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The beta function $B$ (note: $B$ is capital $\beta$ in Greek) is defined by the following formula for $\mathrm{Re}(x)>0$ and $\mathrm{Re}(y)>0$: $$B(x,y)=\displaystyle\int_0^1 t^{x-1}(1-t)^{y-1} \mathrm{d}t.$$


Partial derivative of beta function
Beta in terms of gamma
Beta in terms of sine and cosine
Beta as improper integral
Beta is symmetric
B(x,y)=integral (t^(x-1)+t^(y-1))(1+t)^(-x-y) dt
1/B(n,m)=m((n+m-1) choose (n-1))
1/B(n,m)=n((n+m-1) choose (m-1))
B(x,y)=2^(1-x-y)integral (1+t)^(x-1)(1-t)^(y-1)+(1+t)^(y-1)(1-t)^(x-1) dt
Integral t^(x-1)(1-t)^(y-1)(1+bt)^(-x-y)dt = (1+b)^(-x)B(x,y)
Integral t^(x-1)(1+bt)^(-x-y) dt = b^(-x) B(x,y)
Integral (t-b)^(x-1)(a-t)^(y-1)dt=(a-b)^(x+y-1)B(x,y)
Integral of (t-b)^(x-1)(a-t)^(y-1)/(t-x)^(x+y) dt=(a-b)^(x+y-1)/((a-c)^x(b-c)^y) B(x,y)
Integral of (t-b)^(x-1)(a-t)^(y-1)/(c-t)^(x+y) dt = (a-b)^(x+y-1)/((c-a)^x (c-b)^y) B(x,y)
Integral of (1+bt^z)^(-y)t^x dt = (1/z)*b^(-(x+1)/z) B((x+1)/z,y-(x+1)/z)
Integral of t^(x-1)(1-t^z)^(y-1) dt=(1/z)B(x/z,y)
Integral of (1+t)^(2x-1)(1-t)^(2y-1)(1+t^2)^(-x-y)dt=2^(x+y-2)B(x,y)


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