finding the closed-form of $k$ in the series $\sum_{n=1}^{m}\beta (2n-1)\zeta (2n)=m+k$

Question

finding the closed-form of $k$ in the series $$\sum_{n=1}^{m}\beta (2n-1)\zeta (2n)=m+k$$ when m go to infinity

from some values of $m$, I found the $$k=0.358971008185307705...$$ any help, thanks

Answer 1

Since $$ \beta(2n-1)=\frac{1}{(2n-2)!}\int_{0}^{1}\frac{(\log x)^{2n-2}}{1+x^2}\,dx, $$ $$ \zeta(2n) = \frac{1}{(2n-1)!}\int_{0}^{1}\frac{(-\log x)^{2n-1}}{1-x}\,dx\tag{1}$$ and: $$ \sum_{n\geq 1}\left(\zeta(2n)-1\right) = \frac{3}{4},$$ $$ \sum_{n\geq 1}\left(\beta(2n-1)-1\right) = -\frac{1}{4} \tag{2}$$ we have:

$$ \begin{eqnarray*}\sum_{n\geq 1}\left(\zeta(2n)\cdot\beta(2n-1)-1\right)\\=\sum_{n\geq 1}(\zeta(2n)-1)(\beta(2n)-1)+\sum_{n\geq 1}(\zeta(2n)-1)+\sum_{n\geq 1}(\beta(2n-1)-1)\\=\frac{1}{2}+\sum_{n\geq 1}\frac{1}{(2n-1)!(2n-2)!}\iint_{(0,1)^2}\frac{x^2(\log x)^{2n-2}y(\log y)^{2n-1}}{(1+x^2)(1-y)}\,dx\,dy\\=\frac{1}{2}+\frac{1}{2}\iint_{(0,1)^2}\frac{x^2 y}{(1+x^2)(1-y)}\sqrt{\frac{\log y}{\log x}}\left(I_1(2\sqrt{(\log x)(\log y)})+J_1(2\sqrt{(\log x)(\log y)})\right)\\=\frac{1}{2}+\frac{1}{2}\iint_{(0,+\infty)^2}\frac{e^{-3x}e^{-2y}}{(1+e^{-2x})(1-e^{-y})}\sqrt{\frac{y}{x}}\left(I_1(2\sqrt{xy})+J_1(2\sqrt{xy})\right)\tag{3}\end{eqnarray*}$$