Evaluate $\sum\limits_{k=1}^{\infty}\frac{B(k,k)}{k}$ where $B$ denotes the beta function

Question

How can I evaluate $\displaystyle\sum_{k=1}^{\infty}\frac{B(k,k)}{k}$ Here B is the beta function

Answer 1

Here is a solution:

\begin{align*} \sum_{k=1}^{\infty} \frac{B(k,k)}{k} &= \sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{1} x^{k-1}(1-x)^{k-1} \, \mathrm{d} x \\ &= - \int_{0}^{1} \frac{\log(1 - x(1-x))}{x(1-x)} \, \mathrm{d}x \\ &= -2 \int_{0}^{1} \frac{\log(1 - x + x^2)}{x} \, \mathrm{d}x \\ &= 2 \int_{0}^{1} \frac{\log(1 + x) - \log(1 + x^3)}{x} \, \mathrm{d}x \\ &= \frac{4}{3} \int_{0}^{1} \frac{\log(1 + x)}{x} \, \mathrm{d}x \\ &= \frac{4}{3} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} \\ &= \frac{4}{3} \left(1 - \frac{2}{2^2}\right)\zeta(2) \\ &= \frac{\pi^2}{9}. \end{align*}

Here are explanations for some non-trivial steps:

• In the third step, we exploited $\frac{\log(1 - x(1-x))}{x(1-x)} = \frac{\log(1 - x(1-x))}{x} + \frac{\log(1 - x(1-x))}{1-x}$. By symmetry, integrals of the two terms in the right-hand side coincide.

• $\int_{0}^{1} \frac{\log(1+x^3)}{x} \, \mathrm{d}x = \frac{1}{3} \int_{0}^{1} \frac{\log(1+u)}{u} \, \mathrm{d}u$ by the substitution $u = x^3$.

Answer 2

You can also use the power series of the squared arcsine: $$ \sum \limits_{k=1}^\infty \frac{\operatorname{B}(k,k)}{k} = \sum \limits_{k=1}^\infty \frac{k!^2}{k^3 (2k-1)!} = 2\sum \limits_{k=1}^\infty \frac{1}{k^2 {2k \choose k}} = 4 \arcsin^2 \left(\frac{1}{2}\right) = \frac{\pi^2}{9} \, .$$