For x > 0 x > 0 x > 0 y > 0 y > 0 y > 0
B ( x , y ) = ∫ 0 1 t x − 1 ( 1 − y ) y − 1 d t
\mathbf{B}(x, y) = \int_0^1 t^{x-1} (1-y)^{y-1} \, dt
B ( x , y ) = ∫ 0 1 t x − 1 ( 1 − y ) y − 1 d t (rst Euler integral ) has the value
B ( x , y ) = Γ ( x ) Γ ( y ) Γ ( x + y ) .
\mathbf{B}(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}.
B ( x , y ) = Γ ( x + y ) Γ ( x ) Γ ( y ) . Indeed this integral converges for x > 0 x > 0 x > 0 y > 0 y > 0 y > 0 x ↦ B ( x , y ) x \mapsto \mathbf{B}(x, y) x ↦ B ( x , y ) logarithmically convex for x > 0 x > 0 x > 0
B ( x + 1 , y ) = ∫ 0 1 ( 1 − t ) x + y − 1 ( t 1 − t ) x d t
\mathbf{B}(x+1, y) = \int_0^1 (1-t)^{x+y-1} \left( \frac{t}{1-t} \right)^x \, dt
B ( x + 1 , y ) = ∫ 0 1 ( 1 − t ) x + y − 1 ( 1 − t t ) x d t Hence, on integrating by parts,
B ( x + 1 , y ) = − ( 1 − t ) x + y x + y ( t 1 − t ) x ∣ 0 1 + x x + y ∫ 0 1 ( 1 − t ) x + y ( t 1 − t ) x − 1 d t ( 1 − t ) 2 = x x + y B ( x , y ) .
\mathbf{B}(x+1, y) = -\frac{(1-t)^{x+y}}{x+y} \left( \frac{t}{1-t} \right)^x \Bigg|_0^1
+ \frac{x}{x+y} \int_0^1 (1-t)^{x+y} \left( \frac{t}{1-t} \right)^{x-1} \frac{dt}{(1-t)^2} = \frac{x}{x+y} \mathbf{B}(x, y).
B ( x + 1 , y ) = − x + y ( 1 − t ) x + y ( 1 − t t ) x 0 1 + x + y x ∫ 0 1 ( 1 − t ) x + y ( 1 − t t ) x − 1 ( 1 − t ) 2 d t = x + y x B ( x , y ) . It follows that f ( x ) = B ( x , y ) Γ ( x + y ) f(x) = \mathbf{B}(x, y) \Gamma(x+y) f ( x ) = B ( x , y ) Γ ( x + y ) f ( 1 ) = B ( 1 , y ) Γ ( y + 1 ) f(1) = \mathbf{B}(1, y) \Gamma(y+1) f ( 1 ) = B ( 1 , y ) Γ ( y + 1 ) B ( 1 , y ) = ∫ 0 1 ( 1 − t ) y − 1 d t = 1 / y \mathbf{B}(1, y) = \int_0^1 (1-t)^{y-1} dt = 1/y B ( 1 , y ) = ∫ 0 1 ( 1 − t ) y − 1 d t = 1/ y f ( 1 ) = 1 y Γ ( y + 1 ) = Γ ( y ) f(1) = \frac{1}{y} \Gamma(y+1) = \Gamma(y) f ( 1 ) = y 1 Γ ( y + 1 ) = Γ ( y ) f ( x ) / Γ ( y ) f(x)/\Gamma(y) f ( x ) /Γ ( y ) Γ ( x ) \Gamma(x) Γ ( x )
By the change of variable t = u u + 1 t = \frac{u}{u+1} t = u + 1 u
∫ 0 ∞ t x − 1 ( 1 + t ) x + y d t = Γ ( x ) Γ ( y ) Γ ( x + y )
\int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}} dt = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}
∫ 0 ∞ ( 1 + t ) x + y t x − 1 d t = Γ ( x + y ) Γ ( x ) Γ ( y ) (19)
and by the change of variable t = sin 2 φ t = \sin^2 \varphi t = sin 2 φ
∫ 0 π / 2 sin 2 x − 1 φ cos 2 y − 1 φ d φ = 1 2 Γ ( x ) Γ ( y ) Γ ( x + y ) .
\int_0^{\pi/2} \sin^{2x-1} \varphi \cos^{2y-1} \varphi d\varphi = \frac{1}{2} \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}.
∫ 0 π /2 sin 2 x − 1 φ cos 2 y − 1 φ d φ = 2 1 Γ ( x + y ) Γ ( x ) Γ ( y ) . (20)
If one puts x = y = 1 2 x = y = \frac{1}{2} x = y = 2 1
Γ ( 1 2 ) = π
\Gamma(\frac{1}{2}) = \sqrt{\pi}
Γ ( 2 1 ) = π (21)
whence, by (17),
∫ 0 ∞ e − t 2 d t = 1 2 π .
\int_0^\infty e^{-t^2} dt = \frac{1}{2} \sqrt{\pi}.
∫ 0 ∞ e − t 2 d t = 2 1 π . (22)
From the relation (7) of VII, p. 307, one has the asymptotic expansion
Γ ( x ) = 1 x Γ ( x + 1 ) = 1 x + Γ ′ ( 1 ) + 1 2 ! Γ ′ ′ ( 1 ) x + ⋯ + 1 n ! Γ ( n ) ( 1 ) x n − 1 + O ( x n )
\begin{align*}
\Gamma(x) &= \frac{1}{x} \Gamma(x+1) \\
&= \frac{1}{x} + \Gamma'(1) + \frac{1}{2!} \Gamma''(1)x + \cdots + \frac{1}{n!} \Gamma^{(n)}(1)x^{n-1} + O(x^n)
\end{align*}
Γ ( x ) = x 1 Γ ( x + 1 ) = x 1 + Γ ′ ( 1 ) + 2 ! 1 Γ ′′ ( 1 ) x + ⋯ + n ! 1 Γ ( n ) ( 1 ) x n − 1 + O ( x n ) (23)
for Γ ( x ) \Gamma(x) Γ ( x )
Similarly, for all y y y > 0 > 0 > 0
1 Γ ( x + y ) = 1 Γ ( y ) + D ( 1 Γ ( y ) ) x + 1 2 ! D 2 ( 1 Γ ( y ) ) x 2 + ⋯ + 1 n ! D n ( 1 Γ ( y ) ) x n + O 1 ( x n + 1 )
\begin{align*}
\frac{1}{\Gamma(x+y)} &= \frac{1}{\Gamma(y)} + D \left( \frac{1}{\Gamma(y)} \right) x \\
&\quad + \frac{1}{2!} D^2 \left( \frac{1}{\Gamma(y)} \right) x^2 + \cdots + \frac{1}{n!} D^n \left( \frac{1}{\Gamma(y)} \right) x^n + O_1(x^{n+1})
\end{align*}
Γ ( x + y ) 1 = Γ ( y ) 1 + D ( Γ ( y ) 1 ) x + 2 ! 1 D 2 ( Γ ( y ) 1 ) x 2 + ⋯ + n ! 1 D n ( Γ ( y ) 1 ) x n + O 1 ( x n + 1 ) and the formula (18) then gives, for y y y
B ( x , y ) = 1 x + ( Γ ′ ( 1 ) − Γ ′ ( y ) Γ ( y ) ) + ( Γ ′ ′ ( 1 ) 2 − Γ ′ ( 1 ) Γ ′ ( y ) Γ ( y ) + 2 Γ ′ 2 ( y ) − Γ ( y ) Γ ′ ′ ( y ) 2 Γ 2 ( y ) ) x + O ( x 2 )
\mathbf{B}(x, y) = \frac{1}{x} + \left( \Gamma'(1) - \frac{\Gamma'(y)}{\Gamma(y)} \right)
+ \left( \frac{\Gamma''(1)}{2} - \Gamma'(1) \frac{\Gamma'(y)}{\Gamma(y)} + \frac{2{\Gamma'}^2(y) - \Gamma(y)\Gamma''(y)}{2\Gamma^2(y)} \right) x + O(x^2)
B ( x , y ) = x 1 + ( Γ ′ ( 1 ) − Γ ( y ) Γ ′ ( y ) ) + ( 2 Γ ′′ ( 1 ) − Γ ′ ( 1 ) Γ ( y ) Γ ′ ( y ) + 2 Γ 2 ( y ) 2 Γ ′ 2 ( y ) − Γ ( y ) Γ ′′ ( y ) ) x + O ( x 2 ) on a neighbourhood of
x = 0 x = 0 x = 0 .
Moreover, for x > 0 x > 0 x > 0 y > 0 y > 0 y > 0
B ( x , y ) = ∫ 0 1 ( t x − 1 + t x ( 1 − t ) y − 1 − 1 t ) d t = 1 x + ∫ 0 1 t x ( 1 − t ) y − 1 − 1 t d t .
\mathbf{B}(x, y) = \int_0^1 \left( t^{x-1} + t^x \frac{(1-t)^{y-1} - 1}{t} \right) dt
= \frac{1}{x} + \int_0^1 t^x \frac{(1-t)^{y-1} - 1}{t} dt.
B ( x , y ) = ∫ 0 1 ( t x − 1 + t x t ( 1 − t ) y − 1 − 1 ) d t = x 1 + ∫ 0 1 t x t ( 1 − t ) y − 1 − 1 d t . The function
φ ( t ) = ( 1 − t ) y − 1 − 1 t \varphi(t) = \frac{(1-t)^{y-1} - 1}{t} φ ( t ) = t ( 1 − t ) y − 1 − 1 is continuous on the compact interval
[ 0 , 1 ] [0, 1] [ 0 , 1 ] ; since
t x = e x log t = 1 + x log t + x 2 2 ! ( log t ) 2 + ⋯ + x n n ! ( log t ) n + r n ( x , t )
t^x = e^{x \log t} = 1 + x \log t + \frac{x^2}{2!} (\log t)^2 + \cdots + \frac{x^n}{n!} (\log t)^n + r_n(x, t)
t x = e x l o g t = 1 + x log t + 2 ! x 2 ( log t ) 2 + ⋯ + n ! x n ( log t ) n + r n ( x , t ) with
∣ r n ( x , t ) ∣ ≤ x n + 1 ( n + 1 ) ! ∣ log t ∣ n + 1 |r_n(x, t)| \leq \frac{x^{n+1}}{(n+1)!} |\log t|^{n+1} ∣ r n ( x , t ) ∣ ≤ ( n + 1 )! x n + 1 ∣ log t ∣ n + 1 (since
log t ≤ 0 \log t \leq 0 log t ≤ 0 and
x > 0 x > 0 x > 0 ), formula (25) gives the asymptotic expansion
B ( x , y ) = 1 x + ∫ 0 1 φ ( t ) d t + x ∫ 0 1 φ ( t ) log t d t + ⋯ + x n n ! ∫ 0 1 φ ( t ) ( log t ) n d t + O 2 ( x n + 1 )
\mathbf{B}(x, y) = \frac{1}{x} + \int_0^1 \varphi(t) dt + x \int_0^1 \varphi(t) \log t \, dt + \cdots
+ \frac{x^n}{n!} \int_0^1 \varphi(t) (\log t)^n \, dt + O_2(x^{n+1})
B ( x , y ) = x 1 + ∫ 0 1 φ ( t ) d t + x ∫ 0 1 φ ( t ) log t d t + ⋯ + n ! x n ∫ 0 1 φ ( t ) ( log t ) n d t + O 2 ( x n + 1 ) for
B ( x , y ) \mathbf{B}(x, y) B ( x , y ) on a neighbourhood of 0.
For n = 1 n = 1 n = 1
Γ ′ ( 1 ) − Γ ′ ( y ) Γ ( y ) = ∫ 0 1 ( 1 − t ) y − 1 − 1 t d t .
\Gamma'(1) - \frac{\Gamma'(y)}{\Gamma(y)} = \int_0^1 \frac{(1-t)^{y-1} - 1}{t} dt.
Γ ′ ( 1 ) − Γ ( y ) Γ ′ ( y ) = ∫ 0 1 t ( 1 − t ) y − 1 − 1 d t . Furthermore, the formula (10) gives Γ ′ ( 1 ) = Γ ′ ( 1 ) / Γ ( 1 ) = − γ \Gamma'(1) = \Gamma'(1)/\Gamma(1) = -\gamma Γ ′ ( 1 ) = Γ ′ ( 1 ) /Γ ( 1 ) = − γ Gauss' integral )
Γ ′ ( x ) Γ ( x ) + γ = ∫ 0 1 1 − ( 1 − t ) x − 1 t d t .
\frac{\Gamma'(x)}{\Gamma(x)} + \gamma = \int_0^1 \frac{1 - (1-t)^{x-1}}{t} dt.
Γ ( x ) Γ ′ ( x ) + γ = ∫ 0 1 t 1 − ( 1 − t ) x − 1 d t . Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.