Timeline for Is this Riemann zeta function product equal to the Fourier transform of the von Mangoldt function?

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Apr 13, 2017 at 12:19	history	edited	CommunityBot		replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Mar 6, 2016 at 5:18	comment	added	reuns		and a change of variable $x= e^y$ shows that the Mellin transform of $f(x)$ is the Fourier (bilateral Laplace) transform of $f(e^x)e^{-x}$
Mar 6, 2016 at 5:16	comment	added	reuns		when you write "the Fourier transform of $\Lambda(1) \ldots \Lambda(k)$" you mean the Fourier (Laplace) transform of the weighted logarithmically spaced Dirac comb $\sum_{n=1}^\infty \Lambda(n) \delta(x-\ln n)$ which is $\int_{-\infty}^\infty e^{-sx} \sum_{n=1}^\infty \Lambda(n) \delta(x-\ln n) dx = \sum_{n=1}^\infty \Lambda(n) n^{-s}$, which is also the Mellin transform of the weighted (but this time linearly spaced ) Dirac comb $\sum_{n=1}^\infty \Lambda(n) \delta(x-n)$ which is $\int_0^\infty x^{-s} \sum_{n=1}^\infty \Lambda(n) \delta(x-n) dx = \sum_{n=1}^\infty \Lambda(n) n^{-s}$.
Jan 24, 2016 at 14:48	history	edited	Mats Granvik	CC BY-SA 3.0	added 147 characters in body
S Jul 31, 2015 at 13:08	history	answered	Mats Granvik	CC BY-SA 3.0
S Jul 31, 2015 at 13:08	history	made wiki			Post Made Community Wiki by Mats Granvik