Asymptotics for the Fourier transform of $\frac{1}{|x|} \frac{1}{e^{|x|} -1}$ I am looking at the following (generalized) Fourier transform:
$$
F(y) = \int_{-\infty}^{\infty} dx\ \frac{1}{|x|} \frac{1}{e^{|x|} -1 } e^{- i y x} 
$$
I am unable to evaluate the above Fourier transform. I'm interested in obtaining the asymptotics of the above in the limit $|y| \to \infty$.
As I understand it, the behaviour of $F(y)$ as $|y| \to \infty$ is determined by the behaviour of the integrand as $|x| \to 0$. I know that the integrand has the following behaviour for $|x| \to 0$ (which also happens to be the location of the integrand's only singularity):
$$
\frac{1}{|x|} \frac{1}{e^{|x|} -1 } = \frac{1}{x^2} - \frac{1}{2|x|} + \frac{1}{12} + \mathscr{O}(x)
$$
From Lighthill's "An Introduction to Fourier Analysis and Generalised Functions", I have the following:
$$
\int_{-\infty}^{\infty} dx\ \frac{e^{- i y x}}{x^2} \ = \ - \pi |t| \\
\int_{-\infty}^{\infty} dx\ \frac{e^{- i y x}}{|x|} \ = \ - 2 \log|y| - 2 \gamma \\
$$
Is it so simple to then state the following in the limit $y \to \infty$?
$$
F(y) = - \pi |y| + \log|y| + \gamma + \mathscr{O}\left(\tfrac{1}{y} \right)
$$
My questions are:
(1) are these actually the asymptotics for $F$?
(2) If so, what else do I need to check that the above are the correct aysmptotics? From Lighthill's book I believe I need to check that the $N^{\mathrm{th}}$ derivatives of the integrand are bounded for some $N$, but I am unable to parse exactly what I need to check here. 
 A: I think the idea of the method is sound. First you need to have a definition of $|x|^{-k}$. Then you have to define what is meant by $|x|^{-1} (e^{|x|} - 1)^{-1}$, because distributions cannot be multiplied (even regular ones, and those are singular). What you can do is subtract the singular part and write the distribution as
$$|x|^{-1} (e^{|x|} - 1)^{-1} = \\
|x|^{-2} - \frac 1 2 |x|^{-1} +
 \left( |x|^{-1} (e^{|x|} - 1)^{-1} - |x|^{-2} + \frac 1 2 |x|^{-1} \right),$$
where the last part is a regular distribution $-$ provided that this is indeed the distribution you're working with, as, for instance, $|x|^{-1}/2$ is not the same as $|x|^{-1}\rvert_{x = 2x}$.
Then, for the regular part, you can analyze the Fourier integral of the corresponding ordinary function. For the two singular terms, the Fourier transforms are regular distributions, thus it makes perfect sense to talk about the asymptotic of the ordinary functions that induce them. The constant term that you found is correct as well.
