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 ?
EDIT:
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.