# Find the value of integral

find the value of $\int_{|z|=2}^{}\log\frac{z+1}{z-1}dz$ assume that for $w\in C-\{z;\text{Re}(z)\leq0\}$ we have $-\pi<\text{Im}(\log w)<\pi$

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Note that $(z+1)/(z-1)$ is on the negative real axis only for $z$ on the line segment from $-1$ to $1$. So deform the contour to one consisting of the straight line segments $L_1$ from $-1+\eta-i\epsilon$ to $1-\eta-i\epsilon$ and $L_2$ from $1-\eta+i\epsilon$ to $-1+\eta+i\epsilon$ and two arcs of circles, where $\eta$ is very small with $\eta >> \epsilon > 0$. Note that for $z \in L_1$, $(z+1)/(z-1)$ is in the upper half-plane near the negative real axis, while for $z \in L_2$, it is in the lower half-plane. Thus for $z \in L_1$, $f(z) - f(\overline{z}) \approx 2 \pi i$, so that $$\int_{L_1} f(z)\ dz + \int_{L_2} f(z)\ dz \approx \int_{-1}^{1} (2 \pi i)\ dz = 4 \pi i$$ Thus (after convincing yourself that the integrals over the circular arcs go to $0$) the answer is $4 \pi i$.

EDIT: Another way to do this: note that by Cauchy's theorem the answer should be the same for the integral over $|z|=r$ for any $r > 1$. So take $r$ very large. Note that for $|z|$ large, $$\frac{z+1}{z-1} = 1 + \frac{2}{z} + O\left(\frac{1}{|z|^2}\right)$$ and so $$\log\left( \frac{z+1}{z-1} \right) = \frac{2}{z} + O\left(\frac{1}{|z|^2}\right)$$

Thus $$\oint_{|z|=r} \log\left( \frac{z+1}{z-1} \right) \ dz = 2 \oint_{|z|=r} \frac{2}{z}\ dz + O\left(\frac{1}{r}\right) = 4 \pi i + O\left(\frac{1}{r}\right)$$

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Parametrize the integral using $z=2 e^{i t}$, $dz = 2 i e^{i t} dt$. The integral is then

$$i 2 \int_0^{2 \pi} dt \: e^{i t} \log{\left ( \frac{2 e^{i t}+1}{2 e^{i t}-1} \right )}$$

Use the fact that $\log{z} = \log{|z|} + i \arg{z}$ and we end up with four integrals to evaluate (expressed as two here for reasons of space):

\begin{align}\\ &i 2 \int_0^{2 \pi} dt \: \left [ \frac{1}{2} \cos{t} \log{\left ( \frac{5 + 4 \cos{t}}{5 - 4 \cos{t}}\right )} + \sin{t} \arctan{\left ( \frac{4}{3} \sin{t} \right )}\right ]\\ &- 2 \int_0^{2 \pi} dt \: \left [ \frac{1}{2} \sin{t} \log{\left ( \frac{5 + 4 \cos{t}}{5 - 4 \cos{t}}\right )} - \cos{t} \arctan{\left ( \frac{4}{3} \sin{t} \right )}\right ]\end{align}

The last two integrals are zero. The first integral evaluates to $i 2 \pi$, while the second also evaluates to $i 2 \pi$. The result is then

$$\oint_{|z|=2}dz \: \log{\left( \frac{z+1}{z-1}\right)} = i 4 \pi$$

NB I used Mathematica to evaluate these integrals. I will try to prove these results by hand, as these integrals do look interesting in and of themselves. But in any case, even though I used Mathematica, it still took significant amount of work just to get the integral into a form where Mathematica could be useful.

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