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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.

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    $\begingroup$ 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)? $\endgroup$ Jul 22, 2010 at 14:35
  • $\begingroup$ 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 $\frac{1}{2}\log(2\pi)-1.$ $\endgroup$ Jul 23, 2010 at 21:38

3 Answers 3

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

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    $\begingroup$ $\int_0^{1} \log \sin(\pi z) \ dz=-\log (2)$ is calculated by the method of residues in this note math.ucsb.edu/~mckernan/Teaching/02-03/Autumn/202A/l_22.pdf $\endgroup$ Jul 22, 2010 at 18:10
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    $\begingroup$ 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. $\endgroup$ Jul 22, 2010 at 23:28
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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.

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  • $\begingroup$ @Anixx: can you show me how to obtain the infinite integral of the Gamma function? $\int \Gamma(x){dx}=C e^{\psi(-2,x)-\frac {x}{2}\ln 2\pi}$ $\endgroup$ Oct 9, 2011 at 15:44
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    $\begingroup$ It's multiplicative integral, so the dx should be in superscript (or you should indicate it's multiplicative integral otherwise). $\endgroup$
    – Anixx
    Feb 19, 2012 at 6:13
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$$\int_0^1 \log\Gamma(x)\mathrm dx=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \log\Gamma\bigg(\frac{k}{n}\bigg)=\lim_{n\to\infty}\frac{1}{n}\log\bigg(\prod_{k=1}^n \Gamma\bigg(\frac{k}{n}\bigg)\bigg)=\lim_{n\to \infty}\frac{1}{n}((\sqrt{2\pi})^{n-1}n^{-1/2})=\frac{1}{2}\ln(2\pi)$$Here, Gauss Multiplication Theorem is used for evaluation of the product.

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    $\begingroup$ Thank you, this is the computation I made in the question $\endgroup$ Oct 6, 2021 at 7:43

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