Sine transform of $f(x) = \dfrac{1}{e^{\sqrt{2 \pi} x}-1} - \dfrac{1}{\sqrt{2 \pi} x}$

Question

Quoted from Titchmarsh's book The Theory of the Riemann Zeta-Function:

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Now it is known that the function $$f(x) = \dfrac{1}{e^{\sqrt{2 \pi} x}-1} - \dfrac{1}{\sqrt{2 \pi} x}$$ is self-reciprocal for sine transform, i.e. that $$f(x) = \sqrt{\dfrac{2}{\pi}} \int_0^{\infty} f(y) \sin (xy) dy.$$

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I attempted to prove this but I failed: First I studied the subject from lecture notes available online which included only simple examples like $e^{-\pi x^2}$ but the methods are not practical for this $f(x)$. Then I searched some books to learn more but none helps neither. I asked the question in many different forms on Wolfram Alpha to get a hint but neither full solution nor a single hint I could get from the site.

How the claim of the book holds? A simple clear explanation would be much appreciated.

Answer 1

Trying to perform the integration directly would've been most prudent, at least to me, since one in particular is a famous known integral.

Note that you have to calculate $$ \sqrt{ \frac 2 \pi } \int_0^\infty \frac{1}{e^{\sqrt{2\pi} y} - 1} \sin(xy) \, \mathrm{d}y - \frac 1 \pi \int_0^\infty \frac{1}{y} \sin(xy) \, \mathrm{d}y $$ The latter is easy to calculate with the obvious substitution and knowing that $$ \int_0^\infty \frac{\sin z}{z} \, \mathrm{d} z = \frac \pi 2 $$ as proved, e.g., here.

The former integral seems more difficult, but most likely can be found via the same ideas as the simpler integral, $$ \int_0^\infty \frac{\sin(z)}{e^z - 1} \, \mathrm{d}z = \frac \pi 2 \coth(\pi) - \frac 1 2 $$ as discussed here.