How to calculate $\int_{0}^{1}-\frac{1}{2\pi }\ln(\left \| x-y \right \|)\cdot e^{i2\pi kt_{x}}dt_{x}$

Question

I came across the integral $$\int_{0}^{1}-\frac{1}{2\pi }ln(\sqrt{(r\cdot\cos(2\pi t_{x})-r\cdot \cos(2\pi t_{y}))^{2}+(r\cdot\sin(2\pi t_{x})-r\cdot\sin(2\pi t_{y}))^{2}})\cdot e^{i2\pi kt_{x}}dt_{x}$$ with $k\in \mathbb{N}_0$ but wasn't able to calculate it so far.

I already know and can use that $$\int_{0}^{1}\ln(\sin(\pi t))\sin(2\pi kt)dt=0$$ and $$\int_{0}^{1}\ln(\sin(\pi t))\cos(2\pi kt)dt=\left\{\begin{matrix} -\ln(2), k=0\\ -\dfrac{1}{2k},k\geq 1 \end{matrix}\right.$$

I also know that the result should be something like $$\lambda _k\cdot e^{i2\pi kt},k\in \mathbb{N}_0$$

I tried to use several trigonometric identities but did not succeed.

Answer 1

First, let's clean some things up:

$$-\frac{1}{4\pi }\int_{0}^{1}ln{[(r\cdot cos(2\pi t_{x})-r\cdot cos(2\pi t_{y}))^{2}+(r\cdot sin(2\pi t_{x})-r\cdot sin(2\pi t_{y}))^{2}}]\cdot e^{i2\pi kt_{x}}dt_{x}$$

$$=-\frac{1}{4\pi }\int_{0}^{1}2lnr+ln{[(cos(2\pi t_{x})- cos(2\pi t_{y}))^{2}+(sin(2\pi t_{x})-sin(2\pi t_{y}))^{2}}]\cdot e^{i2\pi kt_{x}}dt_{x}$$

$$=-\frac{1}{4\pi }\int_{0}^{1}2lnr+ln{[cos^2(2\pi t_{x})- 2cos(2\pi t_{x})cos(2\pi t_{y})+ cos^2(2\pi t_{y})+sin^2(2\pi t_{x})- 2sin(2\pi t_{x})sin(2\pi t_{y})+ sin^2(2\pi t_{y})}]\cdot e^{i2\pi kt_{x}}dt_{x}$$

$$=-\frac{1}{4\pi }\int_{0}^{1}2lnr+ln{[2- 2cos(2\pi t_{x})cos(2\pi t_{y})- 2sin(2\pi t_{x})sin(2\pi t_{y})}]\cdot e^{i2\pi kt_{x}}dt_{x}$$

$$=-\frac{1}{4\pi }\int_{0}^{1}2lnr+ln2+ln{[1- cos(2\pi t_{x})cos(2\pi t_{y})- sin(2\pi t_{x})sin(2\pi t_{y})}]\cdot e^{i2\pi kt_{x}}dt_{x}$$

Then we use some trigonometric identities:

$$=-\frac{1}{4\pi }\int_{0}^{1}2lnr+ln2+ln{[1- cos2(\pi t_{x}-\pi t_{y})}]\cdot e^{i2\pi kt_{x}}dt_{x}$$

$$=-\frac{1}{4\pi }\int_{0}^{1}2lnr+ln2+ln{[2cos^2(\pi t_{x}-\pi t_{y})}]\cdot e^{i2\pi kt_{x}}dt_{x}$$

$$=-\frac{1}{4\pi }\int_{0}^{1}2lnr+2ln2+2ln{[cos(\pi t_{x}-\pi t_{y})}]\cdot e^{i2\pi kt_{x}}dt_{x}$$

$$=-\frac{1}{4\pi }\int_{0}^{1}2lnr+2ln2+2ln0.5+2ln{[e^{i(\pi t_{x}-\pi t_{y})})+e^{-i(\pi t_{x}-\pi t_{y})}}]\cdot e^{i2\pi kt_{x}}dt_{x}$$

Things should be much easier to solve from here. :)

EDIT: Next part of the solution:

$$=-\frac{1}{4\pi }\int_{0}^{1}2lnr+2ln2+2ln0.5+2i(\pi t_{x}-\pi t_{y}){{+2ln[1+e^{-2i(\pi t_{x}-\pi t_{y})}}]}\cdot e^{i2\pi kt_{x}}dt_{x}$$

$$=-\frac{1}{2\pi }\int_{0}^{1}lnr+ln2+ln0.5+i\pi t_{x}-i\pi t_{y}{{+ln[1+e^{-2i(\pi t_{x}-\pi t_{y})}}]}\cdot e^{i2\pi kt_{x}}dt_{x}$$

Using the fact that $\int e^x(ln(1+e^x))dx=(e^x+1)ln(1+e^x)-e^x+C$,

$$=-\frac{1}{2\pi }[t_{x}(lnr+ln2+ln0.5-i\pi t_{y} )+\frac{i\pi (t_{x})^2}{2}+e^\frac{-1}{k}(1+e^{-2i(\pi t_{x}-\pi t_{y})})ln(1+e^{-2i(\pi t_{x}-\pi t_{y})})-e^\frac{-1}{k}e^{-2i(\pi t_{x}-\pi t_{y})}]_{0}^{1}$$

And lo and behold:

$$=[lnr+ln2+ln0.5-i\pi t_{y} )+\frac{i\pi}{2}+e^\frac{-1}{k}(1+e^{-2i(\pi-\pi t_{y})})ln(1+e^{-2i(\pi -\pi t_{y})})-e^\frac{-1}{k}e^{-2i(\pi-\pi t_{y})}]-[e^\frac{-1}{k}(1+e^{2i\pi t_{y}}ln(1+e^{2i\pi t_{y}})-e^\frac{-1}{k}e^{2i\pi t_{y}}]$$

Apologies if I've made any errors, the working is just too chaotic for me. :P