I am interested in studying the limit as \(n\) tends to infinity of \[\frac 1 {\sqrt n} \int_{-n\pi}^{+n \pi} e^{n (1-d)(\cos s -1)} \mathrm{d} s.\] In the following R script I will pick values of \(d\) and \(n\) and plot the integrand between the upper and lower limit of the range of integration.

d <- 0.05
n <- 10
s <- seq(from = - n * pi, to = + n * pi, length = 100000)
plot(s, exp(n * (1 - d) * (cos(s) - 1)) / sqrt(n), type = "l")

Now I will approximate the integral in the obvious way.

Nstep <- 10^6
s <- seq(from = - n * pi, to = + n * pi, length = Nstep)
sum((2 * n * pi / Nstep) * exp(n * (1 - d) * (cos(s) - 1)) / sqrt(n))
## [1] 2.607872

We see what we should have thought of before: the integrand is periodic, with period \(2 \pi\). So there are just \(n\) pieces each contributing the same value to the integral. Now I plot just one of those pieces, taking the value of \(n\) a factor \(10\) larger:

d <- 0.05
n <- 100
s <- seq(from = - pi, to = + pi, length = 100000)
plot(s, exp(n * (1 - d) * (cos(s) - 1)) / sqrt(n), type = "l")

I also compute the integral more accurately, as \(n\) times the contribution from one piece:

Nstep <- 10^6
s <- seq(from = - pi, to = + pi, length = Nstep)
sum((2 * pi / Nstep) * exp(n * (1 - d) * (cos(s) - 1)) * sqrt(n))
## [1] 2.575148

Now I’ll increase \(n\) by a factor \(10\). First, the plot

n <- 1000
s <- seq(from = - pi, to = + pi, length = 100000)
plot(s, exp(n * (1 - d) * (cos(s) - 1)) / sqrt(n), type = "l")

Then the integral:

Nstep <- 10^6
s <- seq(from = - pi, to = + pi, length = Nstep)
sum((2 * pi / Nstep) * exp(n * (1 - d) * (cos(s) - 1)) * sqrt(n))
## [1] 2.572082

This should give us ideas for a good strategy for doing the analysis.

First of all, use the periodicity.

Secondly, rescale the argument \(t\) so that we get an integrand which converges pointwise.