{-\delta {\sqrt {n(-f''(x_{0})+\varepsilon )}}}^{\delta {\sqrt {n(-f''(x_{0})+\varepsilon )}}}e^{-{\frac {1}{2}}y^{2}}\,dy\,\cdot {\sqrt {\frac {-f''(x_{0})}{-f''(x_{0})+\varepsilon }}}={\sqrt {\frac {-f''(x_{0})}{-f''(x_{0})+\varepsilon }}}}$

since this is true for arbitrary ${\displaystyle \varepsilon }$ we get the lower bound:

${\displaystyle \lim _{n\to \infty }{\frac {\int _{a}^{b}e^{nf(x)}\,dx}{e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}}}\geq 1}$

Note that this proof works also when ${\displaystyle a=-\infty }$ or ${\displaystyle b=\infty }$ (or both).

Upper bound: The proof is similar to that of the lower bound but there are a few inconveniences. Again we start by picking an ${\displaystyle \varepsilon >0}$ but in order for the proof to work we need ${\displaystyle \varepsilon }$ small enough so that ${\displaystyle f''(x_{0})+\varepsilon <0.}$ Then, as above, by continuity of ${\displaystyle f''}$ and Taylor's Theorem we can find ${\displaystyle \delta >0}$ so that if ${\displaystyle |x-x_{0}|<\delta }$, then

${\displaystyle f(x)\leq f(x_{0})+{\frac {1}{2}}(f''(x_{0})+\varepsilon )(x-x_{0})^{2}.}$

Lastly, by our assumptions (assuming ${\displaystyle a,b}$ are finite) there exists an ${\displaystyle \eta >0}$ such that if ${\displaystyle |x-x_{0}|\geq \delta }$, then ${\displaystyle f(x)\leq f(x_{0})-\eta }$.

Then we can calculate the following upper bound:

${\displaystyle {\begin{aligned}\int _{a}^{b}e^{nf(x)}\,dx&\leq \int _{a}^{x_{0}-\delta }e^{nf(x)}\,dx+\int _{x_{0}-\delta }^{x_{0}+\delta }e^{nf(x)}\,dx+\int _{x_{0}+\delta }^{b}e^{nf(x)}\,dx\\&\leq (b-a)e^{n(f(x_{0})-\eta )}+\int _{x_{0}-\delta }^{x_{0}+\delta }e^{nf(x)}\,dx\\&\leq (b-a)e^{n(f(x_{0})-\eta )}+e^{nf(x_{0})}\int _{x_{0}-\delta }^{x_{0}+\delta }e^{{\frac {n}{2}}(f''(x_{0})+\varepsilon )(x-x_{0})^{2}}\,dx\\&\leq (b-a)e^{n(f(x_{0})-\eta )}+e^{nf(x_{0})}\int _{-\infty }^{+\infty }e^{{\frac {n}{2}}(f''(x_{0})+\varepsilon )(x-x_{0})^{2}}\,dx\\&\leq (b-a)e^{n(f(x_{0})-\eta )}+e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0})-\varepsilon )}}}\end{aligned}}}$

If we divide both sides of the above inequality by

${\displaystyle e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}}$

and take the limit we get:

${\displaystyle \lim _{n\to \infty }{\frac {\int _{a}^{b}e^{nf(x)}\,dx}{e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}}}\leq \lim _{n\to \infty }(b-a)e^{-\eta n}{\sqrt {\frac {n(-f''(x_{0}))}{2\pi }}}+{\sqrt {\frac {-f''(x_{0})}{-f''(x_{0})-\varepsilon }}}={\sqrt {\frac {-f''(x_{0})}{-f''(x_{0})-\varepsilon }}}}$

Since ${\displaystyle \varepsilon }$ is arbitrary we get the upper bound:

${\displaystyle \lim _{n\to \infty }{\frac {\int _{a}^{b}e^{nf(x)}\,dx}{e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}}}\leq 1}$

And combining this with the lower bound gives the result.

Note that the above proof obviously fails when ${\displaystyle a=-\infty }$ or ${\displaystyle b=\infty }$ (or both). To deal with these cases, we need some extra assumptions. A sufficient (not necessary) assumption is that for ${\displaystyle n=1,}$

${\displaystyle \int _{a}^{b}e^{nf(x)}\,dx<\infty ,}$

and that the number ${\displaystyle \eta }$ as above exists (note that this must be an assumption in the case when the interval ${\displaystyle [a,b]}$ is infinite). The proof proceeds otherwise as above, but with a slightly different approximation of integrals:

${\displaystyle \int _{a}^{x_{0}-\delta }e^{nf(x)}\,dx+\int _{x_{0}+\delta }^{b}e^{nf(x)}\,dx\leq \int _{a}^{b}e^{f(x)}e^{(n-1)(f(x_{0})-\eta )}\,dx=e^{(n-1)(f(x_{0})-\eta )}\int _{a}^{b}e^{f(x)}\,dx.}$

When we divide by

${\displaystyle e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}},}$

we get for this term

${\displaystyle {\frac {e^{(n-1)(f(x_{0})-\eta )}\int _{a}^{b}e^{f(x)}\,dx}{e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}}}=e^{-(n-1)\eta }{\sqrt {n}}e^{-f(x_{0})}\int _{a}^{b}e^{f(x)}\,dx{\sqrt {\frac {-f''(x_{0})}{2\pi }}}}$

whose limit as ${\displaystyle n\to \infty }$ is ${\displaystyle 0}$. The rest of the proof (the analysis of the interesting term) proceeds as above.

The given condition in the infinite interval case is, as said above, sufficient but not necessary. However, the condition is fulfilled in many, if not in most, applications: the condition simply says that the integral we are studying must be well-defined (not infinite) and that the maximum of the function at ${\displaystyle x_{0}}$ must be a "true" maximum (the number ${\displaystyle \eta >0}$ must exist). There is no need to demand that the integral is finite for ${\displaystyle n=1}$ but it is enough to demand that the integral is finite for some ${\displaystyle n=N.}$

1. Relative error

The “approximation” in this method is related to the relative error and not the absolute error. Therefore, if we set

${\displaystyle s={\sqrt {\frac {2\pi }{M\left|f''(x_{0})\right|}}},}$

the integral can be written as

${\displaystyle {\begin{aligned}\int _{a}^{b}e^{Mf(x)}\,dx&=se^{Mf(x_{0})}{\frac {1}{s}}\int _{a}^{b}e^{M(f(x)-f(x_{0}))}\,dx\\&=se^{Mf(x_{0})}\int _{\frac {a-x_{0}}{s}}^{\frac {b-x_{0}}{s}}e^{M(f(sy+x_{0})-f(x_{0}))}\,dy\end{aligned}}}$

where ${\displaystyle s}$ is a small number when ${\displaystyle M}$ is a large number obviously and the relative error will be

${\displaystyle \left|\int _{\frac {a-x_{0}}{s}}^{\frac {b-x_{0}}{s}}e^{M(f(sy+x_{0})-f(x_{0}))}dy-1\right|.}$

Now, let us separate this integral into two parts: ${\displaystyle y\in [-D_{y},D_{y}]}$ region and the rest.

2. ${\displaystyle e^{M(f(sy+x_{0})-f(x_{0}))}\to e^{-\pi y^{2}}}$ around the stationary point when ${\displaystyle M}$ is large enough

Let’s look at the Taylor expansion of ${\displaystyle M(f(x)-f(x_{0}))}$ around x₀ and translate x to y because we do the comparison in y-space, we will get

${\displaystyle M(f(x)-f(x_{0}))={\frac {Mf''(x_{0})}{2}}s^{2}y^{2}+{\frac {Mf'''(x_{0})}{6}}s^{3}y^{3}+\cdots =-\pi y^{2}+O\left({\frac {1}{\sqrt {M}}}\right).}$

Note that ${\displaystyle f'(x_{0})=0}$ because ${\displaystyle x_{0}}$ is a stationary point. From this equation you will find that the terms higher than second derivative in this Taylor expansion is suppressed as the order of ${\displaystyle {\tfrac {1}{\sqrt {M}}}}$ so that ${\displaystyle \exp(M(f(x)-f(x_{0})))}$ will get closer to the Gaussian function as shown in figure. Besides,

${\displaystyle \int _{-\infty }^{\infty }e^{-\pi y^{2}}dy=1.}$

3. The larger ${\displaystyle M}$ is, the smaller range of ${\displaystyle x}$ is related

Because we do the comparison in y-space, ${\displaystyle y}$ is fixed in ${\displaystyle y\in [-D_{y},D_{y}]}$ which will cause ${\displaystyle x\in [-sD_{y},sD_{y}]}$; however, ${\displaystyle s}$ is inversely proportional to ${\displaystyle {\sqrt {M}}}$, the chosen region of ${\displaystyle x}$ will be smaller when ${\displaystyle M}$ is increased.

4. If the integral in Laplace's method converges, the contribution of the region which is not around the stationary point of the integration of its relative error will tend to zero as ${\displaystyle M}$ grows.

Relying on the 3rd concept, even if we choose a very large D_y, sD_y will finally be a very small number when ${\displaystyle M}$ is increased to a huge number. Then, how can we guarantee the integral of the rest will tend to 0 when ${\displaystyle M}$ is large enough?

The basic idea is to find a function ${\displaystyle m(x)}$ such that ${\displaystyle m(x)\geq f(x)}$ and the integral of ${\displaystyle e^{Mm(x)}}$ will tend to zero when ${\displaystyle M}$ grows. Because the exponential function of ${\displaystyle Mm(x)}$ will be always larger than zero as long as ${\displaystyle m(x)}$ is a real number, and this exponential function is proportional to ${\displaystyle m(x),}$ the integral of ${\displaystyle e^{Mf(x)}}$ will tend to zero. For simplicity, choose ${\displaystyle m(x)}$ as a tangent through the point ${\displaystyle x=sD_{y}}$ as shown in the figure:

If the interval of the integration of this method is finite, we will find that no matter ${\displaystyle f(x)}$ is continue in the rest region, it will be always smaller than ${\displaystyle m(x)}$ shown above when ${\displaystyle M}$ is large enough. By the way, it will be proved later that the integral of ${\displaystyle e^{Mm(x)}}$ will tend to zero when ${\displaystyle M}$ is large enough.

If the interval of the integration of this method is infinite, ${\displaystyle m(x)}$ and ${\displaystyle f(x)}$ might always cross to each other. If so, we cannot guarantee that the integral of ${\displaystyle e^{Mf(x)}}$ will tend to zero finally. For example, in the case of ${\displaystyle f(x)={\tfrac {\sin(x)}{x}},}$ ${\displaystyle \int _{0}^{\infty }e^{Mf(x)}dx}$ will always diverge. Therefore, we need to require that ${\displaystyle \int _{d}^{\infty }e^{Mf(x)}dx}$ can converge for the infinite interval case. If so, this integral will tend to zero when ${\displaystyle d}$ is large enough and we can choose this ${\displaystyle d}$ as the cross of ${\displaystyle m(x)}$ and ${\displaystyle f(x).}$

You might ask why not choose ${\displaystyle \int _{d}^{\infty }e^{f(x)}dx}$ as a convergent integral? Let me use an example to show you the reason. Suppose the rest part of ${\displaystyle f(x)}$ is ${\displaystyle -\ln x,}$ then ${\displaystyle e^{f(x)}={\tfrac {1}{x}}}$ and its integral will diverge; however, when ${\displaystyle M=2,}$ the integral of ${\displaystyle e^{Mf(x)}={\tfrac {1}{x^{2}}}}$ converges. So, the integral of some functions will diverge when ${\displaystyle M}$ is not a large number, but they will converge when ${\displaystyle M}$ is large enough.

First, use ${\displaystyle x_{0}=0}$ to denote the global maximum, which will simplify this derivation. We are interested in the relative error, written as ${\displaystyle |R|}$,

${\displaystyle \int _{a}^{b}h(x)e^{Mg(x)}\,dx=h(0)e^{Mg(0)}s\underbrace {\int _{a/s}^{b/s}{\frac {h(sy)}{h(0)}}e^{M\left[g(sy)-g(0)\right]}dy} _{1+R},}$

where

${\displaystyle s\equiv {\sqrt {\frac {2\pi }{M\left|g''(0)\right|}}}.}$

So, if we let

${\displaystyle A\equiv {\frac {h(sy)}{h(0)}}e^{M\left[g(sy)-g(0)\right]}}$

and ${\displaystyle A_{0}\equiv e^{-\pi y^{2}}}$, we can get

${\displaystyle \left|R\right|=\left|\int _{a/s}^{b/s}A\,dy-\int _{-\infty }^{\infty }A_{0}\,dy\right|}$

since ${\displaystyle \int _{-\infty }^{\infty }A_{0}\,dy=1}$.

For the upper bound, note that ${\displaystyle |A+B|\leq |A|+|B|,}$ thus we can separate this integration into 5 parts with 3 different types (a), (b) and (c), respectively. Therefore,

${\displaystyle |R|<\underbrace {\left|\int _{D_{y}}^{\infty }A_{0}dy\right|} _{(a_{1})}+\underbrace {\left|\int _{D_{y}}^{b/s}Ady\right|} _{(b_{1})}+\underbrace {\left|\int _{-D_{y}}^{D_{y}}\left(A-A_{0}\right)dy\right|} _{(c)}+\underbrace {\left|\int _{a/s}^{-D_{y}}Ady\right|} _{(b_{2})}+\underbrace {\left|\int _{-\infty }^{-D_{y}}A_{0}dy\right|} _{(a_{2})}}$

where ${\displaystyle (a_{1})}$ and ${\displaystyle (a_{2})}$ are similar, let us just calculate ${\displaystyle (a_{1})}$ and ${\displaystyle (b_{1})}$ and ${\displaystyle (b_{2})}$ are similar, too, I’ll just calculate ${\displaystyle (b_{1})}$.

For ${\displaystyle (a_{1})}$, after the translation of ${\displaystyle z\equiv \pi y^{2}}$, we can get

${\displaystyle (a_{1})=\left|{\frac {1}{2{\sqrt {\pi }}}}\int _{\pi D_{y}^{2}}^{\infty }e^{-z}z^{-1/2}dz\right|<{\frac {e^{-\pi D_{y}^{2}}}{2\pi D_{y}}}.}$

This means that as long as ${\displaystyle D_{y}}$ is large enough, it will tend to zero.

For ${\displaystyle (b_{1})}$, we can get

${\displaystyle (b_{1})\leq \left|\int _{D_{y}}^{b/s}\left[{\frac {h(sy)}{h(0)}}\right]_{\text{max}}e^{Mm(sy)}dy\right|}$

where

${\displaystyle m(x)\geq g(x)-g(0){\text{as}}x\in [sD_{y},b]}$

and ${\displaystyle h(x)}$ should have the same sign of ${\displaystyle h(0)}$ during this region. Let us choose ${\displaystyle m(x)}$ as the tangent across the point at ${\displaystyle x=sD_{y}}$ , i.e. ${\displaystyle m(sy)=g(sD_{y})-g(0)+g'(sD_{y})\left(sy-sD_{y}\right)}$ which is shown in the figure

From this figure you can find that when ${\displaystyle s}$ or ${\displaystyle D_{y}}$ gets smaller, the region satisfies the above inequality will get larger. Therefore, if we want to find a suitable ${\displaystyle m(x)}$ to cover the whole ${\displaystyle f(x)}$ during the interval of ${\displaystyle (b_{1})}$, ${\displaystyle D_{y}}$ will have an upper limit. Besides, because the integration of ${\displaystyle e^{-\alpha x}}$ is simple, let me use it to estimate the relative error contributed by this ${\displaystyle (b_{1})}$.

Based on Taylor expansion, we can get

${\displaystyle {\begin{aligned}M\left[g(sD_{y})-g(0)\right]&=M\left[{\frac {g''(0)}{2}}s^{2}D_{y}^{2}+{\frac {g'''(\xi )}{6}}s^{3}D_{y}^{3}\right]&&{\text{as }}\xi \in [0,sD_{y}]\\&=-\pi D_{y}^{2}+{\frac {(2\pi )^{3/2}g'''(\xi )D_{y}^{3}}{6{\sqrt {M}}|g''(0)|^{\frac {3}{2}}}},\end{aligned}}}$

and

${\displaystyle {\begin{aligned}Msg'(sD_{y})&=Ms\left(g''(0)sD_{y}+{\frac {g'''(\zeta )}{2}}s^{2}D_{y}^{2}\right)&&{\text{as }}\zeta \in [0,sD_{y}]\\&=-2\pi D_{y}+{\sqrt {\frac {2}{M}}}\left({\frac {\pi }{|g''(0)|}}\right)^{\frac {3}{2}}g'''(\zeta )D_{y}^{2},\end{aligned}}}$

and then substitute them back into the calculation of ${\displaystyle (b_{1})}$; however, you can find that the remainders of these two expansions are both inversely proportional to the square root of ${\displaystyle M}$, let me drop them out to beautify the calculation. Keeping them is better, but it will make the formula uglier.

${\displaystyle {\begin{aligned}(b_{1})&\leq \left|\left[{\frac {h(sy)}{h(0)}}\right]_{\max }e^{-\pi D_{y}^{2}}\int _{0}^{b/s-D_{y}}e^{-2\pi D_{y}y}dy\right|\\&\leq \left|\left[{\frac {h(sy)}{h(0)}}\right]_{\max }e^{-\pi D_{y}^{2}}{\frac {1}{2\pi D_{y}}}\right|.\end{aligned}}}$

Therefore, it will tend to zero when ${\displaystyle D_{y}}$ gets larger, but don't forget that the upper bound of ${\displaystyle D_{y}}$ should be considered during this calculation.

About the integration near ${\displaystyle x=0}$, we can also use Taylor's Theorem to calculate it. When ${\displaystyle h'(0)\neq 0}$

${\displaystyle {\begin{aligned}(c)&\leq \int _{-D_{y}}^{D_{y}}e^{-\pi y^{2}}\left|{\frac {sh'(\xi )}{h(0)}}y\right|\,dy\\&<{\sqrt {\frac {2}{\pi M|g''(0)|}}}\left|{\frac {h'(\xi )}{h(0)}}\right|_{\max }\left(1-e^{-\pi D_{y}^{2}}\right)\end{aligned}}}$

and you can find that it is inversely proportional to the square root of ${\displaystyle M}$. In fact, ${\displaystyle (c)}$ will have the same behave when ${\displaystyle h(x)}$ is a constant.

Conclusively, the integral near the stationary point will get smaller as ${\displaystyle {\sqrt {M}}}$ gets larger, and the rest parts will tend to zero as long as ${\displaystyle D_{y}}$ is large enough; however, we need to remember that ${\displaystyle D_{y}}$ has an upper limit which is decided by whether the function ${\displaystyle m(x)}$ is always larger than ${\displaystyle g(x)-g(0)}$ in the rest region. However, as long as we can find one ${\displaystyle m(x)}$ satisfying this condition, the upper bound of ${\displaystyle D_{y}}$ can be chosen as directly proportional to ${\displaystyle {\sqrt {M}}}$ since ${\displaystyle m(x)}$ is a tangent across the point of ${\displaystyle g(x)-g(0)}$ at ${\displaystyle x=sD_{y}}$. So, the bigger ${\displaystyle M}$ is, the bigger ${\displaystyle D_{y}}$ can be.