Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I thought up a curious definite integral. Let $D = \{ z \in \mathbb{C} : |z|<1\}$. Let $A$ denote area measure on $D$, normalized so that $A(D) = \pi$. I claim that $$\iint_D \left|\log \left( \frac{e}{1-z} \right) \right|^2 \ dA = \frac{\pi^3}{6}.$$ My explanation goes as follows. Let $H$ be the Hilbert space of holomorphic functions $f :D \to \mathbb{C}$ such that $\iint |f|^2 \ dA$ is finite (see Bergman space). It's not hard to see that the functions $1,z,z^2,\ldots$ constitute an orthogonal set in $H$ and that their $L^2$ norms are given by $\| z^n\| = \sqrt{\frac{\pi}{n+1}}$. Moreover, their closed linear span is all of $H$ - roughly because holomorphic functions on $D$ have power series expansions about $0$. The function $\log \left( \frac{e}{1-z} \right)$ is chosen to have a particular power series $$ \log \left( \frac{e}{1-z} \right) = 1 - \log(1-z) = 1 + \sum_{n=1}^\infty \frac{z^n}{n}.$$ By Parseval's identity $$ \left\| \log \left( \frac{e}{1-z} \right) \right\|^2 = \|1\|^2 + \sum_{n=1}^\infty \frac{1}{n^2} \|z^n\|^2 = \pi + \pi \sum_{n=1}^\infty \frac{1}{n^2(n+1)}.$$ The series can be evaluated by splitting it into two convergent series $$\sum_{n=1}^\infty \frac{1}{n^2(n+1)} = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{n(n+1)} = \frac{\pi^2}{6} - 1$$ where the 1st sum is classical and the second telescopes. So, we obtain $$\iint_D \left|\log \left( \frac{e}{1-z} \right) \right|^2 \ dA = \pi + \pi \left(\frac{\pi^2}{6} - 1 \right) = \frac{\pi^3}{6}.$$ My somewhat frivolous question is:

Is there a good alternative explanation for why $$\iint_D \left|\log \left( \frac{e}{1-z} \right) \right|^2 \ dA = \frac{\pi^3}{6}?$$

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.