I have the following expression, involving the Meijer G-Function: $$\frac{1}{\sqrt{a} (2 \pi)^{(a-1)/2}} G_{1,a+1}^{a+1,1}\left( \frac{c^a}{a^a} \left|\begin{matrix}0\\ 0,0,\frac{1}{a},\frac{2}{a},\dots,\frac{a-1}{a}\end{matrix}\right.\right)$$ Here $a \in \mathbb{N},~c>0$. I believe this can be simplified, using the misterious ‘duplication formula’ from this paper, top of page 3. It’s really too long to type so I’ll reproduce the relevant parts of the paper on this […]

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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