Beta function

Derivative of Dirichlet Beta function

We prove the following infinite product which turned out to be a modified version of the Dirichlet Beta function derivative. $$\frac{1}{\sqrt[3]{3}}\frac{\sqrt[5]{5}}{\sqrt[7]{7}}\frac{\sqrt[9]{9}}{\sqrt[11]{11}}\frac{\sqrt[13]{13}}{\sqrt[15]{15}}\ldots =\left(\frac{\pi}{e^\gamma \Gamma(\frac{3}{4})^4}\right)^\frac{\pi}{4}$$ Proof

Take logarithm in both sides $$-\frac{\ln(3)}{3}+\frac{\ln(5)}{5}-\frac{\ln(7)}{7}+\frac{\ln(9)}{9}-\frac{\ln(11)}{11}+\frac{\ln(13)}{13}-... = \frac{\pi}{4}\left[\ln(\pi)-\gamma -4\ln\Gamma\left(\frac{3}{4}\right)\right]$$ Then, we want to prove $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1} \ln(2n-1)}{(2n-1)} = \frac{\pi}{4}\left[\ln(\pi)-\gamma -4\ln\Gamma\left(\frac{3}{4}\right)\right]$$ Consider the Dirichlet beta function $$\beta(v) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^{v}}$$ Then $$\beta'(v) = \sum_{n=1}^{\infty} \frac{(-1)^{n}\ln(2n-1)}{(2n-1)^{v}}$$ Therefore we want to find $$-\beta'(1) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}\ln(2n-1)}{(2n-1)}$$ There are various ways to find the derivative, one of the most straigforward is using the Kummer's Fourier series for the log-gamma function: $$\ln \Gamma(t) = \frac{1}{2}\ln\left(\frac{\pi}{\sin \pi t}\right)+\left[\gamma+\ln(2\pi)\right]\left(\frac{1}{2}-t\right) +\frac{1}{\pi}\sum_{n=1}^{\infty} \frac{\ln n}{n} \sin(2\pi n t)$$ It turned out that if $\displaystyle t=\frac{1}{4}$ then $$\sin\left(\frac{\pi n}{2}\right)= \left\{\begin{array}{lr} \sin\left(\frac{\pi}{2}\right) = 1 & \textrm{ if } n=1 \\ \sin\left(\pi \right) = 0 & \textrm{ if } n=2 \\ \sin\left(\frac{3\pi}{2}\right) = -1 & \textrm{ if } n=3 \\ \sin\left(2\pi \right) = 0 & \textrm{ if } n=4 \\ \vdots & \vdots \\ \end{array} \right\}$$ Then $\displaystyle \sum_{n=1}^{\infty} \frac{\ln n}{n}\sin\left(\frac{\pi n}{2}\right) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \ln(2n-1)}{(2n-1)}$ Therefore, $$\begin{align*} \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \ln(2n-1)}{(2n-1)} =& \pi\left[ \ln \Gamma\left(\frac{1}{4}\right)- \frac{1}{2}\ln\left(\frac{\pi}{\sin \frac{\pi}{4} }\right)-\left[\gamma+\ln(2\pi)\right]\left(\frac{1}{4}\right) \right]\\ =& \pi\left[ \ln(\pi)- \frac{1}{2}\ln(2) -\ln\Gamma\left(\frac{3}{4}\right) - \frac{1}{2} \ln(\pi) -\frac{1}{4} \ln(2) -\frac{1}{4} \gamma -\frac{\ln (2\pi)}{4} \right]\\ =& \frac{\pi}{4} \left[\ln(\pi)-\gamma-4\Gamma\left(\frac{3}{4}\right)\right] \end{align*}$$ Therefore $$\boxed{\frac{1}{\sqrt[3]{3}}\frac{\sqrt[5]{5}}{\sqrt[7]{7}}\frac{\sqrt[9]{9}}{\sqrt[11]{11}}\frac{\sqrt[13]{13}}{\sqrt[15]{15}}\ldots =\left(\frac{\pi}{e^\gamma \Gamma(\frac{3}{4})^4}\right)^\frac{\pi}{4}}$$