Consider an integral equation: $$\int_{-\infty}^{+\infty}\frac{e^{-\sigma y}f(y)}{e^{e^{x-y}}+1}dy=0$$, where $\sigma\in(\frac{1}{2},1)$

In https://arxiv.org/abs/2003.00581 there is written that this equation has no bounded solution other than trivial $f(y)=0$ iff Riemann hypothesis is true.

But here is the question:

How does the proof of this equivalence look like?

I would be grateful if someone gave me a proof of this, because i can't find it in the web.


  • 4
    $\begingroup$ Did you take a look at the original article $\endgroup$
    – Momo
    Jan 11, 2022 at 20:20

1 Answer 1


Nothing mysterious. With $u=e^y$ and $\Re(s) >0$ $$\Gamma(s)\eta(s)=\int_0^\infty \frac{u^{s-1}}{e^u+1}du=\int_{-\infty}^\infty \frac{e^{s y}}{e^{e^y}+1}dy$$

  • If $\eta(\sigma+it)=0$ then $$e^{\sigma x}\int_{-\infty}^\infty \frac{e^{-\sigma y}}{e^{e^{x-y}}+1}e^{ity}dy=\int_{-\infty}^\infty \frac{e^{\sigma y}}{e^{e^y}+1}e^{it(x-y)}dy=0$$

  • If $\eta(s)$ has no zero on $\Re(s)=\sigma$ and $f\in L^\infty$ doesn't "vanish almost everywhere" then the Fourier transform of $$g(x)=\int_{-\infty}^\infty \frac{e^{\sigma (x-y)}}{e^{e^{x-y}}+1}f(y)dy$$ is $$\hat{g}(t)=\Gamma(\sigma-it)\eta(\sigma-it) \hat{f}(t)$$

    (where $\hat{f}$ is the Fourier transform in the sense of distributions)

    which is not the zero distribution so $g$ is not identically $0$.


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