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Gamma and Beta Functions

Gamma Function #

$$ \begin{align} \Gamma(n)=\int\limits_{0}^{\infty}e^{-x}x^{n-1}dx=(n-1)!\\ \Gamma(n)=k^n\int\limits_{0}^{\infty}e^{-kx}x^{n-1}dx \end{align} $$

Properties #

  • $\Gamma (\frac{1}{2}) = \sqrt{\pi}$
    • This and $\Gamma(n)=2\int_0^\infty e^{-x^2}x^{2n-1}dx$ can be used to prove
  • $n\Gamma (n)=\Gamma (n+1)=n!$

Beta Function #

$$ \begin{align} \beta(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}dx \end{align} $$

Properties #

  • B(m,n)=B(n,m)
    • Symmetrical because of the way it’s defined and ($\int _a^bf(x)dx = \int_a^bf(a+b-x)dx$)
  • B(m,n) = $2 \cdot \int_0^{\frac{\pi}{2}}\sin^{2m-1}(x) \cos^{2m-1}(x)dx$
    • Useful to evaluate the integral of product of power of sin and cos.
    • Can be proved by substituting x=$\sin^2\theta$ in the main definition.
  • B(m,n) = $\int_0^\infty \frac{x^{m-1}}{(1+x)^{m+n}}dx$
    • Can be proved using substituting x=$\frac{t}{1+t}$ in $\int_0^1 x^{m-1}(1-x)^{n-1}dx$
  • B(m,n)=$\int_0^1 \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}dx$
    • From previous property, B(m,n)=$\int_0^\infty \frac{x^{m-1}}{(1+x)^{m+n}}dx$ = $\int_0^1 \frac{x^{m-1}}{(1+x)^{m+n}}dx + \int_1^\infty \frac{x^{m-1}}{(1+x)^{m+n}}dx$
    • We can prove the second integral, $\int_1^\infty \frac{x^{m-1}}{(1+x)^{m+n}}dx = \int_0^1 \frac{x^{n-1}}{(1+x)^{m+n}}dx$ by using x=1/t.

Relation between Beta and Gamma functions #

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

Proof

We’ll be using $B(m,n)=\int_0^\infty \frac{x^{m-1}}{(1+x)^{m+n}}dx$ and $\Gamma(n)=k^n\int\limits_{0}^{\infty}e^{-kx}x^{n-1}dx$ $$ \begin{align} \Gamma(m)=\int\limits_{0}^{\infty}z^me^{-zx}x^{m-1}dx\\ \times e^{-z}z^{n-1}dz\\ =\int_0^\infty\Gamma(m)e^{-z}z^{n-1}dz=\int_0^\infty\int\limits_{0}^{\infty}z^me^{-zx}x^{m-1}e^{-z}z^{n-1}dzdx\\ \Gamma(m)\Gamma(n)=\int_0^\infty x^{m-1} \left(\int\limits_{0}^{\infty}e^{-z(x+1)}z^{m+n-1}dz\right)dx\\ \Gamma(m)\Gamma(n)=\int_0^\infty x^{m-1}\left(\frac{\Gamma (m+n)}{(x+1)^{m+n}}\right)dx\\ =\Gamma(m+n)\beta(m,n)\\ \\ \therefore \beta(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \end{align} $$

Could also be done like https://youtu.be/E9sc1FnJF9k?t=191

Some more properties that can be proved using this relation #

  • $n\beta(m+1,n)=m\beta(m,n+1)$
  • $\beta(m,n)=\beta(m,n+1)+\beta(m+1,n)$

More properties #

  • $\Gamma(n)\Gamma(1-n)=\frac{\pi}{\sin(n\pi)}$ (proof is out of the scope for now https://youtu.be/XgnPg0Ab6hE)
  • $\Gamma(n)\Gamma(n+0.5)=\frac{\sqrt{\pi} \cdot \Gamma(2n)}{2^{2n-1}}$

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