Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question
Pietro, are you serious about "it would be a remarkable case of an integral that one can only evaluate directly from the definition of Riemann sums"? You just need a right representation of the logarithm of gamma function... (My turn for going to bed.) Have you tried Mathematica or Maple (symbolically)? – Wadim Zudilin Jul 22 '10 at 14:35
I undersand what you mean (I didn't mean to make a strong statement). Actually, following your suggestion, if we integrate on $[0,1]$ by series with this representation: $$\log(x!)=-\gamma\, x + \sum_{k=1}^{\infty}\left(\frac{x}{k}-\log\big(1+ \frac{x}{k}\big)\right)$$ and use the Stirling formula we find again $\frac{1}{2}\log(2\pi)-1.$ – Pietro Majer Jul 23 '10 at 21:38

2 Answers 2

up vote 10 down vote accepted

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 $$2I=\int_{0}^{\pi/2}\log(\frac{\sin 2x}{2}) \ dx=\int_{0}^{\pi/2}\log(\sin 2x) \ dx-\frac{\pi \log 2}{2}=I-\frac{\pi \log 2}{2}$$

share|cite|improve this answer
$\int_0^{1} \log \sin(\pi z) \ dz=-\log (2)$ is calculated by the method of residues in this note – Andrey Rekalo Jul 22 '10 at 18:10
Thanks Andrey and Gjergij! – David Speyer Jul 22 '10 at 18:14
Nice to see an elegant reduction of the latter "non-elementary" integral. I worked it in on my way to the office in a more elementary way: after the change $t=\sqrt{\sin(\pi x)}$ the integral (up to constant) becomes $F'(y)|_{y=0}$ where $$F(y)=\int_0^1t^{1/2+y/2}(1-t)^{1/2}dt,$$ the Euler beta integral. – Wadim Zudilin Jul 22 '10 at 23:28
thanks everybody! everything very nice. – Pietro Majer Jul 23 '10 at 20:42

Multiplicative integral of gamma function is

$$\int \Gamma(x)^{dx}=C e^{\psi^{(-2)}(x)}$$

if to use the popular generalization of polygamma function $\psi^{(p)}(z)$ put forward by Grossman in 1976.

share|cite|improve this answer
@Anixx: can you show me how to obtain the infinite integral of the Gamma function? $\int \Gamma(x){dx}=C e^{\psi(-2,x)-\frac {x}{2}\ln 2\pi}$ – Nguyen Tuan Minh Oct 9 '11 at 15:44
It's multiplicative integral, so the dx should be in superscript (or you should indicate it's multiplicative integral otherwise). – Anixx Feb 19 '12 at 6:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.