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.

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    $\begingroup$ 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. $\endgroup$ – Sangchul Lee May 2 '18 at 7:57
  • $\begingroup$ Oh oops, I fixed that now! I definitely meant the range of integration $\mathbb{R}$! $\endgroup$ – Greg.Paul May 2 '18 at 15:01
  • $\begingroup$ Is your result $F(y) = \log\Gamma(1-iy)$ using the range $(0,\infty)$? Also, how did you derive this result? $\endgroup$ – Greg.Paul May 2 '18 at 15:16
  • $\begingroup$ 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. $\endgroup$ – Sangchul Lee May 2 '18 at 15:23
  • $\begingroup$ @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 $F(y) = \ln\Gamma(1-i y) + \ln\Gamma(1+i y)$. $\endgroup$ – 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 $|x|^{-1}\rvert_{x = 2x}$.

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.

  • $\begingroup$ 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. $\endgroup$ – Greg.Paul May 17 '18 at 6:18
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    $\begingroup$ 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. $\endgroup$ – Maxim May 17 '18 at 7:35

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