In a question of Arfken and Weber's Mathematical Methods for Physicists,
For small values of $x$, $\ln(x!)=-\gamma x+\sum\limits_{n=2}^{\infty}(-1)^n\frac{\zeta(n)}{n}x^n$, where the symbols used have their usual meanings.
Now I need to prove that this can also be written as $$\ln(x!)=\frac{1}{2}\ln\left(\frac{\pi x}{\sin\pi x}\right)-\gamma x-\sum\limits_{n=1}^{\infty}\frac{\zeta(2n+1)}{2n+1}x^{2n+1}$$
I can only do this.
$$\sum\limits_{n=2}^{\infty}(-1)^n\frac{\zeta(n)}{n}x^n=\\ \left(\frac{\zeta(2)}{2}x^2+\frac{\zeta(4)}{4}x^4+\frac{\zeta(6)}{6}x^6+...\right)-\left(\frac{\zeta(3)}{3}x^3+\frac{\zeta(5)}{5}x^5+\frac{\zeta(7)}{7}x^7+...\right) =\\\sum\limits_{n=1}^{\infty}\frac{\zeta(2n)}{2n}x^{2n}-\sum\limits_{n=1}^{\infty}\frac{\zeta(2n+1)}{2n+1}x^{2n+1}$$
Putting this in the original equation and comparing it with what needs to be proven, it can be concluded that
$$\frac{1}{2}\ln\left(\frac{\pi x}{\sin\pi x}\right)=\sum\limits_{n=1}^{\infty}\frac{\zeta(2n)}{2n}x^{2n}$$
I don't know how to prove this.
$\frac{\pi x}{\sin\pi x}=x\Gamma(x)\Gamma(1-x)$ I think this has to do something with it since gamma function and zeta function are related to each other through an integral.