(9) where γ 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 not containing any integer <0 (Weierstrass’ formula).
The function Γ(x) is indefinitely differentiable on ]0,+∞[ and one has
Γ(x)Γ′(x)=−γ−x1+n=1∑∞(n1−x+n1)
and
Dk(logΓ(x))=n=0∑∞(x+n)k(−1)k(k−1)!fork≥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.
Indeed, the proof of prop. 1 of VII, p. 306, shows that
Γ(x)=n→∞limx(x+1)…(x+n−1)nx(n−1)!
whence Gauss’ formula, since x+nn tends to 1 as n tends to +∞. One can also write
logn−1n=n−11+(logn−1n−n−11).
so (in the notation of prop. 1)
exp(un(x))=e(logn−1n−n−11)1+n−1xex/(n−1)
and the series with general term logn−1n−n−11 is absolutely convergent and has sum −γ, where γ denotes Euler’s constant (V, p. 242), whence we obtain Weierstrass’ formula.
For ∣x∣≤a one has ∣1/(x+n)k∣≤1/(n−a)k once n>a, so the series on the right-hand side in formula (11) is absolutely and uniformly convergent on every compact interval of R not containing any integer ≤0, for any integer k≥2; the same argument applies to the right-hand side of (10), since n1−x+n1≤n(n−a)a for ∣x∣≤a and n>a. Since these series are obtained by differentiating the series
logΓ(x)=−γx−logx+n=1∑∞(nx−log(1+nx))
term-by-term, and this converges for every x>0, the series with general term nx−log(1+nx) 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, the expression nx−log(1+nx) 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.
The function Γ(x), defined for x>0, can be extended to the whole set of points x different from the integers ≤0 so as to satisfy equation (7) of VII, p. 307, on this set: it suffices, for −(n+1)<x<−n, to put
Γ(x)=x(x+1)…(x+n)1Γ(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)∣) replacing Dk(logΓ(x)) in (11), remain valid on this set. Formula (9) (VII, p. 307) shows that Γ(x)∼1/x as x tends to 0, whence, by (7) of VII, p. 307,
Γ(x)∼n!(x+n)(−1)n
as x tends to −n (n an integer ≥0). The function 1/Γ(x) can then be extended by continuity to all of R, assigning it the value 0 at integers ≤0; then, for all x∈R
Γ(x)1=n→∞limnnn!x(x+1)…(x+n)
and
Γ(x)1=ey1xxn=1∏∞(1+nx)e−y1/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.
Since Γ(x)>0 for x>0, equation (7) of VII, p. 307, shows that Γ(x)<0 for −(2n−1)<x<−(2n−2) and Γ(x)>0 for
−2n<x<−(2n−1)
(n an integer ≥1); also Γ(x) has right limit +∞ at the points −2n and −∞ at the points −(2n+1), and has left limit −∞ at the points −2n and +∞ at the points −(2n+1) (for all n∈N). Formula (11) of VII, p. 308, shows that, for k=2, the right-hand side is always ≥0 when it is defined, so
Γ′′(x)Γ(x)−(Γ′(x))2≥0,
and consequently Γ′′(x) has the same sign as Γ(x); thus Γ is convex for x>0 and for −(2n+2)<x<−(2n+1), and concave for −(2n+1)<x−2n(n∈N); one deduces from this that, on the intervals where Γ is convex, Γ′(x) increases from −∞ to +∞, and on the intervals where Γ is concave, Γ′(x) decreases from +∞ to −∞. Whence the graph of Γ (fig. 1).
Fig. 1
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.