Using calculus of residues, prove that $\int z^2log[(z+1)/(z-1)]dz$

Question

Using calculus of residues, how can it be proven that $$ \int z^2\log\left[\frac{z+1}{z-1}\right]\;dz $$ taken round the circle $\left\vert z\right\vert=2$ has the value $\frac{4\pi i}{3}$?

Answer 1

Note: I'm assuming that $\log \frac{z+1}{z-1}$ is defined on $\mathbb{C}\setminus [-1,1]$, or at least that a branch-cut that lies entirely inside the disk $\{ z : \lvert z\rvert < 2\}$ is used. For arbitrary branch-cuts intersecting the contour, the integral depends on the exact choice of the branch-cut.

One way is to substitute

$$w = \frac{z+1}{z-1}.$$

The circle $\lvert z\rvert = 2$ becomes $\left\lvert w - \frac{5}{3}\right\rvert = \frac{4}{3}$, and since

$$z = \frac{w+1}{w-1},$$

we have $dz = -\frac{2dw}{(w-1)^2}$, but the circle $\left\lvert w - \frac{5}{3}\right\rvert = \frac{4}{3}$ is traversed clockwise, so overall we get the integral

$$2\int_{\lvert w-5/3\rvert = 4/3} \frac{(w+1)^2\log w}{(w-1)^4}\,dw,$$

which has a pole of order $3$ (the numerator has a simple zero in $1$ from the $\log$ factor) in $1$ as the only singularity enclosed by the contour. That integral can be evaluated by the residue theorem in the usual manner.

Taking the principal branch of the logarithm, we can expand the numerator

$$\bigl((w-1)^2 + 4(w-1) + 4\bigr)\left((w-1) - \frac{(w-1)^2}{2} + \frac{(w-1)^3}{3} + O\left((w-1)^4\right)\right)\\ = \dotsc + (1 - 4\frac{1}{2} + 4\frac{1}{3})(w-1)^3+ \dotsc,$$

so the residue is $\frac{1}{3}$. Choosing a different branch of the logarithm doesn't affect the residue, since only the coefficients of $(w-1)^k$ for $k\in \{-4,-3,-2\}$ change.

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Another way to evaluate the integral is to rewrite and expand the logarithm

$$\log \frac{z+1}{z-1} = \log \frac{1+z^{-1}}{1-z^{-1}} = 2m\pi i + 2 \sum_{k=0}^\infty \frac{1}{(2k+1)z^{2k+1}}.$$

Since the Laurent series converges uniformly on the contour, we can interchange summation and integration, and obtain

$$\int_{\lvert z\rvert = 2} z^2\log \frac{z+1}{z-1}\,dz = 2m\pi i \int_{\lvert z\rvert = 2} z^2\,dz + 2\sum_{k=0}^\infty \frac{1}{2k+1}\int_{\lvert z\rvert = 2} z^{1-2k}\,dz = \frac{2}{3}\int_{\lvert z\rvert = 2} \frac{dz}{z}.$$