A question on the approximation of $\ln(x!)$ for small x

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.

• Using $\zeta(2n)=(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$ and $\sum_{n=0}^\infty \frac{B_n}{n!} x^n = \frac{x}{e^x-1}$ you obtain $f(x)=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2n}x^{2n}=\frac12 (\frac{2\pi ix}{e^{2\pi ix}-1}+ \frac{-2\pi ix}{e^{-2\pi ix}-1})$ and you can compare $f(x)/x$ with the derivative of $\frac{1}{2}\ln\left(\frac{\pi x}{\sin\pi x}\right)$ – reuns Jan 6 at 21:15
• Using $\sum\limits_{n=1}^{\infty}\frac{B_{2n}}{(2n)!}x^{2n}=\frac{x}{e^x-1}+\frac{x}{2}-1$, I got $\sum\limits_{n=1}^{\infty}(-1)^n\frac{B_{2n}}{(2n)!}(2\pi x)^{2n}=\frac{1}{2}[\frac{2\pi\iota x}{e^{2\pi\iota x}-1}+\frac{-2\pi\iota x}{e^{-2\pi\iota x}-1}]-1$. When zeta function is expressed in terms of Bernoulli numbers, $f(x)=\frac{-1}{2}\sum\limits_{n=1}^{\infty}(-1)^n\frac{B_{2n}}{(2n)! (2n)}(2\pi x)^{2n}$. $2n$ appears in the denomiantor. How to proceed? – Asit Srivastava Jan 6 at 21:56
• Differentiate both side of $\frac{1}{2}\ln\left(\frac{\pi x}{\sin\pi x}\right)=\sum\limits_{n=1}^{\infty}\frac{\zeta(2n)}{2n}x^{2n}$ – reuns Jan 6 at 22:19
• Thanks. It worked. – Asit Srivastava Jan 6 at 22:51