# Commutation of limit and integration

I'm a physics student and perhaps do not have an in depth exposure to analysis. In the following calculation $$\tilde{\Phi}(\Omega)= \lim_{\omega \rightarrow0} \int_{-\infty}^{\infty} d\tau\:e^{-i\Omega\tau} \text{cos}\big(\omega g^{-1}e^{-g \tau}\big)$$ it turns out that when I carry out the integral and evaluate $$|\tilde{\Phi}(\Omega)|^2$$, then take the limit $$\omega$$ going to zero, I get a finite result (The finite result is well verified, it is from a paper with whose author I have gone through the calculation). However, if I take the limit $$\omega$$ going to zero first then $$\tilde{\Phi}(\Omega)=\delta(\Omega)$$ and $$|\tilde{\Phi}(\Omega)|^2$$ is the Dirac delta function squared. Pertaining to the corresponding interpretation in terms of physics, the finite result is what is the 'correct' computation. However, I wanted to know if this is the correct way mathematically too, what exactly is the subtelty that has gone into this, and are there similar examples of the same? For instance I know that when I just have two limits, the order can be important. For instance if $$L = \lim_{ x,y \rightarrow{0}} \frac{f(x)}{g(y)}$$ where both $$f(x)$$ and $$g(y)$$ approach zero when $$x,y$$ approach zero then suppose I took the $$x$$ limit first. Then that would lead to $$L$$ being zero. But the commutation of a limit and an integral is something that I can't quite wrap my head around.

• This is in theory more related to infinite sums than to integrals, but you may be interested in checking en.wikipedia.org/wiki/Uniform_convergence .It is true, though, that not in all circumstances can limits and integrals, or order of limits can be swapped. Still, under some regularity conditions, you can get away with it – David Jul 16 '19 at 7:38