Beta Function
The beta function (also known as euler integral of the first kind) is important in calculus and analysis due to its close connection to the gamma function
Contents
Definition
The beta function, denoted by \(B(x,y)\) and is defined as \[B(x,y) = \int_0^1 t^{x1}(1t)^{y1} \, dt .\] this is also the Euler's integral of the first kind. \(_\square\)
Symmetry
\[B(x,y)=B(y,x)\]
Because of the convergent property of definite integrals \[\int_0^a f(x) \, dt = \int_0^a f(ax) \, dt,\] we can rewrite the above integral as \[B(x,y) = \int_0^1 t^{y1}(1t)^{x1} \, dt.\]Thus, we get that the beta function is symmetric, \(B(x,y) = B(y,x).\)
Relation with Gamma function
\[B(x,y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.\] And for positive integers \(x,\ y\), we can define the beta function as \[B(x,y) = \dfrac{(x1)!(y1)!}{(x+y1)!}.\]
Recall the definition of gamma function \[\displaystyle \Gamma(s) = \int _0^\infty x^{s1}e^{x}\, dx \] Now one can write \[\displaystyle \Gamma(m)\Gamma(n)=\int _0^\infty x^{m1}e^{x}\, dx \int _0^{\infty} y^{n1}e^{y}\, dy . \] Then we can rewrite it as a double integral: \[ \displaystyle \Gamma(m)\Gamma(n)=\int _0^\infty \int _0^\infty x^{m1}y^{n1}e^{(x+y)}\, dx \, dy. \]
Applying the substitution \( x=vt\) and \(y=v(1t) ,\) we have \[ \displaystyle \Gamma(m)\Gamma(n) = \int _0^1 t^{m1}(1t)^{n1}\, dt\int _0^\infty v^{ m+n1 }e^{v}\, dv. \] Using definition of gamma and beta function, 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 \[B(5,7) = \int_0^1 t^4(1t)^{6} \, dt,\] 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
\[B(x,y)=\int_0^{\pi/2} 2\sin^{2x1}(t)\cos^{2y1}(t) dt\]
Consider \[B(x,y) = \int_0^1 u^{y1}(1u)^{x1} \, du\] using the sub \(u=\sin^2(t)\to du= 2\sin(t)\cos(t), \mid_0^1\to \mid_0^{\pi/2}\) we have \[B(x,y)=\int_0^{\pi/2} 2\sin^{2x1}(t)\cos^{2y1}(t)dt\]
Find \[\int_0^{\pi/2} \sin^{9}(x)\cos(x)dx\]
The given integral is simply \[B(5,1)=\dfrac{\Gamma(5)\Gamma(1)}{\Gamma(6)}=\dfrac{4!}{5!}=\dfrac{1}{5}\]
Recurrence Relation
The recurrence relation of beta function is given by \[B(x+1,y) = B(x,y)\dfrac{x}{x+y}.\]
We have \[B(x+1,y) = \dfrac{x!(y1)!}{(x+y)!} = \dfrac{(x1)!(y1)!}{(x+y1)!}\times \dfrac{x}{x+y} = B(x,y)\dfrac{x}{x+y}.\] From the above relation and because of symmetry of 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 \[B(n,n+1) = \dfrac{1}{n{2n \choose n}}. \ _\square\] 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 \[S=\sum_{n=1}^\infty B(n+1,n)=\sum_{n=1}^\infty\int_0^1 x^n(1x)^{n1}dx\] We can interchange the summation and integral sign due to absolute convergence of the integrand. \[S=\int_0^1\sum_{n=1}^\infty x^n(1x)^{n1}dx= \int_0^1 \dfrac{x}{x^2x+1} dx\] The geometric progression sum sum was used. Now we can easily solve the indefinite integral and then put in the limits. we have \[S=\dfrac{\sqrt{3} \pi}{9}\]
Derivative of the Beta Function
The derivative of the beta function is a great way to solve some integrals. \[ \dfrac{\partial}{\partial x} B(x,y)= B(x,y)(\psi(x)\psi(x+y))\] \[\dfrac{\partial^2 }{\partial x\partial y} B(x,y)=B(x,y)((\psi(x)\psi(x+y))(\psi(y)\psi(x+y))\psi'(x+y))\] Lets see how to use this, but first read the digamma function wiki
Find \[\int_0^1 \ln(t)\ln(1t) dt\]
consider \[B(x,y)= \int_0^1 t^{x1} (1t)^{y1}dx\] differentiate with respect to x first then y to get \[\dfrac{\partial^2 }{\partial x\partial y} B(x,y)=\int_0^1 \ln(t)\ln(1t)t^{x1} (1t)^{y1}dx\] put x=1 and y=1 to have \[B(1,1) ((\psi(1)\psi(2))^2 \psi^{(1)} (2))= B(1,1)((1)^2  \zeta(2)+1) = 2\dfrac{\pi^2}{6}\]
Approximation
Using Stirling's Formula, we can easily define the asymptotic approximation of beta function as
\[B(x,y) = \dfrac{(x1)!(y1)!}{(x+y1)!} \large\text{~} \sqrt{2\pi}\dfrac{x^{x1/2}y^{y1/2}}{(x+y)^{x+y1/2}}\]
for large \(x\) and large \(y\)
Implementation in Mathematica
Beta function can be implemented in mathematica as follows:
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See Also
\(1.\) Gamma Function
\(2.\) Kishlaya's Identity