Vtyjkyuk

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 $xinmathbbR$, $lim_Ntoinftyg_N(x)/h_N(x)=1$.

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]?

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.

1. 
   

   A simple way is to consider a constant sequence of Gaussian functions,
   and a sequence of "moving" Gaussians,

   
   
   

   $$f(x)=1$$ $$g_N(x)=frac1sqrt2pimathrme^-fracx^22$$ $$h_N(x)=frac1sqrt2pimathrme^-fracx^22+frac1sqrt2pimathrme^-frac(x-N)^22$$
   which leads to $I_1=1$, $I_2=2$ ($N$-independent).

   
   


2. 
   

   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

   
   
   

   $$h_N(x)=frac1sqrt2pimathrme^-fracx^22+sqrtfracN2pimathrme^-Nfracx^22.$$ Again, $I_1=1$, $I_2=2$, while
   $lim_Ntoinftyg_N(x)/h_N(x)=1;forall xneq0$.

   
   


3. 
   

   Finally, consider
   $$f(x)=theta(x),$$
   $$g_N(x)=mathrme^-N x,$$
   $$h_N(x)=logleft(1+mathrme^-N xright).$$

   
   
   

   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 $x=0$.