From: Robin Chapman <rjc@maths.ex.ac.uk>
Subject: Re: Integral of log of Gamma
Date: Mon, 26 Jul 1999 08:03:04 GMT
Newsgroups: sci.math
Keywords: Raabe's formula

In article <wlsufihcz1hd@forum.swarthmore.edu>,
  qqquet@hotbot.com (Leroy Quet) wrote:
> I know this must be known, but I'm wondering if I'm right.
> integral_0^m [ln(Gamma(x+1))]dx=
> m(ln(2pi)-1-m)+sum_{k=1}^m[k lnk],
> where m is a positive integer and Gamma(n)=(n-1)!.

Let's first consider
I = integral_0^1 log(Gamma(x)) dx.
Then I = integral_0^1 log(Gamma(1-x)) dx and so
2I = integral_0^1 log(Gamma(x) Gamma(1-x)) dx
   = integral_0^1 log(pi/log(sin pi x)) dx
   = log pi - integral_0^1 log(sin pi x)) dx
   = log pi + log 2
   = log (2pi)
since it's well-known that int_0^pi log(sin x) dx = - pi log 2.
Now let
I_m = integral_m^{m+1} log(Gamma(x)) dx.
Since Gamma(x+1) = x Gamma(x) it follows that
I_{m+1} = I_m + integral_m^{m+1} log x dx.
We then get for positive integers m,
I_m = I + integral_0^m log x dx
    = (1/2) log(2pi) + m log m - m.
Leroy's integral is I_1 + ... + I_m = (m/2) log(2pi) - m(m+1)/2
+ 2 log 2 + 3 log 3 + ... + m log m, so his formula is correct except
it lacks a denominator of 2 on the first bracket.

--
Robin Chapman
 http://www.maths.ex.ac.uk/~rjc/rjc.html
  "They did not have proper palms at home in Exeter."
    Peter Carey, _Oscar and Lucinda_


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From: Raymond Manzoni <raymman@club-internet.fr>
Subject: Re: Integral of log of Gamma
Date: Sun, 25 Jul 1999 23:34:12 +0200
Newsgroups: sci.math

Leroy Quet wrote:

> I know this must be known, but I'm wondering if I'm right.
> integral_0^m [ln(Gamma(x+1))]dx=
> m(ln(2pi)-1-m)+sum_{k=1}^m[k lnk],
> where m is a positive integer and Gamma(n)=(n-1)!.
> Thanks,
> Leroy Quet

    Hi,

    There's a formula (of Raabe I think) that'll probably answer your
question:

    integral_m^(m+1) [ln(Gamma(x))]dx = m*(ln(m)-1) + ln(2*Pi)/2

    So that your integral reads in fact (you simply forgot the /2 under
m):

    integral_0^m [ln(Gamma(x+1))]dx = m/2(ln(2pi)-1-m) + sum_{k=1}^m[k
lnk]

    All this looks numerically right.
    With my best feelings,

                Raymond Manzoni