I was asked recently to show that $F(t) := \int_0^{\infty} \cos(tx)e^{-x^2/2}dx$ is differentiable. This is for a class on Lebesgue integrals. We also covered Lebesgue's FTOC (not sure if that matters).
Here is my attempt at a solution: We have to compute the limit: $$ \lim_{h\to 0} \frac{F(t+h) - F(t)}{h} $$ and see if it exists. Substituting in the formula for $F$, we get: $$ \lim_{h\to 0} \int_0^{\infty} e^{-x^2/2}\frac{\cos((t+h)x)-\cos(tx)}{h}dx $$ Take an arbitrary sequence $(h_n)_1^{\infty}$ converging to 0. We get that the above is equal to: $$ \lim_{n\to\infty} \int_0^{\infty} e^{-x^2/2}\frac{\cos((t+h_n)x) - \cos(tx)}{h_n}dx $$ Note $\cos(tx)e^{-x^2/2}\chi_{[0,n]}\to \cos(tx)e^{-x^2/2}$ and $|\cos(tx)e^{-x^2/2}\chi_{[0,n]}|\leq e^{-x^2/2}$ which is certainly integrable on $[0,\infty)$. Thus, we use DCT to get: $$ \int_0^{\infty} e^{-x^2/2}(\lim_{n\to\infty} \frac{\cos((t+h_n)x) - \cos(tx)}{h_n})dx $$ which is equivalent to saying: $$ \int_0^{\infty} e^{-x^2/2}(\lim_{h\to 0} \frac{\cos((t+h)x) - \cos(tx)}{h})dx $$ We know the derivative of $\cos(tx)$ taken with respect to $x$ is simply $(-t)\sin(tx)$. Thus, we get: $$ F'(t) = \lim_{h\to 0} \frac{F(t+h) - F(t)}{h} = -\int_0^{\infty} t\sin(tx)e^{-x^2/2}dx $$ Thus $F'(t)$ existing comes down to showing that $t\sin(tx)e^{-x^2/2}$ is Lebesgue integrable on $[0,\infty)$. This is true so $F'(t)$ exists. Thus $F$ is differentiable.
I used some assumptions here that I am not sure work out. For example, I presume since I kept $h_n$ to be an arbitrary sequence covering to 0, I can swap $\lim_{h\to 0}$ with $\lim_{n\to \infty}$. Any criticism is welcome. Thank you in advance.