How to prove Raabe's Formula

Question

For quite some time, I've been trying to prove Raabe's Formula, or in other words:

$$\int_a^{a+1} \ln\bigg(\Gamma(t)\bigg)dt=\dfrac{1}{2}\ln(2\pi)+a\ln(a)-a$$

This is how I tried: $$I(s)=\int_a^{a+1}\ln\bigg(s\Gamma(t)\bigg)dt$$Differentiating with respect to $s,$ $$I'(s)=\int_a^{a+1}\dfrac{\Gamma(t)}{s\Gamma(t)}dt=\int_a^{a+1} \dfrac{dt}{s}$$ However, at this point I stopped thinking I must have made a mistake because I was told that proving Raabe's Formula was really difficult, and this seemed too simple a method to prove Raabe's Formula. $$$$ I would be grateful if somebody would be so kind as to tell me how to prove this result, as well as what went wrong with my method. Many, many thanks in advance!

This seems correct to me at first glance, and appears to lead you to $I^\prime(s)=\frac{1}{s}$; how do you continue from there? — Clement C., Jun 25 '15 at 12:59
Sir, integrating $I'(s)=\int_a^{a+1}\dfrac{dt}{s},$ we get $$I(s)=\dfrac{t}{s}\bigg|^{a+1}_{a}+C$$ — User1234, Jun 25 '15 at 13:02
You have $I^\prime(s) = \dfrac{1}{s} \int_a^{a+1} dt = \dfrac{1}{s} \cdot 1 = \dfrac{1}{s}$. Now, you integrate with regard to $s$ to get $I(s) = I(\alpha) + \ln \dfrac{s}{\alpha}$, with which point $\alpha$? — Clement C., Jun 25 '15 at 13:03
Sir, could you please use dfrac?I'm unable toread the fractions:( — User1234, Jun 25 '15 at 13:06
Sir $\alpha =1,$ right Sir? Sir, could you please tell me what to do next? — User1234, Jun 25 '15 at 13:14
So $I(s) = I(1) + \ln s$. But then, since you want to compute $I(1)$ (i.e., take $s=1$ at the end), this just proves the tautology $I(1)=I(1) + \ln 1$. — Clement C., Jun 25 '15 at 13:20
But Sir then we can't use my method to finally evaluate $\int_a^{a+1}\ln(\Gamma(x))dx$, right? — User1234, Jun 25 '15 at 13:33
That's the point, yes -- your method does not give the desired result. — Clement C., Jun 25 '15 at 13:33
Proven in addendum to this answer: http://math.stackexchange.com/questions/1260270/evaluating-lim-limits-n-rightarrow-infty-frac1n2-ln-left-fracnn/1260315#1260315 — Ron Gordon, Jun 25 '15 at 13:47

Jack D'Aurizio · Accepted Answer · 2015-06-25T12:55:35.810

5

Who told you that the proof is difficult lied.

The proof is simple: take the integral as the limit of the Riemann sums

$$\int_{a}^{a+1}\log\Gamma(t)\,dt = \lim_{n\to +\infty}\sum_{k=0}^{n-1}\log\Gamma\left(a+\frac{k}{n}\right) = \lim_{n\to +\infty}\frac{1}{n}\log\prod_{k=0}^{n-1}\Gamma\left(a+\frac{k}{n}\right)$$ then apply the multiplication theorem for the $\Gamma$ function to the RHS.

This is one of the few cases in which it is faster to compute a Riemann integral from its definition, rather than from the fundamental theorem of Calculus or from integration by parts. Not by chance, the same happens for: $$ \int_{0}^{\pi/2}\log\sin t\,dt = -\frac{\pi}{2}\log 2.$$

edited Jun 25 '15 at 12:55

answered Jun 25 '15 at 12:50

Jack D'Aurizio

353,855

Thanks a lot Sir! Sir, could the proof also be done the way I've tried it? – User1234 Jun 25 '15 at 12:54
1

You may use differentiation under the integral sign if you define $I(s)$ as $\int_{a}^{a+1}\log\Gamma(st),dt.$ Then you have $I'(s)=\int_{a}^{a+1} t\cdot\psi(st),dt $, but that is the same as integrating by parts. What you did is just to take $I(s)$ as $\log(s)+I(1)$, not really useful. – Jack D'Aurizio Jun 25 '15 at 13:00
@JackD'Aurizio, please can you explain what do you mean by the paragraph in end? Great answer by the way! – Aditya Agarwal Dec 30 '15 at 08:22
@AdityaAgarwal: just this: we are used to compute integrals through the fundamental theorem of calculus, rather than through the definition of the Riemann integral. However, in some special cases (like the ones depicted above) to go through the definition is a faster way. – Jack D'Aurizio Dec 30 '15 at 09:33
Ohkk, can we use $\Gamma$ function to compute the integral in the end? – Aditya Agarwal Dec 30 '15 at 09:39
@AdityaAgarwal: well, sure, since $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$. – Jack D'Aurizio Dec 30 '15 at 09:40
Yea, but aint it only for $0<x<1$? – Aditya Agarwal Dec 30 '15 at 09:42
@AdityaAgarwal: such functional identity holds for every $x\not\in\mathbb{Z}$. – Jack D'Aurizio Dec 30 '15 at 11:11
Okay, and how would we integrate $\log\sin t$ as a limit of a sum? Because the first term gives $\log 0$. – Aditya Agarwal Dec 31 '15 at 04:34
This (well-known) method seems work only if $a=0$. – Liam Aug 31 '23 at 08:23

score 3 · Answer 2 · answered Jun 25 '15 at 14:31

First of all note that $$\int_{0}^{1}\log\left(\Gamma\left(x\right)\right)dx=\log\left(\sqrt{2\pi}\right). $$ In fact by Euler reflection formula we have $$\int_{0}^{1}\log\left(\Gamma\left(x\right)\right)dx=\frac{1}{2}\left(\log\left(\pi\right)-\int_{0}^{1}\log\left(\sin\left(\pi x\right)\right)dx\right)=\frac{1}{2}\left(\log\left(\pi\right)-\frac{1}{\pi}\int_{0}^{\pi}\log\left(\sin\left(x\right)\right)dx\right) $$ and the last integral can be calculated using the formula $$\int_{a}^{b}\log\left(\sin\left(x\right)\right)dx=-\left(b-a\right)\log\left(2\right)-\sum_{k\geq1}\frac{\sin\left(2kb\right)-\sin\left(2ka\right)}{2k^{2}}. $$ Now consider $$F\left(a\right)=\int_{a}^{a+1}\log\left(\Gamma\left(x\right)\right)dx $$ then by the fundamental theorem of calculus we have $$F'\left(a\right)=\log\left(\Gamma\left(a+1\right)\right)-\log\left(\Gamma\left(a\right)\right)=\log\left(a\right) $$ then $$F\left(a\right)=\int F'\left(a\right)da=a\log\left(a\right)-a+c $$ and now if we take $a=0 $ we find (remeber that $\lim_{a\rightarrow0}a\log\left(a\right)=0 $), using the first part, that $$c=\log\left(\sqrt{2\pi}\right) $$ and this complete the proof.

Wouldn't we need Raabe's formula to know the value of $\int_0^1 \log\left(\Gamma(x) \right)dx$? Seems a bit circular — Sam, Nov 01 '22 at 13:25
@SameerAbbas Why? See my anser, you only need the Euler's reflection formula and how to integrate $\log(\sin(x))$, which can be done using its Fourier expansion. — Marco Cantarini, Nov 01 '22 at 17:11

How to prove Raabe's Formula

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