Skip to content
Sahithyan's S1
Sahithyan's S1 — Mathematics

Beta function

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

B(m,n)=∫01xm−1(1−x)n−1 dxB(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}\,\text{d}x

Aka. Eulerian integral of the first kind.

Properties

Symmetrical

From the definition:

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

Relation with gamma function

∀m,n>0\forall m,n \gt 0.

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

Transformations

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

Form 0, 6

∫ab(x−a)m−1(b−x)n−1 dx=(b−a)m+n−1B(m,n)\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=0a=0 and b=1b=1, .

Form 1, 3

∫0∞xm−1(ax+b)m+n dx=B(m,n)ambn\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=1a=b=1.

Form 2

B(m,n)=∫01xm−1+xn−1(1+x)m+n dxB(m,n) = \int_0^1 \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}}\,\text{d}x

Form 4

∫0π2sin⁡2m−1(x)cos⁡2n−1(x)(asin⁡2x+bcos⁡2x)m+n dx=B(m,n)2ambn\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}

Form 5, 7

∫01xm−1(1−x)n−1(a+bx)m+n dx=B(m,n)an(a+b)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}

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

Transformations Proofs

Form 1

B(m,n)=∫0∞xn−1(x+1)m+n dx=∫0∞xm−1(x+1)m+n dxB(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

Form 2

B(m,n)=∫01xm−1+xn−1(1+x)m+n dxB(m,n) = \int_0^1 \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}}\,\text{d}x

Form 3

∫0∞xm−1(ax+b)m+n dx=B(m,n)ambn\int_0^\infty \frac{x^{m-1}}{(ax + b)^{m+n}}\,\text{d}x = \frac{B(m,n)}{a^m b^n}

Form 4

∫0π2sin⁡2m−1(x)cos⁡2n−1(x)(asin⁡2x+bcos⁡2x)m+n dx=B(m,n)2ambn\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}

Form 5

∫01xm−1(1−x)n−1(x+a)m+n dx=B(m,n)an(1+a)m\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}

Form 6

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

Form 7

∫01xm−1(1−x)n−1(a+(b−a)x)m+n dx=B(m,n)anbm\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} ∫01xm−1(1−x)n−1(a+bx)m+n dx=B(m,n)an(a+b)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}