How to show that this function is differentiable? 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?
 A: 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 $s=0$.
