# expressing $\log(\left \lfloor x \right \rfloor!)$ in terms of zeta-zeros

let $\psi(x)$ be the second Chebyshev Function. By the definition of this summatory function, and the fundamental theorem of arithmetic, we have the identity: $$\log(\left \lfloor x \right \rfloor!)=\sum_{n=1}^{\infty}\psi\left(\frac{x}{n}\right)$$

Is there a way to utilize the well-known explicit formula :

$$\psi(x)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}-\log(2\pi)-\frac{1}{2}\log\left(1-x^{-2}\right)\;\;\;\;\;\zeta(\rho)=0\;(0<\Re(\rho)<1)$$ to express $\log(\left \lfloor x \right \rfloor!)$ in terms of zeta-zeros ($\rho$)!?

please notice that $\psi\left(\frac{x}{n}\right)=0$ for $n>\frac{x}{2}$. Thus, the summation might be truncated at $\left \lfloor \frac{x}{2} \right \rfloor$ .

• Apart from substituting your second formula into the first, you mean? I suspect this isn't a route to understanding much, because in a naive expansion the four terms of the explicit formula for $\psi$ would give you four divergent series in the expression for $\log (\lfloor x \rfloor ! )$; and the role of the zeta-zeros is pretty well obscured by the need to balance all the series to reach a finite sum. A further problem is that $(1 - (x/n)^{-2})$ is negative when $n$ is large, for any real $x$, so you will meet issues over the complex logarithm. – HTFB Nov 27 '13 at 10:58
• I agree with HTFB. At the very least you should manually truncate the sum at $n=(\log x)/\log 2$. The resulting formula will be valid but, I suspect, unenlightening. – Greg Martin Nov 27 '13 at 17:17