The definition of the beta function is
B(p,q)=∫01xp−1(1−x)q−1dx
where p>0,q>0.
A characteristics of the beta function is
B(p,q)=Γ(p+q)Γ(p)Γ(q)
Let's prove this characteristics.
Γ(p)Γ(q)=∫0∞xp−1e−xdx∫0∞yq−1e−ydy=∫0∞∫0∞xp−1yq−1e−(x+y)dxdy
Let x=u2, y=v2 , dx=2udu , dy=2vdv , u:0→∞ , v:0→∞.
=∫0∞∫0∞u2(p−1)v2(q−1)e−(u2+v2)2udu2vdv=4∫0∞∫0∞u2p−1v2q−1e−(u2+v2)dudv
Let u=rcos(θ) , v=rsin(θ) , r:0→∞ , θ:0→2π.
The Jacorbian is
du/drdv/drdu/dθdv/dθ=cos(θ)sin(θ)−rsin(θ)rcos(θ)=rcos2(θ)+rsin2(θ)=r
=4∫0∞∫02π(rcos(θ))2p−1(rsin(θ))2q−1e−((rcos(θ))2+(rsin(θ))2)rdrdθ=4∫0∞∫02πr2(p+q)−1cos2p−1(θ)sin2q−1(θ)e−r2drdθ=4∫0∞r2(p+q)−1e−r2dr∫02πcos2p−1(θ)sin2q−1(θ)dθ
Let t=r2 , x=cos2(θ) , t:0→∞ , x:1→0, dt=2rdr , dx=−2cos(θ)sin(θ)dθ ,
=−∫0∞r2(p+q)−2e−r22rdr∫02πcos2p−2(θ)sin2q−2(θ)−2cos(θ)sin(θ)dθ=−∫0∞t(p+q)−1e−tdt∫10xp−1(1−x)q−1dx=∫0∞t(p+q)−1e−tdt∫01xp−1(1−x)q−1dx=Γ(p+q)B(p,q)
Here, we have gotten,
Γ(p)Γ(q)=Γ(p+q)B(p,q)⟺B(p,q)=Γ(p+q)Γ(p)Γ(q)
Discussion