Let $\phi(q)$ be the Euler's function given by the infinite product : $$\phi(q)=\prod_{n=1}^{\infty}\frac{1}{1-q^{n}}\;\;\;\;|q|<1$$ And let $\mu(k)$ be the $\text{M}\ddot{\text{o}}\text{bius}$ function. We have the integral : $$z\int_{0}^{1}\sum_{k=1}^{\infty}\frac{\mu(k)}{k}e^{z\omega/k}\log\phi\left(e^{-sz\omega}\right)d\omega$$ Where $z\in \mathbb{R}^{+}$ and $s\in \mathbb{C}:\Re(s)>0$.

My attempt : Using the fact that : $$\log\phi(q)=\sum_{n=1}^{\infty}\frac{\sigma(n)}{n}q^{n}$$ Where $\sigma(n)$ is the divisor function, we obtain : $$z\int_{0}^{1}\sum_{k=1}^{\infty}\frac{\mu(k)}{k}e^{z\omega/k}\log\phi\left(e^{-sz\omega}\right)d\omega=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\mu(k)\sigma(n)\frac{1-e^{z\left(\frac{1}{k}-ns \right )}}{n(kns-1)}$$ But this is hardly useful ! any insight on how to obtain a simpler expression ?


We have : $$\sum_{k=1}^{\infty}\frac{\mu(k)}{k}e^{z\omega/k}=\sum_{k=1}^{\infty}\frac{\mu(k)}{k}\sum_{j=0}^{\infty}\frac{(z\omega)^{j}}{j!k^{j}}=\sum_{j=1}^{\infty}\frac{(z\omega)^{j}}{j!\zeta(j+1)}$$ And using the fact that : $$\int_{0}^{1}\omega^{j}\log\phi\left(e^{-sz\omega}\right)d\omega=\sum_{n=1}^{\infty}\frac{\sigma(n)}{n}\int_{0}^{1}\omega^{j}e^{-nsz\omega}d\omega=\sum_{n=1}^{\infty}\frac{\sigma(n)}{n}(nsz)^{-j-1}\left(j!-\Gamma(j+1,nsz)\right)$$ our integral now reads : $$\sum_{j=1}^{\infty}\sum_{n=1}^{\infty}\frac{(ns)^{-j-1}}{\zeta(j+1)}\frac{\sigma(n)}{n}\left(1-\frac{\Gamma(j+1,nsz)}{j!}\right)$$ This is more manageable than the previous expression, but i believe it can be simplified even further.

  • $\begingroup$ Do you have some reason for believing there's a simpler expression? $\endgroup$ – Gerry Myerson Apr 30 at 9:13
  • $\begingroup$ If we Taylor expand $\sum_{k=1}^{\infty}\frac{\mu(k)}{k}e^{z\omega/k}$ around $\omega=0$, the integral becomes a sum over the incomplete gamma function multiplied by $\sigma(n)$, which looks tantalizingly like an Eichler integral !! hence, I am a bit confused trying to find the simplest expression for the integral $\endgroup$ – Mohammad Al Jamal Apr 30 at 9:25
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    $\begingroup$ OK. I hope you get an answer here, but if a few days go by with no resolution, I'd suggest posting to MathOverflow. $\endgroup$ – Gerry Myerson Apr 30 at 14:06
  • $\begingroup$ It is quite obvious it won't simplify. First of all why do you want to consider $f(z)g(z),\int_0^z f(u)g(u)du,\Re(z) > 0$ where $f(z) = \sum_k \mu(k)k^{-1} e^{-z/k}, g(z) = \sum_n \sigma(n) n^{-1} e^{-nz}$. $\endgroup$ – reuns Apr 30 at 16:49
  • $\begingroup$ I am looking for the 'simplest' expression. At the end of the day, there is a huge space of the ways you can express any function, but the simplest expression is what we usually seek, right ? and this integral came up in the study of a certain arithmetical function, $\endgroup$ – Mohammad Al Jamal Apr 30 at 16:53

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