Simple counterexample on integral asymptotics Let
$$
I_1(N)=\int_{-\infty}^\infty dx\ f(x)g_N(x)
$$
and
$$
I_2(N)=\int_{-\infty}^\infty dx\ f(x)h_N(x)
$$
be two well-defined and finite integrals depending on a parameter $N$. Assume that $g_N(x)\sim h_N(x)$ for large $N$, meaning that for all $x\in\mathbb{R}$, $\lim_{N\to\infty}g_N(x)/h_N(x)=1$.
Can you exhibit the simplest counterexample to show that the statement
$$
I_1(N)\sim I_2(N)
$$
for large $N$ may be untrue [that is to say, one cannot naively replace (part of) the integrand with its large-$N$ asymptotics to obtain the large-$N$ asymptotics of the full integral]?
 A: This can be done using sequences $g_N, h_N$ that converge pointwise to the same function, but not uniformly in $x$. Here are a few examples.


*

*A simple way is to consider a constant sequence of Gaussian functions,
and a sequence of "moving" Gaussians,
$$f(x)=1$$ $$g_N(x)=\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\frac{x^2}{2}}$$ $$h_N(x)=\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\frac{x^2}{2}}+\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\frac{(x-N)^2}{2}}$$
which leads to $I_1=1$, $I_2=2$ ($N$-independent).

*Another similar example, which technically does not satisfy the
convergence hypothesis in $x=0$, but could be interesting, has to do
with the Dirac measure: take $f$ and $g_N$ same as above, and 
$$h_N(x)=\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\frac{x^2}{2}}+\sqrt{\frac{N}{2\pi}}\mathrm{e}^{-N\frac{x^2}{2}}.$$ Again, $I_1=1$, $I_2=2$, while
$\lim_{N\to\infty}g_N(x)/h_N(x)=1\;\forall x\neq0$.

*Finally, consider
$$f(x)=\theta(x),$$
$$g_N(x)=\mathrm{e}^{-N x},$$
$$h_N(x)=\log\left(1+\mathrm{e}^{-N x}\right).$$
Then $I_1=1/N$, $I_2=\pi^2/(12 N)$. Naively (wrongly) using the asymptotics for $h_N$ in the integral gives you the right scaling, but with the wrong coefficient. Note again that this does not satisfy your hypothesis at $x=0$.
