# 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.

• Are you sure that the domain of integration is $(0,\infty)$ and not all of $\mathbb{R}$? This is because (1) On $(0,\infty)$ you do not need absolute values, and (2) The results you are referring to are not true, and it is justified when the domain of integration is all of $\mathbb{R}$. Anyway, formally we have $$F(y) = \log\Gamma(1-iy) = -\frac{\pi}{2}\log|y| + \frac{1}{2}\log|y| + \frac{1}{2}\log(2\pi) + o(1)$$ as $y\to\infty$, although the constant part is not important in application. – Sangchul Lee May 2 '18 at 7:57
• Oh oops, I fixed that now! I definitely meant the range of integration $\mathbb{R}$! – Greg.Paul May 2 '18 at 15:01
• Is your result $F(y) = \log\Gamma(1-iy)$ using the range $(0,\infty)$? Also, how did you derive this result? – Greg.Paul May 2 '18 at 15:16
• Yes, so with the current version of your $F(y)$, the answer in my comment should be doubled. And this follows by computing $$ \, F''(y) = \text{''} \int_{-\infty}^{\infty} \frac{|x|}{e^{|x|}-1}e^{-iyx}\,dx$$ and then integrating twice. – Sangchul Lee May 2 '18 at 15:23
• @Sangchul_Lee I think when you doubled your range of integration you forgot a minus sign switch in an exponenential. Using your trick, I get the result $F(y) = \ln\Gamma(1-i y) + \ln\Gamma(1+i y)$. – Greg.Paul May 11 '18 at 11:49

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 $|x|^{-1}\rvert_{x = 2x}$.
• Thanks for your feedback - I definitely overlooked the fact that you can't multiply the distributions. Do you have any insight how to show more rigorously why the above function seems to evaluate precisely to $\ln\Gamma(1-iy)+\ln\Gamma(1+iy)$? Although the arguments of the functions are complex, overall this function is purely real. Can $\ln\Gamma(1-iy)+\ln\Gamma(1+iy)$ be understood as a distribution, and if so, is there a way to see why my Fourier transform evaluates to this function? I am mystified why Sangchul's trick worked. – Greg.Paul May 17 '18 at 6:18
• We still don't know how you define the thing that you're taking the Fourier transform of. Also, the transform of $|x| / (e^{|x|} - 1)$ is $1/y^2 - \pi^2 \operatorname{csch}^2 \pi y$, I have no idea where the gamma function comes from. Besides, even if we know $\mathcal F[x^2 f(x)]$, we can't determine the $\delta(x)$ and $\delta'(x)$ terms in $f$ and therefore the constant and linear terms in its transform; first you need to know that your regularization doesn't contain the $\delta$ and $\delta'$ terms. – Maxim May 17 '18 at 7:35