I am looking for help with doing the following integral : $$\frac{1}{2\pi i}\int_{1}^{\infty}\ln\left(\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}} \right )\frac{dx}{x\left(\ln x+z\right)}\;\;\;\;z\in \mathbb{C}$$ i tried to transform it into a complex integral along a 'keyhole contour', with a branch cut along the +ive real line $\left[1,\infty\right)$. but then $\;\ln x\;$ would be transformed into $\;\ln x+2\pi i\;$ when doing the integral along the segment parallel to and below $\left(\infty,1\right]\;$ which doesn't add up nicely to the portion along the segment parallel to and above $\left[1,\infty\right)$ .

It can be shown that the above integral is equivalent to: $$\int_{1}^{\infty}\sum_{n=1}^{\infty}\frac{\sin(2\pi n x)}{n\pi}\left(\frac{1}{x\left(\ln x+z\right)}\right)dx$$ And $$\int_{1}^{\infty}\left(\frac{1}{2}-x+\left \lfloor x \right \rfloor \right )\left(\frac{1}{x\left(\ln x+z\right)}\right)dx$$ EDIT: Another way to think of it is to take the Taylor expansion of the $ \log $: $$\frac{1}{2\pi i}\left(\ln\left(1-e^{-2\pi i x}\right)-\ln(1-e^{2\pi i x})\right)=\frac{1}{2\pi i}\sum_{k=1}^{\infty}\frac{e^{2\pi i k x}-e^{-2\pi i kx}}{k}=\sum_{n=1}^{\infty}\frac{\sin(2\pi k x)}{k\pi}$$ Which in turn is the Fourier expansion of $\frac{1}{2}-x+\left \lfloor x \right \rfloor $.

Also, using the definition of the zeta function : $$\zeta(s)=\frac{s}{s-1}-s\int_{1}^{\infty}(x-\left \lfloor x \right \rfloor )x^{-s-1}dx\;\;\;\;\;\Re(s)>0$$ We have: $$\frac{\zeta(s)}{s}+\frac{1}{2s}-\frac{1}{s-1}=\int_{1}^{\infty}\left(\frac{1}{2}-x+\left \lfloor x \right \rfloor\right)x^{-s-1}dx$$ And: $$\int_{1}^{\infty}\left(\frac{1}{2}-x+\left \lfloor x \right \rfloor \right )\left(\frac{1}{x\left(\ln x+z\right)}\right)dx=\int_{0}^{\infty}\left(\frac{\zeta(s)}{s}+\frac{1}{2s}-\frac{1}{s-1}\right)e^{-zs}ds$$

I posted this question on MS, and got no answers, even with a bounty !! any insights are appreciated.