expressing $\log(\left \lfloor x \right \rfloor!)$ in terms of zeta-zeros referring to a question i posted on MS,  I post it here, as I didn't get an answer:
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 - $\rho$ - !?
 A: It is conceivable that ways do exist to express $\log (\lfloor x\rfloor !)$ in an expression involving the zeros, but any apparent relationship will be superfluous because the "explicit formula" for this function comes from $\zeta'(s)$, not something involving $1/\zeta(s)$. 
When $x$ is not an integer, you have 
$$\log (\lfloor x\rfloor !)=\sum_{n\leq x}\log n=-\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}\frac{\zeta'(s)x^sds}{s}.$$
The double pole at $s=1$ gives you a residue of $x\log x-x$ and the integrand has no other poles in $\sigma>0$. Thus, you have 
$$\log (\lfloor x\rfloor !)=x\log x-x-\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{\zeta'(s)x^sds}{s}$$
for any $0<\sigma<1$. (You can't cross the imaginary axis because the integral doesn't converge when $\sigma<0$. One way to see this is to consider Stirling's approximation $$\log ((x-1)!)=(x-1/2)\log x-x +(1/2)\log 2\pi +o(1)$$
because, if it were valid to cross the imaginary axis, picking up an $O(1)$ contribution from the residue of the pole at $s=0$, we would have a contradiction). 
To evaluate the $O(\log x)$ terms coming from the integral, you have to do a little more work. I think Binet gave an alternative expression as a Laplace transform, and there are both convergent (Macdonald) and asymptotic (Stirling) series expansions. Try wikipedia as a starting point.  
