Prove that $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\ln(2n+1)}{2n+1} = \frac{\pi}{4}(\gamma - \ln \pi) + \pi \ln \Gamma(\frac{3}{4})$

$$\sum_{n=1}^\infty(-1)^{n+1}\frac{\ln(2n+1)}{2n+1} =\frac\pi4(\gamma-\ln\pi)+\pi\ln\Gamma(3/4)$$ This sum was obtained by Malmsten in 1842. I researched this series and found no proof. I also saw that this series is the derivative of Dirichlet's beta function at point 1. I tried to turn the series into an integral from 0 to 1, but I did not succeed.

Thanks for any help.

• en.wikipedia.org/wiki/Gamma_function#Fourier_series_expansion – metamorphy Jul 4 at 15:38
• Why do you reject the proof you have given: it is the derivative of Dirichlet's beta function evaluated at $1$? – Eric Towers Jul 4 at 15:38
• I would like a direct proof. – Mathsource Jul 4 at 15:42
• With Mathematica I have:$\frac{1}{4} \left(-\gamma _1\left(\frac{1}{4}\right)+\gamma _1\left(\frac{3}{4}\right)-\pi \ln (4)\right)$, where: $\gamma _1\left(\frac{3}{4}\right)$ and $\gamma _1\left(\frac{1}{4}\right)$ is generalized Stieltjes constant. – Mariusz Iwaniuk Jul 4 at 18:05