Proposition 3

Tag 08JE, permanent. Cite it as [Bourbaki, Tag 08JE].

Functions of a Real Variable, chapter VII, § 2. THE GAMMA FUNCTION IN THE COMPLEX DOMAIN

For every real $x > 0$ one has

\int_1^{x+1} \log \Gamma(t)\ dt = x(\log x - 1) + \frac{1}{2} \log 2\pi

(Raabe’s integral).

First we establish formula (13) for $x = 0$ . Since $\log \Gamma(x) \sim \log \frac{1}{x}$ as $x$ tends to 0, the integral $\int_0^1 \log \Gamma(x)\ dx$ converges. Further, the function $\log \Gamma(x)$ decreases on $]0, 1]$ (VII, p. 310); for every $\alpha > 0$ one thus has

\frac{1}{n} \sum_{k=1}^q \log \Gamma \left( \frac{k}{n} \right) \leqslant \int_0^\alpha \log \Gamma(x)\ dx,

$q$ being the largest integer such that $q/n \leqslant \alpha$ . Since $\int_0^\alpha \log \Gamma(x) dx$ tends to 0 with $\alpha$ and also $\frac{1}{n} \sum_{k=q+1}^n \log \Gamma \left( \frac{k}{n} \right)$ tends to $\int_\alpha^1 \log \Gamma(x) dx$ as $n$ tends to $+\infty$ (II, p. 57, prop. 5) one has

\int_0^1 \log \Gamma(x) dx = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \log \Gamma \left( \frac{k}{n} \right).

But, by (12), the right-hand side of this formula is the limit of

\frac{n-1}{2n} \log 2\pi - \frac{1}{2} \frac{\log n}{n},

whence

\int_0^1 \log \Gamma(x) dx = \frac{1}{2} \log 2\pi.

(14)

Next we remark that from the identity

\log \Gamma(x+1) = \log \Gamma(x) + \log x

one deduces, on integrating, that for $x > 0$

\int_0^x \log \Gamma(t+1) dt = \int_0^x \log \Gamma(t) dt + \int_0^x \log t dt.

But the integral on the left-hand side is also equal to $\int_1^{x+1} \log \Gamma(t) dt$ . Thus, by (14),

\int_1^{x+1} \log \Gamma(t) dt = \int_0^x \log t dt + \frac{1}{2} \log 2\pi = x (\log x - 1) + \frac{1}{2} \log 2\pi.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Proposition 3

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