What's about $e\zeta(3)-\zeta(6)=2e^2\int_1^2\frac{(3\zeta'(3x)-\zeta(3x))\sinh (x)}{e^{2x}-1}dx$?

Question

After I was exploring a kind of integrals, I asked to Wolfram Alpha the following code

$$\int_1^2 \frac{a^{-x}sinh(x)}{e^{2x}-1}dx$$

for its online calculator, to obtain a closed form for $$\int_1^2\frac{a^{-x}\sinh (x)}{e^{2x}-1}dx,$$ where I think that holds for $a\geq 1$. Thus I did the specialization $a=n^{\frac{6}{2}}=n^3$ and after a summation from $$2e^2(\log(a)+1 )\int_1^2\frac{a^{-x}\sinh(x)}{e^{2x}-1}dx=\frac{e}{a}-\frac{1}{a^2},$$ if there are no mistakes, and all my calculations can be justified one gets $$e\zeta(3)-\zeta(6)=2e^2\int_1^2\frac{(3\zeta'(3x)-\zeta(3x))\sinh (x)}{e^{2x}-1}dx,$$ where $\zeta(s)$ is the Riemann Zeta function

Question. Was right my identity? What's about a generalization for which my identity is a case of your generalization (you can take different specialization or parameters... to write your generalization of my integral)?

Thanks in advance.

Answer 1

If I have understood correctly, essentially you want to know if holds $$2e^{2}\sum_{n\geq1}\left(3\log\left(n\right)+1\right)\int_{1}^{2}\frac{n^{-3x}\sinh\left(x\right)}{e^{2x}-1}dx=2e^{2}\int_{1}^{2}\sum_{n\geq1}\frac{\left(3\log\left(n\right)+1\right)}{n^{3x}}\frac{\sinh\left(x\right)}{e^{2x}-1}dx. $$ Since $1<x<2$ for all $N\in\mathbb{N} $ holds $$\sum_{n\leq N}\frac{\left(3\log\left(n\right)+1\right)}{n^{3x}}<\sum_{n\geq1}\frac{\left(3\log\left(n\right)+1\right)}{n^{3x}}=-3\zeta'\left(3x\right)+\zeta\left(3x\right)<C $$ for some $C>0 $ and obviously $$\int_{1}^{2}\frac{\sinh\left(x\right)}{e^{2x}-1}dx=\frac{1}{2}\int_{1}^{2}e^{-x}=\frac{e^{-1}-e^{-2}}{2} $$ so by the Dominated Convergence Theorem we can exchange the series with the integral and so $$\begin{align} 2e^{2}\sum_{n\geq1}\left(3\log\left(n\right)+1\right)\int_{1}^{2}\frac{n^{-3x}\sinh\left(x\right)}{e^{2x}-1}dx= & 2e^{2}\int_{1}^{2}\frac{\left(-3\zeta'\left(3x\right)+\zeta\left(3x\right)\right)\sinh\left(x\right)}{e^{2x}-1}dx \\ = & e\zeta\left(3\right)-\zeta\left(6\right). \end{align}$$

For a generalization, We can take every $n^{b},\,b>1$ and with the same calculations we get $$2e^{2}\int_{1}^{2}\frac{\left(-b\zeta'\left(bx\right)+\zeta\left(bx\right)\right)\sinh\left(x\right)}{e^{2x}-1}dx=e\zeta(b)-\zeta(2b).$$ If we take the $p_{n}^{b},\,b>1$ where $p_{n}$ are prime numbers we get $$2e^{2}\int_{1}^{2}\frac{\left(-bP'\left(bx\right)+P\left(bx\right)\right)\sinh\left(x\right)}{e^{2x}-1}dx=eP(b)-P(2b)$$ where $P(s)$ is the prime zeta function.

Answer 2

If we return to your original integral, $$ \int_1^2 \frac{a^{-x}sinh(x)}{e^{2x}-1}dx$$ using that $$\frac{sinh(x)}{e^{2x}-1}=\frac{1}{2}e^{-x}$$ we have that your integral is equal to: $$\frac{1}{2} \int_1^2 a^{-x}e^{-x} dx = \frac{1}{2} \int_1^2 (ea)^{-x} dx.$$ Which has elementary antiderivative, $$\frac{-1}{2} \frac{(ea)^{-x}}{ln(ae)}|_1^2=\frac{-(ea)^{-2}}{2(ln(a)+1)}-\frac{-(ea)^{-1}}{2(ln(a)+1)}.$$ Simplifying a little more gives, $$\frac{-1}{2ea(ln(a)+1)}(\frac{1}{ea}-1).$$ I'm not entirely sure how the Riemann zeta function became involve. I would be very interested to see the details of "if there are no mistakes, and all my calculations can be justified one gets".