A recent question on the notion and notation of multiplicative integrals
( What is the standard notation for a multiplicative integral? ) induced me to play with the Riemann products of the Gamma function, in order to evaluate the multiplicative integral of $\Gamma(x)$, exploiting the multiplicative formula. I will, however, put the question mainly in terms of a standard integral; and I will also use the factorial function $x!=\Gamma(x+1)$ instead (that seems to be more appreciated here). Consider the multiplicative formula for $x!$:
$$x!=(2\pi)^{-\frac{m-1}{2}}\, m^{x+\frac{1}{2}}\, \Big( \frac{x}{m} \Big)!\,\Big( \frac{x-1}{m} \Big)!\dots \Big( \frac{x-m+1}{m} \Big)!\, \,$$
For $x=m\in\mathbb{N}$ we get, using the Stirling asymptotics for $m!$:
$$\prod_{k=1}^{m}\Big(\frac{k}{m} \Big)!\sim (2\pi)^{\frac{m}{2}}e^{-m} $$
Take a logarithm; divide by $m$ and let $m\to\infty$: one finds
$$\int_0^1\log(x!)\, dx=\frac{1}{2}\log(2\pi )-1,$$
or, as a multiplicative integral
$$\prod_0^1 (x!\, dx)=\frac{\sqrt{2\pi}}{e}.$$
Now the question: How to evaluate the above integral
by means of standard integral
calculus?
I guess it's feasible, but how? Otherwise, it would be a remarkable case of an integral that one can only (edit: or say "more easily") evaluate directly from the definition of Riemann sums, like one does e.g. with $x^2$ in introductory calculus courses.
Best Answer
Here a start:
We have the reflection formula $$z! (1-z)! = \frac{\pi z (1-z)}{\sin (\pi z)}.$$ Taking $\log$'s, $$\log (z!) + \log ((1-z)!) = \log \pi + \log z + \log (1-z) - \log \sin (\pi z).$$ Split our integral in half and rearrange it $$\int_0^1 \log (z!) \ dz = \int_0^{1/2} \left( \log (z!) + \log ((1-z)!) \right) dz.$$
So we have three elementary integrals to deal with, plus $$\int_0^{1/2} \log \sin(\pi z) \ dz. \quad (*)$$ According to Mathematica, $(*) = - \log(2)/2$. So, if we can find a clean proof of this fact, we will have evaluated the integral. This may be difficult, because the indefinite integral $\int \log \sin(\pi z) \ dz$ involves dilogarithms. To me, $(*)$ looks like a good target for residues. Anyone want to finish it off?
Edit: Here is a way to calculate $(*)$: Denote $I=\int_{0}^{\pi/2}\log(\sin x) \ dx=\int_{0}^{\pi/2}\log(\cos x) \ dx$ and so $$2I=\int_{0}^{\pi/2}\log(\frac{\sin 2x}{2}) \ dx=\int_{0}^{\pi/2}\log(\sin 2x) \ dx-\frac{\pi \log 2}{2}=I-\frac{\pi \log 2}{2}$$