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!

Answer 1

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: $$ \int_{0}^{\pi/2}\log\sin t\,dt = -\frac{\pi}{2}\log 2.$$

Answer 2

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.