Help with an irregular integral

1. Dec 9, 2012

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.

2. Dec 9, 2012

haruspex

Doesn't this simplify fairly easily?

3. Dec 10, 2012

mmzaj

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

4. Dec 10, 2012

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
$$\int_{1}^{\infty}\sum_{n=1}^{\infty}\frac{\sin(2 \pi n x)}{n\pi}\left(\frac{1}{x\left(\ln x+z\right)}\right)dx$$

5. Dec 10, 2012

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?

6. Dec 12, 2012

mmzaj

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

Last edited: Dec 12, 2012