In the application of Laplace method (or steepest descent) it is often assumed that the dependence on the factor N, on which we are expanding the integral, is only in the argument of the exponential. What if we have an expression to expand for large $N$ of the type:

$\int_{-\infty}^{+\infty} f(N,t) e^{Ng(t) dt}$

where the function $f$ is only mildly depending on $N$. Would the method change? Are there examples of functions of this form?


The Laplace method can indeed handle some integrals like this, and broad strokes of the procedure are essentially the same.

First you would establish that the main contribution to the size of the integral comes from a neighborhood of some $t = t_0$ (often a critical point of $g$), where the size of the neighborhood potentially depends on $N$. Then you would show that, in this neighborhood, $f(N,t) \approx f(N,t_0)$ in some sense. The details of these two steps depend on the growth and decay properties of $f$ and $g$.

In my answer to this question I outlined these steps for the integral

$$ I(N) = \int_0^\infty f(N,t) e^{Ng(t)}\,dt $$


$$ f(N,t) = N^{-t} \qquad \text{and} \qquad g(t) = \log\!\left(\frac{t}{1+t^2}\right). $$

In this case the function $g$ has a critical point at $t=1$, and in the answer I end up showing that

$$ I(N) \approx f(N,1) \int_{1-\epsilon}^{1+\epsilon} e^{Ng(t)}\,dt. $$

If $f$ and $g$ are nice enough this will be the usual outcome of the method.

It should be noted, of course, that not all integrals of the form

$$ \int_0^\infty f(N,t) e^{Ng(t)}\,dt $$

can be approximated in this way. Take, for example, $f(N,t) = (t+1)^N$ and $g(t) = -t^2$. In this case $g$ has a critical point at $t=0$ but

$$ \int_0^\infty f(N,t) e^{Ng(t)}\,dt \not\approx f(N,0) \int_{0}^{\epsilon} e^{Ng(t)}\,dt. $$

Here the growth of $f(N,t)$ competes with the decay of $e^{Ng(t)}$ as $t \to \infty$, creating a critical point for the whole integrand which does not belong to $g$ alone.

  • $\begingroup$ Really useful thanks a lot, and very useful also the answer linked! $\endgroup$ – tirrel Feb 4 '16 at 11:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.