I'm interested in the sum, $$\sum_{n=1}^\infty\frac{\zeta(2n)\Gamma(2n)}{\Gamma(2k+2n+2)}x^{2n}$$ Otherwise written as $$\sum_{n=1}^\infty\frac{\zeta(2n)}{(2n)(2n+1)\cdots(2n+2k+1)}x^{2n}$$

I am looking for some values of $k\gt0\in\mathbb{Z}$ and $x\ne0\in\mathbb{R}$ for which the expression equals a value in terms of known constants preferably in a closed form.


You “only” have to know, how to integrate $2k+2$ times the functions $\displaystyle \frac{1}{x}$ and $\cot x$ or $H(x)$

with $\enspace\displaystyle H(x):=\sum_{k=1}^\infty \frac{x}{k(x+k)}=\int_0^1 \frac{1-t^x}{1-t}dt$ .

The base is $\enspace\displaystyle \sum_{n=1}^\infty\zeta(2n)x^{2n-1}=\frac{1}{2x}-\frac{\pi}{2}\cot(\pi x)=\frac{1}{2}(H(x)-H(-x))$ .


Calculating with $\enspace H(x)\enspace$ we get

$\displaystyle \int_0^{x_0}…\int_0^{x_{n-1}}\frac{H(x_n)-H(-x_n)}{2} dx_n= \frac{1}{2}\int_0^1 \frac{1}{1-t}\left(\frac{t^{-x}-t^x}{\ln^n t}+\sum_{j=0}^n\frac{x^j(1-(-1)^{n-j})}{j!(\ln t)^{n-j}}\right)dt $ .   In your case it's $\enspace n=2k+2$ .


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