证明：
Γ(a)=∫(0,∞)[x^a*e^-x]dx
我们令x=ty(t>0)
有：Γ(a)/(t^a)=∫(0,∞)[y^(a-1)*e^(-ty)]dy
于是：x^k=[1/Γ(-k)]*∫(0,∞)[z^(-k-1)*e^(-zx)]dz
有根据余元公式Γ(p)Γ(1-p)=π/sinpπ
则x^k=(Γ(k+1)sin-kπ)/π*(∫(0,∞)[z^(-k-1)*e^(-zx)]dz)
因而∫(0,∞)[(x^k)sinx]dx
=(Γ(k+1)sin-kπ)/π*∫(0,∞)sinxdx*∫(0,∞)[z^(-k-1)*e^(-zx)]dz
由一致收敛性,交换积分顺序得：
∫(0,∞)[(x^k)sinx]dx
=(Γ(k+1)sin-kπ)/π*∫(0,∞)z^(-k-1)dz*∫(0,∞)[sinx*e^(-zx)]dx
容易计算∫(0,∞)[sinx*e^(-zx)]dx=1/(1+z^2)
(只要用两次分部积分就行,会发现相同形式的)
从而∫(0,∞)[(x^k)sinx]dx
=(Γ(k+1)sin-kπ)/π*∫(0,∞)[z^(-k-1)/(1+z^2)]dz
在此我们再令z^2=t,则dz=(1/2)*z^(-1/2)dz
∫(0,∞)[(x^k)sinx]dx
=(Γ(k+1)sin-kπ)/(2π)*∫(0,∞)[z^(-k/2-1)/(1+k)]dz
考虑到β函数的另外一种变形
即B(a,b)=∫(0,∞)[y^(a-1)/(1+y)^(a+b)]
此时a=-k/2,b=1+k/2,a+b=1,又可以用余元公式了
记B(-k/2,1+k/2)=π/sin(-kπ/2)
故∫(0,∞)[(x^k)sinx]dx
=(Γ(k+1)sin-kπ)/(2π)*B(-k/2,1+k/2)
=(Γ(k+1)sin-kπ)/(2π)*π/sin(-kπ/2)
=Γ(k+1)cos(-kπ/2) （二倍角公式）
=Γ(k+1)cos(kπ/2).
证毕
