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To expand on Lucia's comment, we note that we can assume without loss of generality that each coefficient of $h$ is bounded by $1$, in which case $S(x)$ is largest when $h(p^k) = 1$ for all $k \geq 2$ …

We first observe that as $\frac{\mu(p^k)}{\sigma(p^k)}$ is equal to $1$ if $k = 0$, $-\frac{1}{p + 1}$ if $k = 1$, and $0$ otherwise, its Dirichlet series is \[\sum_{n = 1}^{\infty} \frac{\mu(n)}{\sig …

This is an interesting question. I don't think anyone has worked out what the distribution of the error term $$\frac{E(x)}{x} = \frac{1}{x}\left(\sum_{n \leq x}{\phi(n)} - \frac{3x^2}{\pi^2}\right)$$ …

Expanding on Frank's answer: by partial summation, we have that $$\sum_{p \leq x} \frac{(\log p)^k}{p} = \frac{(\log x)^k}{x} \pi(x) - \int_{2}^{x} \pi(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t …

As implicitly stated in Igor Rivin's answer, the standard way to show the vanishing of $\zeta(s)$ at negative even integers, and that such zeroes are simple, would be to use the functional equation, a …

I think this paper gives a pretty good answer: http://projecteuclid.org/euclid.em/1317758092 In the appendix, Languasco and Zaccagnini give a proof of Norton that if $q \geq 2$ and $1 \leq a < q$ wit …

It's best to split this up into two cases. Case 1: $\chi(a) = 1$. Then for $\Re(s) > 1$, $$\prod_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \prod_{p \equiv a \pmod{q}} \left(1 …

Expanding on Matt's answer, it is possible to show without too much difficulty (see here, Exercise 3 of section 18.2.1) that if $(a,q) = 1$, then $$\sum_{n \leq x}{\mu(n) e^{2\pi i an/q}} = \sum_{d \m …

In all cases, the analytic conductor is defined as the usual arithmetic conductor $q_{\pi}$ times a product of terms coming from the archimedean places. The issue is the definition of the terms at th …

This is a well-studied problem. The number of zeroes of a Dirichlet $L$-function $L(s,\chi)$ associated to a primitive even Dirichlet character $\chi$ modulo $q > 1$ up to height $T \geq 1$, $$N(T,\ch …

We have that \[\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} = \frac{1}{2\pi i} \int_{\sigma_0 - i\infty}^{\sigma_0 + i\infty} \frac{1}{\zeta(s + 1)} \frac{x^s}{s^2} \, ds\] for $\sigma_0$ suffici …

Yes it does. To see this, note that by partial summation, $$\frac{1}{\zeta(s + 1)} = s\int_{1}^{\infty}\sum_{n \leq x} \frac{\mu(n)}{n} x^{-s} \, \frac{dx}{x}$$ for all $\Re(s) > 0$. Now let $\Theta$ …

The Gelbart-Jacquet lift of a $\mathrm{GL}_2$ cuspidal automorphic representation $\pi$ is an automorphic representation $\Pi$ of $\mathrm{GL}_3$. More generally, the Gelbart-Jacquet lift $\Pi$ is a $ …

For unconditional results, see Theorem 11.4 of Montgomery-Vaughan, which states the following. Let $\chi$ be a primitive Dirichlet character modulo $q > 1$. Then there exists a constant $c > 0$ such t …

Explicit upper and lower bounds for $L(1,\chi)$, conditional on the generalised Riemann hypothesis, are given in Theorem 1.5 of the paper "Conditional bounds for the least quadratic non-residue and re …