Evaluate $\zeta'{(3)}$ and $\zeta'{\left(3,\frac{1}{4}\right)}$

Question

I am trying to evaluate this sum: $$\sum_{k=0}^{\infty}\frac{(-1)^k\log{(2k+1)}}{(2k+1)^3}$$ After doing the rest of works, there are still these two terms that I can not find the closed form: $\zeta'{(3)}$ and $\zeta'{\left(3,\frac{1}{4}\right)}$ where $\zeta{(s)}$ is the Riemann zeta function and $\zeta{(s,a)}$ is the Hurwitz zeta function . Can I ask for help from every one? Thank you so much.

Answer 1

Just a beginning.

Using the inverse Laplace transform of $\frac{\log(n)}{n^3}$ we have that

$$ \sum_{k\geq 0}(-1)^k\frac{\log(2k+1)}{(2k+1)^3} = \int_{0}^{+\infty}\frac{x^2}{4\cosh x}\left(\frac{3}{2}-\gamma-\log(x)\right)\,dx $$ equals $$ \frac{\pi^3}{64}(3-2\gamma)-\frac{1}{4}\int_{0}^{+\infty}\frac{x^2\log(x)}{\cosh x}\,dx $$ where (Gradshteyn-Rizhyk 4.371) $$ \int_{0}^{+\infty}\frac{\log(x)}{\cosh(x)}\,dx =\frac{\pi}{2}\log\left(\frac{4\pi^3}{\Gamma^4\left(\frac{1}{4}\right)}\right)$$ such that $\int_{0}^{+\infty} x^2\frac{\log(x)}{\cosh(x)}\,dx$ is at least expected to be of the same form, i.e. a linear combination of $\pi\log 2, \pi\log\pi$ and $\pi\log\Gamma\left(\frac{1}{4}\right)$. I still have to check if integration by parts leads us somewhere.

Answer 2

Starting from @Jack D'Aurizio's answer, two integrals of interest

$$I_n=\int_0^{ + \infty } {\frac{{x^n\, \log (x)}}{{\cosh(x)}}\,dx}$$ $$\color{blue}{I_1 = -2 C (\gamma +2 \log (2))+\frac 18\Bigg[\left[ {\frac{d}{{ds}}\gamma _1(s)} \right]_{s = \frac 14}-\left[ {\frac{d}{{ds}}\gamma _1(s)} \right]_{s = \frac 34}\Bigg]}$$

$$\color{blue}{I_2=-\frac{\pi ^3}{8} (\gamma +2 \log (2))-\frac 1{32}\Bigg[\left[ {\frac{d^2}{{ds^2}}\gamma _1(s)} \right]_{s = \frac 14}-\left[ {\frac{d^2}{{ds^2}}\gamma _1(s)} \right]_{s = \frac 34}\Bigg]}$$

$$\color{red}{S=\sum_{k= 0}^\infty(-1)^k\frac{\log(2k+1)}{(2k+1)^3} =\frac{\pi ^3}{64} (3+4 \log (2))+\frac 1{128}\Bigg[\left[ {\frac{d^2}{{ds^2}}\gamma _1(s)} \right]_{s = \frac 14}-\left[ {\frac{d^2}{{ds^2}}\gamma _1(s)} \right]_{s = \frac 34}\Bigg]}$$