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Kishlaya Jaiswal, Aareyan Manzoor, Travis Mcknight, and
- Ronak Agarwal
- Scott Lee
- Mahindra Jain
- Samir Khan
- Couper Leo
- A Former Brilliant Member
- Kaustubh Miglani
- Ishan Singh
- Sravanth C.
- Jimin Khim
contributed
The beta function (also known as Euler's integral of the first kind) is important in calculus and analysis due to its close connection to the gamma function, which is itself a generalization of the factorial function. Many complex integrals can be reduced to expressions involving the beta function.
Contents
- Definition
- Symmetry
- Relation with Gamma Function
- Trigonometric Representation of the Beta Function
- Recurrence Relation
- Relationship with Central Binomial Coefficient
- Derivative of the Beta Function
- Approximation
- Beta Distribution
- Implementation in Mathematica
- See Also
Definition
The beta function, denoted by \(B(x,y)\), is defined as
\[B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \, dt .\]
This is also the Euler's integral of the first kind.
Symmetry
Symmetry of the Beta Function
\[B(x,y)=B(y,x)\]
Because of the convergent property of definite integrals\[\int_0^a f(t) \, dt = \int_0^a f(a-t) \, dt,\]
so we can rewrite the above integral as
\[B(x,y) = \int_0^1 t^{y-1}(1-t)^{x-1} \, dt.\]
Thus, we get that the beta function is symmetric, \(B(x,y) = B(y,x).\) \(_\square\)
Relation with Gamma Function
We have
\[B(x,y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.\]
See AlsoWhat is a beta level?5.17: The Beta DistributionBeta functionBeta Distribution — Intuition, Derivation, and Examples | Chen XingFor positive integers \(x\) and \(y\), we can define the beta function as
\[B(x,y) = \dfrac{(x-1)!(y-1)!}{(x+y-1)!}.\]
Recall the definition of gamma function
\[\displaystyle \Gamma(s) = \int _0^\infty x^{s-1}e^{-x}\, dx . \]
Now one can write
\[\displaystyle \Gamma(m)\Gamma(n)=\int _0^\inftyx^{m-1}e^{-x}\, dx \int _0^{\infty} y^{n-1}e^{-y}\, dy . \]
Then we can rewrite it as a double integral:
\[ \displaystyle \Gamma(m)\Gamma(n)=\int _0^\infty \int _0^\infty x^{m-1}y^{n-1}e^{-(x+y)}\, dx \, dy. \]
Applying the substitution \( x=vt\) and \(y=v(1-t) ,\) we have
\[ \displaystyle \Gamma(m)\Gamma(n) = \int _0^1 t^{m-1}(1-t)^{n-1}\, dt\int _0^\infty v^{ m+n-1 }e^{-v}\, dv. \]
Using the definitions of gamma and beta functions, we have
\[ \Gamma(m)\Gamma(n) = B(m,n)\Gamma (m+n).\]
Hence proved. \(_\square\)
Compute \(B(5,7).\)
If we go by the definition of beta function to compute \(B(5,7)\), we will have to solve the following integral
\[B(5,7) = \int_0^1 t^4(1-t)^{6} \, dt,\]
which is a very tedious work. Here is when the relation of beta function with gamma function comes in handy:
\[B(5,7) = \frac{4!6!}{11!} = \frac{4!}{11\times 10\times \cdots \times 7} = \frac{1}{2310}. \ _\square\]
Trigonometric Representation of the Beta Function
\[B(x,y)=\int_0^{\pi/2} 2\sin^{2x-1}(t)\cos^{2y-1}(t)\, dt\]
Consider
\[B(x,y) = \int_0^1 u^{y-1}(1-u)^{x-1} \, du.\]
Using the substitution \(u=\sin^2(t)\to du= 2\sin(t)\cos(t)\) and \(\Big|_0^1\to \Big|_0^{\pi/2}\), we have
\[B(x,y)=\int_0^{\pi/2} 2\sin^{2x-1}(t)\cos^{2y-1}(t)\, dt. \ _\square\]
Find
\[\int_0^{\pi/2} \sin^{9}(x)\cos(x)\, dx.\]
The given integral is simply
\[\dfrac{1}{2}B(5,1)=\dfrac{1}{2}\cdot\dfrac{\Gamma(5)\Gamma(1)}{\Gamma(6)}=\dfrac{1}{2}\cdot\dfrac{4!}{5!}=\dfrac{1}{10}. \ _\square\]
Evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\tan x}}{(\cos x+\sin x)^{2}} \, \mathrm{d} x\]
Using the substitution \(t=\tan x\longrightarrow \, \mathrm{d}x=\cos^2(x)\, \mathrm{d}t\) therefore,
\[\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\tan x}}{(\cos x+\sin x)^{2}} \, \mathrm{d} x= \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\tan x}}{\cos ^{2} x(1+\tan x)^{2}} \, \mathrm{d} x=\int_{0}^{\infty} \frac{t^{\frac{1}{2}}}{(1+t)^{2}} \, \mathrm{d} t\]
And by using the alternate definition of the beta function we can rewrite the above integral to fit the definition
\[\int_{0}^{\infty} \frac{t^{\frac{3}{2}-1}}{(1+t)^{\frac{3}{2}+t}} \, \mathrm{d} t=\mathrm{B}\left(\frac{3}{2}, \frac{1}{2}\right)\]
then by switching to gamma functions
\[\frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma(2)}=\frac{\frac{1}{2} \Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{1 !}=\frac{\frac{1}{2} \sqrt{\pi} \sqrt{\pi}}{1}=\frac{\pi}{2}\]
Recurrence Relation
The recurrence relation of the beta function is given by
\[B(x+1,y) = B(x,y)\dfrac{x}{x+y}.\]
We have
\[B(x+1,y) = \dfrac{x!(y-1)!}{(x+y)!} = \dfrac{(x-1)!(y-1)!}{(x+y-1)!}\times \dfrac{x}{x+y} = B(x,y)\dfrac{x}{x+y}.\]
From the above relation and because of symmetry of the beta function, the following two results follow immediately:
\[\begin{align}B(x,y+1) &= B(x,y)\dfrac{y}{x+y}\\B(x+1,y)+B(x,y+1) &= B(x,y). \ _\square\end{align}\]
Relationship with Central Binomial Coefficient
We observe the reciprocal of central binomial coefficient:
\[\begin{align}\dfrac{1}{{2n \choose n}} & = \dfrac{n!n!}{(2n)!} \\& = \dfrac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+1)} \\& = \dfrac{n\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)} \\& = nB(n,n+1).\end{align}\]
Then
\[B(n,n+1) = \dfrac{1}{n{2n \choose n}}.\]
This is a really useful relation, especially when solving summations.
Find
\[S=\sum_{n=1}^\infty \dfrac{1}{n{2n \choose n}}.\]
Using the relation to the beta function, this is
\[S=\sum_{n=1}^\infty B(n+1,n)=\sum_{n=1}^\infty\int_0^1 x^n(1-x)^{n-1}dx.\]
We can interchange the summation and integral signs due to absolute convergence of the integrand:
\[S=\int_0^1\sum_{n=1}^\infty x^n(1-x)^{n-1}dx= \int_0^1 \dfrac{x}{x^2-x+1} dx.\]
The geometric progression sum was used. Now we can easily solve the indefinite integral and then put in the limits. So we have
\[S=\dfrac{\sqrt{3} \pi}{9}. \ _\square\]
\[\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }\left( \begin{matrix} 2n \\ n \end{matrix} \right) } } =\frac { A }{ B } \zeta \left( C \right) \]
In the equation above, \(A,B,C\) are positive integers such that \(A\) and \(B\) are coprime.
Find \(A+B+C.\)
Derivative of the Beta Function
The derivative of the beta function is a great way to solve some integrals:
\[\begin{align}\dfrac{\partial}{\partial x} B(x,y)&= B(x,y)\big(\psi(x)-\psi(x+y)\big)\\\dfrac{\partial^2 }{\partial x\partial y} B(x,y)&=B(x,y)\Big(\big(\psi(x)-\psi(x+y)\big)\big(\psi(y)-\psi(x+y)\big)-\psi'(x+y)\Big).\end{align}\]
Let's see how to use this, but first read the Digamma Function wiki.
Find
\[\int_0^1 \ln(t)\ln(1-t)\, dt.\]
Consider
\[B(x,y)= \int_0^1 t^{x-1} (1-t)^{y-1}dt.\]
Differentiate with respect to \(x\) first and then \(y\) to get
\[\dfrac{\partial^2 }{\partial x\partial y} B(x,y)=\int_0^1 \ln(t)\ln(1-t)t^{x-1} (1-t)^{y-1}dt.\]
Put \(x=1\) and \(y=1\) to have
\[B(1,1) \Big(\big(\psi(1)-\psi(2)\big)^2 -\psi^{(1)} (2)\Big)= B(1,1)\big((-1)^2 - \zeta(2)+1\big) = 2-\dfrac{\pi^2}{6}. \ _\square\]
Approximation
Using Stirling's formula, we can easily define the asymptotic approximation of the beta function as
\[B(x,y) = \dfrac{(x-1)!(y-1)!}{(x+y-1)!} \sim \sqrt{2\pi}\dfrac{x^{x-1/2}y^{y-1/2}}{(x+y)^{x+y-1/2}}\]
for large \(x\) and large \(y.\)
Beta Distribution
In probability theory, the beta distribution, written as \(beta(r, s)\) for \(r, s >0\), is defined by the density
\[\frac{1}{B(r, s)}x^{r-1}(1-x)^{s-1},\]
where \(0 < x < 1\) and \(B(r, s)\) = \(\int_0^1 x^{r-1}(1-x)^{s-1} \, dx\) as defined by the beta function.
\(B(r, s)\) is also known as the normalizing constant because it makes the integral equal to 1.
The \(k^\text{th}\)-order statistic of \(n\) i.i.d. Uniform\((0, 1)\) distributions is modeled by \(beta(k, n - k +1)\).
Implementation in Mathematica
The beta function can be implemented in Mathematica as follows:
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See Also
- Gamma Function
Cite as: Beta Function. Brilliant.org. Retrieved from https://brilliant.org/wiki/beta-function/
