Beta Function | Sahithyan's Notes

Beta function is defined as below, for $m,n\gt 0$ :

B(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}\,\text{d}x

Aka. Eulerian integral of the first kind.

Properties

From the definition:

B(m,n) = B(n,m)

$\forall m,n \gt 0$ .

B(m,n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}

This section is intended to be exam-focused. Proofs for the transformations are included in a separate section.

\int_a^b {(x-a)^{m-1}(b-x)^{n-1}}\,\text{d}x = (b-a)^{m+n-1} B(m,n)

Form 0 (definition) is derived by setting $a=0$ and $b=1$ , .

\int_0^\infty \frac{x^{m-1}}{(ax + b)^{m+n}}\,\text{d}x = \frac{B(m,n)}{a^m b^n}

Form 1 is derived by setting $a=b=1$ .

B(m,n) = \int_0^1 \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}}\,\text{d}x

\int_0^\frac{\pi}{2} \frac{\sin^{2m-1}(x) \cos^{2n-1}(x)}{(a\sin^2 x + b\cos^2 x)^{m+n}}\,\text{d}x = \frac{B(m,n)}{2a^m b^n}

\int_0^1 \frac{x^{m-1}(1-x)^{n-1}}{(a + bx)^{m+n}}\,\text{d}x = \frac{B(m,n)}{a^n(a+b)^m}

Form 5 is derived by setting $b=1$ .

B(m,n)= \int_0^{\infty} \frac{x^{n-1}}{(x+1)^{m+n}}\,\text{d}x= \int_0^{\infty} \frac{x^{m-1}}{(x+1)^{m+n}}\,\text{d}x

B(m,n) = \int_0^1 \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}}\,\text{d}x

\int_0^\infty \frac{x^{m-1}}{(ax + b)^{m+n}}\,\text{d}x = \frac{B(m,n)}{a^m b^n}

\int_0^\frac{\pi}{2} \frac{\sin^{2m-1}(x) \cos^{2n-1}(x)}{(a\sin^2 x + b\cos^2 x)^{m+n}}\,\text{d}x = \frac{B(m,n)}{2a^m b^n}

\int_0^1 \frac{x^{m-1} (1-x)^{n-1}}{(x+a)^{m+n}}\,\text{d}x = \frac{B(m,n)}{a^n (1+a)^m}

\int_a^b {(x-a)^{m-1}(b-x)^{n-1}}\,\text{d}x = (b-a)^{m+n-1} B(m,n)

\int_0^1 \frac{x^{m-1}(1-x)^{n-1}}{(a + (b-a)x)^{m+n}}\,\text{d}x = \frac{B(m,n)}{a^nb^m}

\int_0^1 \frac{x^{m-1}(1-x)^{n-1}}{(a + bx)^{m+n}}\,\text{d}x = \frac{B(m,n)}{a^n(a+b)^m}