For all x > 0 x > 0 x > 0
Γ ( x ) = ∫ 0 ∞ e − t t x − 1 d t
\Gamma(x) = \int_0^\infty e^{-t} t^{x-1} dt
Γ ( x ) = ∫ 0 ∞ e − t t x − 1 d t (second Euler integral).
Now the function g ( x ) = ∫ 0 ∞ e − t t x − 1 d t g(x) = \int_0^\infty e^{-t} t^{x-1} dt g ( x ) = ∫ 0 ∞ e − t t x − 1 d t x > 0 x > 0 x > 0 logarithmically convex on ]0 , + ∞ [ 0, +\infty[ 0 , + ∞ [
g ( x + 1 ) = ∫ 0 ∞ e − t t x d t = − e − t t x ∣ 0 ∞ + x ∫ 0 ∞ e − t t x − 1 d t = x g ( x ) .
g(x+1) = \int_0^\infty e^{-t} t^x \, dt = -e^{-t} t^x \Big|_0^\infty + x \int_0^\infty e^{-t} t^{x-1} dt = x \, g(x).
g ( x + 1 ) = ∫ 0 ∞ e − t t x d t = − e − t t x 0 ∞ + x ∫ 0 ∞ e − t t x − 1 d t = x g ( x ) . Other words, g g g
g ( 1 ) = ∫ 0 ∞ e − t d t = 1 ;
g(1) = \int_0^\infty e^{-t} \, dt = 1;
g ( 1 ) = ∫ 0 ∞ e − t d t = 1 ; proposition therefore follows from prop. 1 of VII, p. 306.
By the change of variable e − t = u e^{-t} = u e − t = u
Γ ( x ) = ∫ 0 1 ( log 1 t ) x − 1 d t .
\Gamma(x) = \int_0^1 \left( \log \frac{1}{t} \right)^{x-1} dt.
Γ ( x ) = ∫ 0 1 ( log t 1 ) x − 1 d t . Similarly, from the change of variable u = t 1 / x u = t^{1/x} u = t 1/ x
x Γ ( x ) = ∫ 0 ∞ e − t 1 / x d t
x \, \Gamma(x) = \int_0^\infty e^{-t^{1/x}} \, dt
x Γ ( x ) = ∫ 0 ∞ e − t 1/ x d t again, taking account of (7) (VII, p. 3),
Γ ( 1 + 1 x ) = ∫ 0 ∞ e − t x d t
\Gamma \left( 1 + \frac{1}{x} \right) = \int_0^\infty e^{-t^x} \, dt
Γ ( 1 + x 1 ) = ∫ 0 ∞ e − t x d t In particular, for x = 2 x = 2 x = 2
Γ ( 3 2 ) = 1 2 Γ ( 1 2 ) = ∫ 0 ∞ e − t 2 d t .
\Gamma \left( \frac{3}{2} \right) = \frac{1}{2} \Gamma \left( \frac{1}{2} \right) = \int_0^\infty e^{-t^2} \, dt.
Γ ( 2 3 ) = 2 1 Γ ( 2 1 ) = ∫ 0 ∞ e − t 2 d t . Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.