How to show that this function is differentiable?

Question

Let $$\phi: \mathbb{R} \rightarrow \mathbb{\mathbb{C}}, s \mapsto \int_2^{\infty} \frac{e^{isx}}{x^2\ln(x)}dx$$, I want to show that this function is differentiable everywhere. Unfortunately, it appears to me that it is impossible to use the standard dominated convergence theorem to show the differentiability, so I wanted to ask you for stronger methods?

Answer 1

We expect that $$ \frac{d}{ds} \int_2^{\infty} \frac{e^{isx}}{x^2\ln(x)}\,dx = i \int_2^{\infty} \frac{e^{isx}}{x\ln(x)}\,dx \tag{1} $$ but justification is problematic because the integral on the right does not converge absolutely (i.e., it does not converge as a Lebesgue integral).

Here's an idea: integrate by parts first. $$ \int_2^{\infty} \frac{e^{isx}}{x^2\ln(x)}\,dx = \frac{e^{2is}}{8 is \ln(2)} + \frac{1}{is} \int_2^{\infty} \frac{e^{isx} (2\ln x+1)}{x^3\ln^2(x)}\,dx \tag{2} $$ Now, formal differentiation with respect to $s$ produces $$ \frac{d}{ds} \int_2^{\infty} \frac{e^{isx} (2\ln x+1)}{x^3\ln^2(x)}\,dx = i \int_2^{\infty} \frac{e^{isx} (2\ln x+1)}{x^2\ln^2(x)}\,dx \tag{3} $$ which is absolutely convergent. So, you can use Fubini's theorem to show that integration of the right hand side in (3) with respect to $s$ indeed brings you back to the function you differentiated.

The above only shows differentiability at $s\ne 0$. Then you should consider the limit of the derivative as $s\to 0$. If it exists, mean value theorem completes the job. If it does not exist (I somehow suspect the imaginary part will blow up), then the function is not differentiable at $s=0$.