Alternating Beta sum

[Tolaso J Kos]

Evaluate the sum:

$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2n}{\rm B}\left ( \frac{1}{2}, \frac{n}{2} \right )$$

Answer

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$\mathcal{S}=\frac{\pi^2}{8}$

[r9m]

\begin{align}\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}B\left(\frac{1}{2},\frac{n}{2}\right) &= \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\int_0^{1} (1-x)^{-1/2} x^{\frac{n}{2} - 1}\,dx\\&= \int_0^{1}\frac{(1-x)^{-1/2} \log \left(1+\sqrt{x}\right)}{x}\,dx\\&= 2\int_0^1 \frac{\log (1+x)}{x\sqrt{1-x^2}}\,dx\\&= 2\int_0^1 \frac{\log \left(1+\frac{2t}{1+t^2}\right)}{t}\,dt\\&= 4\int_0^1 \frac{\log (1+t)}{t}\,dt - 2\int_0^1 \frac{\log (1+t^2)}{t}\,dt\\&= 3\int_0^1 \frac{\log (1+t)}{t}\,dt = \frac{\pi^2}{4}\end{align}

Justifications:

$(2)$ We used $x\mapsto x^2$, $(3)$ we made the change in variable $\displaystyle x = \frac{2t}{1+t^2}$, $(5)$ We made the change of variable $t^2 \mapsto t$ in the second integral.