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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 $\frac{4\pi i}{3}$?

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  • $\begingroup$ How about the branch of $\log$? $\endgroup$ – user64494 Jun 18 '14 at 19:55
  • $\begingroup$ @user64494 the given integrand shall have singularities at $z=1,-1$. but they are not poles. how do i use residue theorem ? $\endgroup$ – Aman Mittal Jun 18 '14 at 19:57
  • $\begingroup$ @ Aman Mittal : My question is about the definition of the integrand and its branch cuts. $\endgroup$ – user64494 Jun 18 '14 at 20:04
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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.


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

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

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  • $\begingroup$ @ Daniel Fisher: Can you explain your choice of the branch of $\log w$ in detail, especially the branch cut?? $\endgroup$ – user64494 Jun 18 '14 at 20:20
  • $\begingroup$ Since in the transformed integral, everything lives in the right half-plane, any branch of the logarithm defined there will do (as mentioned, the integral is the same for all choices). I mentioned the principal branch, so I'd say the branch-cut is $(-\infty,0]$ if necessary, which is good, because the image of $[-1,1]$ under $z\mapsto \frac{z+1}{z-1}$ is $\{\infty\} \cup (-\infty,0]$, and $[-1,1]$ is the natural choice of the branch cut for the original integrand. Any branch cut from $-1$ to $1$ that doesn't intersect the contour will work the same, however. Yes, it is not explicitly stated $\endgroup$ – Daniel Fischer Jun 18 '14 at 20:28
  • $\begingroup$ that the branch cut shall not intersect the contour of integration, but from time to time I make reasonable assumptions beyond the explicitly stated facts. $\endgroup$ – Daniel Fischer Jun 18 '14 at 20:29

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