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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.

Gary
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OnTheWay
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  • Like for $\zeta(3)$, there is likely no closed form expression for $\zeta'(3)$. I guess the same is true for $\zeta'(3,1/4)$. – Gary Jun 28 '22 at 10:13
  • @Gary Hello, thank you for your comment, i make some searches but also found nothing. – OnTheWay Jun 28 '22 at 10:18
  • Please include in question the steps to express your displayed sum in terms of the two zeta-derivatives you ask about. – coffeemath Jun 28 '22 at 10:19

2 Answers2

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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.

Jack D'Aurizio
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  • $\int_0^{ + \infty } {\frac{{x^2 \log x}}{{\cosh x}}dx} = \left[ {\frac{d}{{ds}}\int_0^{ + \infty } {\frac{{x^s }}{{\cosh x}}dx} } \right]_{s = 2} $ and see 25.11.31 – Gary Jun 29 '22 at 12:10
  • Well, we already know that the series is related to a value of $\beta'$, the unknown point is if such value has a closed form in terms of the values of the $\Gamma$ function. – Jack D'Aurizio Jun 29 '22 at 13:12
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Starting from @Jack D'Aurizio's answer, two integrals of interest

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

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

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

Claude Leibovici
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