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 - $\rho$ - !?

  • $\begingroup$ In what way is plugging in the explicit formula into the sum not already an expression in terms of zeta zeros? $\endgroup$ Nov 26, 2013 at 0:04
  • $\begingroup$ because the sum of the individual terms is divergent. for instance, the sum of the $\frac{x}{n}$ term is the divergent harmonic series. $\endgroup$ Nov 26, 2013 at 0:09
  • $\begingroup$ So, to clarify, you want to reverse the order of nested summation and still get something meaningful? $\endgroup$ Nov 26, 2013 at 1:26
  • $\begingroup$ For $n > x$ we have $\psi(n/x) = 0$ so can't you restrict the sum to go only up to $x$ and then use an Euler-Maclaurin type formula to compute $\sum_{j = 1}^{x} n^{-\rho}$ ? Note that $\sum_{\rho} x^{\rho} \rho^{-1}$ is only conditionally convergent, so I would use a version of the explicit formula with an error term and the sum over $\rho$ restricted to $|\rho| < T$ with some parameter $T$ depending on $x$. But yeah, certainly one can find the formula you want. $\endgroup$
    – blabler
    Nov 26, 2013 at 3:00
  • 1
    $\begingroup$ Let me add that the whole thing trivializes after the computation and the zeros disappear, or more precisely they only contribute a constant $\sum_{\rho} \rho^{-1} (1 - \rho)^{-1}$ which can be computed directly by other means. $\endgroup$
    – blabler
    Nov 26, 2013 at 3:08

1 Answer 1


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.

  • $\begingroup$ Very nice! Let me spell out the connection to the formula in the OP. We have $\psi(x) = \sum_{n \geq 1} \mu(n) \log(\lfloor x/n\rfloor !) = - \frac{1}{2 \pi i} \int_{2 - i \infty}^{2+i \infty} \frac{\zeta'(s) ds}{s} \sum_{n=1} \mu(n)\frac{x^s}{n^s} = - \frac{1}{2 \pi i} \int_{2 - i \infty}^{2+i \infty} \frac{\zeta'(s) ds}{\zeta(s) s}$. The formula for $\psi$ as a sum over the roots comes from the $\zeta(s)$ in the denominator of this formula. You are saying that, if you want $\log( \lfloor x \rfloor !)$ in the first place, then doing and undoing this Mobius summation is a bad idea. $\endgroup$ Oct 1, 2014 at 13:03

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