# Mellin transforms with zeros on the critical line

Are there examples of Mellin transforms of functions $\int_{0}^{\infty}f(x)x^{s-1}\mathrm dx$ that have only real zeroes or have only zeroes on the critical line?

For example, the Mellin transform of the Hermite polynomials has zeroes on the critical line $\mathrm{Re}(s)=1/2$. Can this fact be extended to the Mellin transform of any function $f(x)$ which is its own Fourier transform ? $F(f(x))=kf(x)$

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## 1 Answer

The 'modified' Bessel function, or MacDonald's function, is more or less a Mellin transform of the order $s$: $$K_s(x)=\frac12(\frac x2)^s\int_0^\infty\exp(-t-(x^2/4t))t^{-s}\frac{dt}{t}$$ once you swap $s$ for $-s$ and move the leading term to the other side. And $K_s(x)=K_{-s}(x)$, so $K_{it}(x)$ is real for real $x$. I believe all the zeros of $K_s$, as a function of $s$, are on the imaginary axis, but I can't find the reference. Polya did some work with this in connection to the Riemann Hypothesis.

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This mentions a few facts about the zeroes of the modified Bessel function of the second kind. This talks a bit about "imaginary order" modified Bessel functions. –  Ｊ. Ｍ. Oct 20 '11 at 22:30
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