Finding the Laurent series expansion for $1/(z\cdot \log z)$

Question

I'm trying to find the Laurent series around the point $z=1$ for the function $$ f(z) = \frac{1}{z \log z}\,, $$ to calculate the integral $$ \int_{|z-1|=1/2} f(z) \ dz\,. $$

Wolfram Alpha gives $$ \frac{1}{z-1}-\frac{1}{2}+\frac{5 (z-1)}{12}-\frac{3}{8} (z-1)^2+\frac{251}{720} (z-1)^3-\frac{95}{288} (z-1)^4+O\left((z-1)^5\right)\,. $$

As the coefficient for $(z-1)^{-1}$ is 1, I know the integral equals $2\pi i$, however, I've been unable to find the coefficients myself using the 'regular' techniques I've been taught.

I know the expansion of the logarithm around $z=1$ equals $$ (z-1)-\frac{1}{2}(z-1)^2 + \frac{1}{3}(z-1)^3 + O\left((z-1)^4\right)\,, $$ but substitution in $f(z)$ and trying to rewrite to some form of a geometric series or seperating the fractions is not yielding any useful results. I do not see what approach I am supposed to take here.

Answer 1

The function $z\log z$ has a simple zero at $1,$ so $f(z)$ has a simple pole at $1.$ It follows that the residue of $f$ at $1$ is

$$\tag 1\lim_{z\to 1}\,(z-1)\frac{1}{z\log z}.$$

Now $(z-1)/\log z = (z-1)/(\log z - \log 1),$ which has limit $1/\log'(1) = 1.$ It follows that the limit in $(1)$ is $1.$

Answer 2

Hint: You have $$ \frac{1}{z}=\frac{1}{1-(1-z)}=\sum_{k=0}^\infty (1-z)^k. $$ Now multiply by the expansion og the logarithm around $z=1$, and find the answer by long division.

Answer 3

Since you want to calculate the integral of $f$ around a closed curve around $z=1$ you only need the $\frac{1}{z-1}$ term.

Setting $w = z-1$ we get $$\begin{align} \frac{1}{z \log z} & = \frac{1}{(1+w) \log(1+w)} = \frac{1}{1+w} \cdot \frac{1}{\log(1+w)} \\ & = \left( 1 - w + O(w^2)\right) \frac{1}{-w + O(w^2)} \\ & = \left( 1 - w + O(w^2)\right) \frac{-1}{w}\frac{1}{1+O(w)} \\ & = \left( 1 - w + O(w^2)\right) \frac{-1}{w} \left(1-O(w)\right) \\ & = -\frac{1}{w} \left(1 + O(w) \right) \\ & = -\frac{1}{z-1} \left(1 + O(z-1) \right) \end{align}$$