I understand that if $M(N)=O(N^\sigma)$, then $\sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}$ and therefore $$ \frac{1}{s\zeta(s)} = \int_0^\infty M(x) x^{-(s+1)} dx $$ for $s>\sigma$, and that having $\sigma=1/2+\epsilon$ for every $\epsilon>0$ will thus prove the RH. The Wikipedia article on the Mertens Conjecture states that the reverse also holds, but I don’t understand the details of argument: Using the […]

Let $\mathfrak{M}\left(*\right)$ the Mellin transform. We know that holds this identity$$\mathfrak{M}\left(\underset{k\geq1}{\sum}\lambda_{k}g\left(\mu_{k}x\right),\, s\right)=\underset{k\geq1}{\sum}\frac{\lambda_{k}}{\mu_{k}^{s}}\mathfrak{M}\left(g\left(x\right),s\right)$$ so for example if we want a closed form of$$\underset{k\geq1}{\sum}\frac{\sin\left(nx\right)}{n^{3}}=x^{3}\underset{k\geq1}{\sum}\frac{\sin\left(nx\right)}{n^{3}x^{3}}\,\,(1)$$ we are in the case $$\lambda_{k}=1,\thinspace\mu_{k}=k,\, g\left(x\right)=\frac{\sin\left(x\right)}{x^{3}}$$ and so our Mellin transform is $$\zeta\left(s\right)\mathfrak{M}\left(g\left(x\right),s\right)$$ and then we find the sum of $(1)$ by the residue theorem. I tried to use this argument for […]

Introduction. I computed two Mellin transforms while browsing / working on the problem at this MSE link. No solution was found, but some interesting auxiliary results appeared. I am writing to ask for independent confirmation of these results, not necessarily using Mellin transforms. Problem statement. Introduce $$S(x) = \sum_{k\ge 1} \frac{1}{(2k-1)} \frac{1}{\sinh((2k-1)x)} \quad\text{and}\quad T(x) = […]

Give a function \begin{align} f(s)= 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}, \end{align} can we find and inverse Mellin transform for $f(s)$ that is \begin{align} \frac{1}{2 \pi i}\int_{- i\infty}^{ i\infty} 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)} x^{-s-1 } ds \end{align} for $x>0$. I was wondering if the integral […]

Inspired by this answer, I’m trying to show that $$\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1} = \frac{1}{24} – \frac{1}{8 \pi}$$ using the inverse Mellin transform. But the answer I get is twice as much as it should be. EDIT: Thanks to Marko Riedel, the evaluation below is now correct. Since $$ \left\{ \mathcal{M} \ \frac{x}{e^{2\pi x}-1} \right\}(s) […]

With $z=\sigma + x \, i;\,\, \sigma,x \in \mathbb{R}$, numerical evidence strongly suggests that: $$\displaystyle \int_{0}^{\infty} \,\zeta(z)\,\zeta(\overline{z})\,\Gamma(z)\,\Gamma(\overline{z}) \,dx =\pi\,\Gamma(2\,\sigma)\,\big(\zeta(2\,\sigma-1)-\zeta(2\,\sigma)\big)$$ for all $\sigma > 1$. Could this be proven? Just to share that the equation can be extended towards $0<\sigma<1$ by starting from: $$\Gamma(s)\, \zeta(s) = \int_0^\infty x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}\right)\,dx$$ and following the same logic as in the […]

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