Beta Function

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.

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 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, $B(x,y) = B(y,x).$ $_\square$

We have

$$B(x,y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.$$

For 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

$$\displaystyle \Gamma(m)\Gamma(n)=\int _0^\infty x^{m-1}e^{-x}\, dx \int _0^{\infty} y^{n-1}e^{-y}\, dy .$$

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 $B(5,7).$

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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:

$$B(5,7) = \frac{4!6!}{11!} = \frac{4!}{11\times 10\times \cdots \times 7} = \frac{1}{2310}. \ _\square$$

$$B(x,y)=\int_0^{\pi/2} 2\sin^{2x-1}(t)\cos^{2y-1}(t)\, dt$$

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

$$\int_0^{\pi/2} \sin^{9}(x)\cos(x)\, dx.$$

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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 $$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\tan x}}{(\cos x+\sin x)^{2}} \, \mathrm{d} x$$

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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

$$\frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma(2)}=\frac{\frac{1}{2} \Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{1 !}=\frac{\frac{1}{2} \sqrt{\pi} \sqrt{\pi}}{1}=\frac{\pi}{2}$$

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:

$$\begin{aligned} B(x,y+1) &= B(x,y)\dfrac{y}{x+y}\\ B(x+1,y)+B(x,y+1) &= B(x,y). \ _\square \end{aligned}$$

We observe the reciprocal of central binomial coefficient:

$$\begin{aligned} \dfrac{1}{{2n \choose n}} & = \dfrac{n!n!}{(2n)!} \\ & = \dfrac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+1)} \\ & = \dfrac{n\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)} \\ & = nB(n,n+1). \end{aligned}$$

Then

$$B(n,n+1) = \dfrac{1}{n{2n \choose n}}.$$

This is a really useful relation, especially when solving summations.

Find

$$S=\sum_{n=1}^\infty \dfrac{1}{n{2n \choose n}}.$$

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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

$$S=\dfrac{\sqrt{3} \pi}{9}. \ _\square$$

The derivative of the beta function is a great way to solve some integrals:

$$\begin{aligned} \dfrac{\partial}{\partial x} B(x,y)&= B(x,y)\big(\psi(x)-\psi(x+y)\big)\\ \dfrac{\partial^2 }{\partial x\partial y} B(x,y)&=B(x,y)\Big(\big(\psi(x)-\psi(x+y)\big)\big(\psi(y)-\psi(x+y)\big)-\psi'(x+y)\Big). \end{aligned}$$

Let's see how to use this, but first read the Digamma Function wiki.

Find

$$\int_0^1 \ln(t)\ln(1-t)\, dt.$$

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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

$$B(1,1) \Big(\big(\psi(1)-\psi(2)\big)^2 -\psi^{(1)} (2)\Big)= B(1,1)\big((-1)^2 - \zeta(2)+1\big) = 2-\dfrac{\pi^2}{6}. \ _\square$$

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 $y.$

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 $beta(k, n - k +1)$.

The beta function can be implemented in Mathematica as follows:

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Beta[a,b]

• Gamma Function