How to Solve This Irregular Integral?

[mmzaj]

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 \left(\infty,1\right]\; which doesn't add up nicely to the portion along \left[1,\infty\right). any insights are appreciated.

 

[haruspex]

Doesn't this simplify fairly easily?

\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}}

 

[mmzaj]

you would think ! but no, it doesn't ...

 

[mmzaj]

The integral above is equivalent to :

\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
And
\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

 

[haruspex]

\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}} = \frac{e^{-2\pi i x}(e^{2\pi i x}-1)}{1-e^{2\pi i x}} = -e^{-2\pi i x}
I end up with \int_0^∞\frac{1-2e^u}{2(u+z)}du
which surely doesn't converge?

 

[mmzaj]

\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}} = \frac{e^{-2\pi i x}(e^{2\pi i x}-1)}{1-e^{2\pi i x}} = -e^{-2\pi i x}
I end up with \int_0^∞\frac{1-2e^u}{2(u+z)}du
which surely doesn't converge?

you missed the fact that the inverse of the complex exponential - the complex \log function- is multivalued. namely :
\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}}=-e^{-2\pi i x}=e^{-2\pi i \left(x-1/2\right)}=e^{-2\pi i \left(\left \{ x \right \}-1/2\right)}
Where \left \{ x \right \} is the fractional part of x. Thus:
\frac{1}{2\pi i}\ln\left(\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}}\right)=\frac{1}{2}-\left \{ x \right \}
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}-\left \{ x \right \}
Using these facts, we can prove that the integral in question is equal to the limit:
e^{-z}\text{Ei}(z)+\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left(n+\frac{1}{2} \right )\ln\left(\frac{\ln(n+1)+z}{\ln(n)+z} \right )-e^{-z}\text{Ei}(z+\ln N)
Where \text{Ei}(z) is the exponential integral function. But I'm stuck with this cumbersome limit