How to prove $\displaystyle\prod_{n=1}^{\infty}n^{\frac{(-1)^{n+1}}{n^2}}=\left(\frac{A^{12}}{4\pi e^{\gamma}}\right)^{\eta(2)}$

Question

In this page of section 8 given

$$\large\prod_{n=1}^{\infty}n^{\frac{1}{n^2}}=\left(\frac{A^{12}}{2\pi e^{\gamma}}\right)^{\zeta(2)}\tag8$$

I was able to derive $(1)$ via guessing $$\large\prod_{n=1}^{\infty}n^{\frac{(-1)^{n+1}}{n^2}}=\left(\frac{A^{12}}{4\pi e^{\gamma}}\right)^{\eta(2)}\tag1$$

I have no idea how to prove it.

Can anyone show how to prove, if $(1)$ is correct or not.

Where $\eta(2)=\dfrac{\pi^2}{12}$

Answer 1

$$\prod_{n=1}^{\infty}n^{\frac{(-1)^{n+1}}{n^2}}\\ \exp\left[\ln\left(\prod_{n=1}^{\infty}n^{\frac{(-1)^{n+1}}{n^2}}\right)\right]\\ \exp\left[\sum_{n=1}^{\infty}\ln\left(n^{\frac{(-1)^{n+1}}{n^2}}\right)\right]\\ \exp\left[\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\ln\left(n\right)\right] $$ We know that $$\sum_{k=1}^{\infty}\frac{x^n}{n^s}:=\text{Li}_s(x)$$ So $$\sum_{k=1}^{\infty}\frac{x^n}{n^s}\ln(n):=-\frac{\mathrm{d}}{\mathrm{d}s}\text{Li}_s(x)\\ \sum_{k=1}^{\infty}\frac{(-1)^n}{n^s}\ln(n):=-\frac{\mathrm{d}}{\mathrm{d}s}\text{Li}_s(-1)\\ $$ Where $$\text{Li}_s(-1)=-(1 - 2^{1 - s}) \zeta(s)$$ So $$\sum_{k=1}^{\infty}\frac{(-1)^{n+1}}{n^s}\ln(n)=\frac{\mathrm{d}}{\mathrm{d}s}(1 - 2^{1 - s}) \zeta(s)=-\zeta'(s) - 2^{1 - s} (\ln(2) \zeta(s) - \zeta'(s))$$ At $s=2$ $$\sum_{k=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\ln(n)=-\zeta'(2) - 2^{1 - 2} (\ln(2) \zeta(2) - \zeta'(2))\\ =-\frac{1}{2}\zeta'(2) - \frac{\pi^2}{12}\ln(2)$$ In conclusion: $$\prod_{n=1}^{\infty}n^{\frac{(-1)^{n+1}}{n^2}}=\exp\left[\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\ln\left(n\right)\right]=\exp\left[-\frac{1}{2}\zeta'(2) - \frac{\pi^2}{12}\ln(2)\right]$$ Where
$\zeta'(2)=\zeta(2)\left(\gamma+\ln(2\pi)-12\ln(A)\right)=\dfrac{\pi^2}{6}\left(\gamma+\ln(2\pi)-12\ln(A)\right)$
So we have $$\prod_{n=1}^{\infty}n^{\frac{(-1)^{n+1}}{n^2}}=\exp\left[\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\ln\left(n\right)\right]=\exp\left[-\dfrac{\pi^2}{12}\left(\gamma+\ln(2\pi)-12\ln(A)\right) - \frac{\pi^2}{12}\ln(2)\right]\\ =\exp\left[-\frac{\pi^{2}}{12}\left(\gamma+\ln\left(4\pi\right)-12\ln(A)\right)\right]=\left(\frac{A^{12}}{4\pi e^{\gamma}}\right)^{\frac{\pi^2}{12}}$$