Articles of mellin transform

How to simplify this special case of Meijer G-Function?

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 […]

How do you prove that $M(N)=O(N^{1/2+\epsilon})$ from the Riemann Hypothesis?

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 […]

About Mellin transform and harmonic series

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 […]

A functional equation relating two harmonic sums.

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) = […]

Inverse Mellin transform of $f(s)= 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}$

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 […]

Evaluating $\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1}$ using the inverse Mellin transform

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) […]

How can this integral expression for the difference between two $\zeta(s)$s be explained?

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 […]