Bourbaki

Proposition 2

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

Functions of a Real Variable, chapter VII, § 1. THE GAMMA FUNCTION IN THE REAL DOMAIN

For every x>0x > 0 one has

Γ(x)=lim⁡n→∞nxn!x(x+1)…(x+n) \Gamma(x) = \lim_{n \to \infty} \frac{n^x n!}{x(x+1)\ldots(x+n)}
(Gauss' formula), and
Γ(x)=e−γx1x∏n=1∞ex/n1+xn \Gamma(x) = e^{-\gamma x} \frac{1}{x} \prod_{n=1}^{\infty} \frac{e^{x/n}}{1 + \frac{x}{n}}
(9) where γ\gamma denotes Euler’s constant, and the infinite product on the right hand side of (9) is absolutely and uniformly convergent on every compact interval of R\mathbf{R} not containing any integer <0< 0 (Weierstrass’ formula).

The function Γ(x)\Gamma(x) is indefinitely differentiable on ]0,+∞[]0, +\infty[ and one has

Γ′(x)Γ(x)=−γ−1x+∑n=1∞(1n−1x+n) \frac{\Gamma'(x)}{\Gamma(x)} = -\gamma - \frac{1}{x} + \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{x+n} \right)

and

Dk(log⁡Γ(x))=∑n=0∞(−1)k(k−1)!(x+n)kfork≥2, \mathrm{D}^k (\log \Gamma(x)) = \sum_{n=0}^{\infty} \frac{(-1)^k (k-1)!}{(x+n)^k} \quad \text{for} \quad k \geq 2,

the series on the right-hand sides of (10) and (11) being absolutely and uniformly convergent on every compact interval not containing any integer ≤0\leq 0.

Indeed, the proof of prop. 1 of VII, p. 306, shows that

Γ(x)=lim⁡n→∞nx(n−1)!x(x+1)…(x+n−1) \Gamma(x) = \lim_{n \to \infty} \frac{n^x (n-1)!}{x(x+1)\ldots(x+n-1)}

whence Gauss’ formula, since nx+n\frac{n}{x+n} tends to 1 as nn tends to +∞+\infty. One can also write

log⁡nn−1=1n−1+(log⁡nn−1−1n−1). \log \frac{n}{n-1} = \frac{1}{n-1} + \left( \log \frac{n}{n-1} - \frac{1}{n-1} \right).

so (in the notation of prop. 1)

exp⁡(un(x))=e(log⁡nn−1−1n−1)ex/(n−1)1+xn−1 \exp(u_n(x)) = e^{(\log \frac{n}{n-1} - \frac{1}{n-1})} \frac{e^{x/(n-1)}}{1 + \frac{x}{n-1}}

and the series with general term log⁡nn−1−1n−1\log \frac{n}{n-1} - \frac{1}{n-1} is absolutely convergent and has sum −γ-\gamma, where γ\gamma denotes Euler’s constant (V, p. 242), whence we obtain Weierstrass’ formula.

For ∣x∣≤a|x| \leq a one has ∣1/(x+n)k∣≤1/(n−a)k|1/(x+n)^k| \leq 1/(n-a)^k once n>an > a, so the series on the right-hand side in formula (11) is absolutely and uniformly convergent on every compact interval of R\mathbf{R} not containing any integer ≤0\leq 0, for any integer k≥2k \geq 2; the same argument applies to the right-hand side of (10), since ∣1n−1x+n∣≤an(n−a)\left| \frac{1}{n} - \frac{1}{x+n} \right| \leq \frac{a}{n(n-a)} for ∣x∣≤a|x| \leq a and n>an > a. Since these series are obtained by differentiating the series

log⁡Γ(x)=−γx−log⁡x+∑n=1∞(xn−log⁡(1+xn)) \log \Gamma(x) = -\gamma x - \log x + \sum_{n=1}^{\infty} \left( \frac{x}{n} - \log \left( 1 + \frac{x}{n} \right) \right)

term-by-term, and this converges for every x>0x > 0, the series with general term xn−log⁡(1+xn)\frac{x}{n} - \log \left( 1 + \frac{x}{n} \right) is absolutely and uniformly convergent on every compact interval contained in [0, +∞[, and one has the relations (10) and (11) of VII, p. 308, for every x > 0 (II, p. 52, th. 1). Moreover, for every x ∈ R\mathbf{R}, the expression xn−log⁡(1+xn)\frac{x}{n} - \log \left( 1 + \frac{x}{n} \right) is defined once n is large enough, so th. 1 of II, p. 52, again shows that the infinite product in the right-hand side of (9) (VII, p. 307) is absolutely and uniformly convergent on every compact interval containing no integer ≤0\leq 0.

The function Γ(x)\Gamma(x), defined for x>0x > 0, can be extended to the whole set of points x different from the integers ≤0\leq 0 so as to satisfy equation (7) of VII, p. 307, on this set: it suffices, for −(n+1)<x<−n-(n+1) < x < -n, to put

Γ(x)=1x(x+1)…(x+n)Γ(x+n+1). \Gamma(x) = \frac{1}{x(x+1)\ldots(x+n)} \Gamma(x+n+1).
By prop. 2 of VII, p. 307, the formulae (8), (9), (10) and (11) of VII, p. 307 and 308, with Dk(log⁡∣Γ(x)∣)D^k(\log |\Gamma(x)|) replacing Dk(log⁡Γ(x))D^k(\log \Gamma(x)) in (11), remain valid on this set. Formula (9) (VII, p. 307) shows that Γ(x)∼1/x\Gamma(x) \sim 1/x as x tends to 0, whence, by (7) of VII, p. 307,
Γ(x)∼(−1)nn!(x+n) \Gamma(x) \sim \frac{(-1)^n}{n! (x+n)}
as x tends to −n-n (nn an integer ≥0\geq 0). The function 1/Γ(x)1/\Gamma(x) can then be extended by continuity to all of R\mathbf{R}, assigning it the value 0 at integers ≤0\leq 0; then, for all x∈Rx \in \mathbf{R}
1Γ(x)=lim⁡n→∞x(x+1)…(x+n)nnn! \frac{1}{\Gamma(x)} = \lim_{n \to \infty} \frac{x(x+1)\ldots(x+n)}{n^n n!}
and
1Γ(x)=ey1xx∏n=1∞(1+xn)e−y1/n \frac{1}{\Gamma(x)} = e^{y_1 x} x \prod_{n=1}^{\infty} \left( 1 + \frac{x}{n} \right) e^{-y_1/n}
and one shows as in prop. 2 of VII, p. 307, that the infinite product on the right of (13) is absolutely and uniformly convergent on every compact interval of R\mathbf{R}.

Since Γ(x)>0\Gamma(x) > 0 for x>0x > 0, equation (7) of VII, p. 307, shows that Γ(x)<0\Gamma(x) < 0 for −(2n−1)<x<−(2n−2)-(2n-1) < x < -(2n-2) and Γ(x)>0\Gamma(x) > 0 for

−2n<x<−(2n−1) -2n < x < -(2n-1)
(nn an integer ≥1\geq 1); also Γ(x)\Gamma(x) has right limit +∞+\infty at the points −2n-2n and −∞-\infty at the points −(2n+1)-(2n+1), and has left limit −∞-\infty at the points −2n-2n and +∞+\infty at the points −(2n+1)-(2n+1) (for all n∈Nn \in \mathbf{N}). Formula (11) of VII, p. 308, shows that, for k=2k = 2, the right-hand side is always ≥0\geq 0 when it is defined, so
Γ′′(x)Γ(x)−(Γ′(x))2≥0, \Gamma''(x) \Gamma(x) - (\Gamma'(x))^2 \geq 0,

and consequently Γ′′(x)\Gamma''(x) has the same sign as Γ(x)\Gamma(x); thus Γ\Gamma is convex for x>0x > 0 and for −(2n+2)<x<−(2n+1)-(2n+2) < x < -(2n+1), and concave for −(2n+1)<x−2n (n∈N)-(2n+1) < x - 2n \ (n \in \mathbf{N}); one deduces from this that, on the intervals where Γ\Gamma is convex, Γ′(x)\Gamma'(x) increases from −∞-\infty to +∞+\infty, and on the intervals where Γ\Gamma is concave, Γ′(x)\Gamma'(x) decreases from +∞+\infty to −∞-\infty. Whence the graph of Γ\Gamma (fig. 1).

Fig. 1

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

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Cites

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