# Proof of Salem's reformulation of Riemann hypothesis.

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.

Regards

• Did you take a look at the original article
– Momo
Jan 11, 2022 at 20:20

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 $$0$$.