The Beta Function

Beta Function Calculator

Type $m$ and $n$ to find $B(m,n)\text{:}$ integers, decimals or fractions, for example 2, 0.5, -1.5 or 1/3. The calculator shows the solution step by step. It finds the exact value when $m$ or $n$ is an integer, or when $m+n$ is an integer (for numbers up to $40$), and it always gives an approximate value.

What Is the Beta Function?

Many functions have no elementary antiderivative, and yet some of their definite integrals can be found exactly. A large family of such integrals reduces to one integral with two parameters, the Beta function. Leonhard Euler studied it together with the Gamma function, so the two are also called the Euler integrals of the first and of the second kind. The Beta function is used in calculus, in probability (the Beta distribution) and in combinatorics, where it is a close relative of the binomial coefficients.

Definition of the Beta Function $B(m,n)$

The letter $B$ is the capital Greek letter beta. The variable $t$ disappears after the integration, so the value depends only on the two parameters $m$ and $n.$ Some books write $B(x,y),$ $B(p,q)$ or $B(a,b).$

Why $m\gt 0$ and $n\gt 0?$ When $m\ge 1$ and $n\ge 1,$ the integrand is continuous on $[0,1]$ and the integral is an ordinary definite integral. When $0\lt m\lt 1,$ the factor $t^{m-1}=\frac{1}{t^{1-m}}$ grows without bound as $t\to 0,$ and the integral is improper. Near $t=0$ the other factor $(1-\nobreak t)^{n-1}$ is close to $1,$ so the integral behaves like $\int_0^{1/2}\frac{dt}{t^{p}}$ with $p=1-m,$ which converges exactly when $p\lt 1,$ that is, when $m\gt 0.$ The same holds at $t=1$ for the factor $(1-\nobreak t)^{n-1},$ with $p=1-n.$ So the integral converges when $m\gt 0$ and $n\gt 0,$ and diverges when $m\le 0$ or $n\le 0.$

On this page: Calculator · First examples · Symmetry · Recurrence relations · Gamma function · Beta and Gamma · Other forms · Table of values · Beta distribution · Practice

Computing $B(m,n)$ from the Definition

When $m$ and $n$ are positive integers, the integrand is a polynomial, and the integral can be computed directly.

Example 1. Find $B(1,1)$ and $B(2,1).$

$B(1,1)=\int_0^1 1\,dt=1$

${B(2,1)}={\int_0^1 t\,dt}={\frac{t^2}{2}\Big\vert_0^1}={\frac{1}{2}}$

In the same way, ${B(m,1)}={\int_0^1 t^{m-1}\,dt}={\frac{1}{m}}$ for every $m\gt 0.$

Example 2. Find $B(2,3).$

Here $m-1=1$ and $n-1=2.$ Expand $(1-\nobreak t)^2$ and integrate term by term:

${B(2,3)}={\int_0^1 t(1-t)^2\,dt}={\int_0^1\left(t-2t^2+t^3\right)dt}={\frac{1}{2}-\frac{2}{3}+\frac{1}{4}}={\frac{1}{12}}$

Geometrically, $B(m,n)$ is the area under the graph of $y=t^{m-1}(1-\nobreak t)^{n-1}$ between $t=0$ and $t=1.$

Example 3. Find $B\left(\frac{1}{2},1\right).$

The integrand $t^{-1/2}=\frac{1}{\sqrt{t}}$ is unbounded near $0,$ but the improper integral converges:

${B\left(\frac{1}{2},1\right)}={\int_0^1\frac{dt}{\sqrt{t}}}={2\sqrt{t}\,\Big\vert_0^1}={2}$

This agrees with $B(m,1)=\frac{1}{m}.$ For most non-integer $m$ and $n,$ the antiderivative is not an elementary function; for example, $\int t^{1/2}(1-\nobreak t)^{1/3}\,dt$ cannot be written with elementary functions. Such values are found with the Gamma function (see below).

Symmetry

Proof. Substitute $t=1-u$ in the definition. Then $dt=-du,$ the value $t=0$ becomes $u=1,$ and $t=1$ becomes $u=0\text{:}$

${B(m,n)}={\int_1^0 (1-u)^{m-1}u^{n-1}\,(-du)}={\int_0^1 u^{n-1}(1-u)^{m-1}\,du}={B(n,m)}$

The graphs of $y=t(1-\nobreak t)^4$ and $y=t^4(1-\nobreak t)$ are mirror images of each other in the line $t=\frac{1}{2},$ so the areas under them are equal: $B(2,5)=B(5,2).$

Recurrence Relations

Proof. Since $t+(1-\nobreak t)=1,$ the integral $B(m,n)$ splits into two:

${B(m,n)}={\int_0^1 t^{m-1}(1-t)^{n-1}\bigl(t+(1-t)\bigr)\,dt}={B(m+1,n)}+{B(m,n+1)}$

Now integrate $B(m,n+\nobreak 1)$ by parts with $u=(1-\nobreak t)^n$ and $dv=t^{m-1}\,dt,$ so that $v=\frac{t^m}{m}\text{:}$

${B(m,n+1)}={\frac{t^m(1-t)^n}{m}\Big\vert_0^1}+{\frac{n}{m}\int_0^1 t^m(1-t)^{n-1}\,dt}={\frac{n}{m}\,B(m+1,n)}$

The first term is $0,$ because $t^m=0$ at $t=0$ and $(1-\nobreak t)^n=0$ at $t=1.$ Put this into the first equation: ${B(m,n)}={B(m+1,n)}+{\frac{n}{m}\,B(m+1,n)}={\frac{m+n}{m}\,B(m+1,n).}$ This is the first formula. The second one follows from the first by symmetry.

For example, $B(1,3)=B(3,1)=\frac{1}{3},$ so ${B(2,3)}={\frac{1}{1+3}\,B(1,3)}={\frac{1}{4}\cdot\frac{1}{3}}={\frac{1}{12},}$ the same as in Example 2.

When $n$ Is a Positive Integer

Start from $B(m,1)=\frac{1}{m}$ and apply the second formula $n-1$ times. For every $m\gt 0$ and every positive integer $n,$

When $m$ is a positive integer too, the denominator equals $\frac{(m+n-1)!}{(m-1)!},$ so

Example 4. ${B(3,4)}={\frac{2!\cdot 3!}{6!}}={\frac{2\cdot 6}{720}}={\frac{1}{60}.}$

Example 5. Find $\int_0^1 x^4(1-\nobreak x)^3\,dx.$

This is $B(m,n)$ with $m-1=4$ and $n-1=3\text{:}$

${\int_0^1 x^4(1-x)^3\,dx}={B(5,4)}={\frac{4!\cdot 3!}{8!}}={\frac{24\cdot 6}{40320}}={\frac{1}{280}}$

Expanding $(1-\nobreak x)^3$ gives the same answer with more work.

Example 6. ${B\left(\frac{1}{2},3\right)}={\frac{2!}{\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}}}={\frac{2}{\frac{15}{8}}}={\frac{16}{15}.}$

Compare this with the binomial coefficient $\binom{N}{k}=\frac{N!}{k!\,(N-k)!}\text{:}$

${B(k+1,N-k+1)}={\frac{k!\,(N-k)!}{(N+1)!}}={\frac{1}{(N+1)\binom{N}{k}}}$

So for positive integers every value $B(m,n)$ is the reciprocal of an integer, and the Beta function extends the binomial coefficients to non-integer numbers.

The Gamma Function

To find $B(m,n)$ when neither $m$ nor $n$ is an integer, we need the Gamma function:

Its main properties:

• $\Gamma(1)=\int_0^\infty e^{-t}\,dt=1.$
• $\Gamma(x+\nobreak 1)=x\,\Gamma(x)$ (integrate $\int_0^\infty t^{x}e^{-t}\,dt$ by parts).
• For positive integers, $\Gamma(n)=(n-\nobreak 1)!\text{:}$ $\Gamma(2)=1,$ $\Gamma(3)=2,$ $\Gamma(4)=6,$ $\Gamma(5)=24.$ So the Gamma function extends the factorial.
• $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ (proved in Example 7 below). Then $\Gamma\left(\frac{3}{2}\right)=\frac{1}{2}\sqrt{\pi},$ $\Gamma\left(\frac{5}{2}\right)=\frac{3}{2}\cdot\frac{1}{2}\sqrt{\pi}=\frac{3}{4}\sqrt{\pi},$ $\Gamma\left(\frac{7}{2}\right)=\frac{15}{8}\sqrt{\pi},$ and in general $\Gamma\left(n+\frac{1}{2}\right)=\frac{(2n)!}{4^n\,n!}\sqrt{\pi}.$
• The formula $\Gamma(x)=\frac{\Gamma(x+1)}{x}$ extends $\Gamma$ to negative numbers, step by step: first to the interval $(-1,0),$ then to $(-2,-\nobreak 1),$ and so on. For example, ${\Gamma\left(-\frac{1}{2}\right)}={\frac{\Gamma\left(\frac{1}{2}\right)}{-\frac{1}{2}}}={-2\sqrt{\pi}.}$ At $0,$ $-1,$ $-2,$ ... the Gamma function is not defined: $|\Gamma(x)|$ grows without bound near these points.
• The reflection formula (we use it without proof): $\Gamma(x)\,\Gamma(1-\nobreak x)=\frac{\pi}{\sin\pi x}$ for every non-integer $x.$ With $x=\frac{1}{2}$ it gives $\Gamma\left(\frac{1}{2}\right)^2=\pi$ again.

The Beta Function and the Gamma Function

Proof. Write $\Gamma(m)$ with the variable $x$ and $\Gamma(n)$ with the variable $y.$ Their product is a double integral over the first quadrant:

${\Gamma(m)\,\Gamma(n)}={\int_0^\infty\!\!\int_0^\infty x^{m-1}y^{n-1}e^{-(x+y)}\,dx\,dy}$

Substitute $x=uv$ and $y=u(1-\nobreak v),$ where $u\gt 0$ and $0\lt v\lt 1$ (that is, $u=x+y$ and $v=\frac{x}{x+y}$). The Jacobian is

$$\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix} v & u\\ 1-v & -u\end{vmatrix}$$

$=-uv-u(1-\nobreak v)=-u,$

so ${dx\,dy}={|-u|\,du\,dv}={u\,du\,dv.}$ Then ${x^{m-1}y^{n-1}}={u^{m+n-2}v^{m-1}(1-v)^{n-1},}$ $x+y=u,$ and the double integral splits into two:

${\Gamma(m)\,\Gamma(n)}={\int_0^\infty u^{m+n-1}e^{-u}\,du}\cdot {\int_0^1 v^{m-1}(1-v)^{n-1}\,dv}={\Gamma(m+n)\,B(m,n)}$

For positive integers, $\Gamma(k)=(k-\nobreak 1)!$ turns this formula into the one for integers above.

Example 7. Find $B\left(\frac{1}{2},\frac{1}{2}\right)$ and $\Gamma\left(\frac{1}{2}\right).$

Here the antiderivative is elementary, because ${\left(\arcsin(2t-1)\right)'}={\frac{1}{\sqrt{t(1-t)}}\text{:}}$

${B\left(\frac{1}{2},\frac{1}{2}\right)}={\int_0^1\frac{dt}{\sqrt{t(1-t)}}}={\arcsin(2t-1)\Big\vert_0^1}={\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)}={\pi}$

By the formula, $\pi=\frac{\Gamma\left(\frac{1}{2}\right)^2}{\Gamma(1)}=\Gamma\left(\frac{1}{2}\right)^2.$ Since $\Gamma(x)\gt 0$ for $x\gt 0,$ $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}.$

Example 8. Find $B\left(\frac{3}{2},\frac{5}{2}\right).$

${B\left(\frac{3}{2},\frac{5}{2}\right)}={\frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{5}{2}\right)}{\Gamma(4)}}={\frac{\frac{1}{2}\sqrt{\pi}\cdot\frac{3}{4}\sqrt{\pi}}{3!}}={\frac{\frac{3\pi}{8}}{6}}={\frac{\pi}{16}}$

Example 9. When $m+n=1,$ the reflection formula gives ${B(p,1-p)}={\Gamma(p)\,\Gamma(1-p)}={\frac{\pi}{\sin\pi p}}$ for $0\lt p\lt 1.$ For example,

${B\left(\frac{1}{3},\frac{2}{3}\right)}={\frac{\pi}{\sin\frac{\pi}{3}}}={\frac{2\pi}{\sqrt{3}}}={\frac{2\sqrt{3}\,\pi}{3}}\approx{3.6276}$

Negative Values of $m$ or $n$

When $m\le 0$ or $n\le 0,$ the integral diverges. But the right side of $B(m,n)=\frac{\Gamma(m)\,\Gamma(n)}{\Gamma(m+n)}$ still makes sense when $m$ and $n$ are not $0,$ $-1,$ $-2,$ ... (the case when $m+n$ is one of these numbers is below). This formula is used as the definition of $B(m,n)$ for such numbers, with the values of $\Gamma$ at negative numbers found from $\Gamma(x)=\frac{\Gamma(x+1)}{x}.$ Symmetry and the recurrence relations still hold; the integral forms do not.

Example 10. ${B\left(-\frac{1}{2},\frac{3}{2}\right)}={\frac{\Gamma\left(-\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)}{\Gamma(1)}}={\left(-2\sqrt{\pi}\right)\cdot\frac{\sqrt{\pi}}{2}}={-\pi.}$

The formula for a positive integer $n$ also holds for negative non-integer $m\text{:}$ for example, ${B\left(-\frac{1}{2},2\right)}={\frac{1!}{\left(-\frac{1}{2}\right)\cdot\frac{1}{2}}}={-4.}$ When $m+n$ is $0$ or a negative integer (and $m,$ $n$ are not integers), $\Gamma(m+\nobreak n)$ is not defined; there $\frac{1}{\Gamma(m+n)}$ is taken as $0$ (near these points $|\Gamma(x)|$ grows without bound, so $\frac{1}{\Gamma(x)}\to 0$), and $B(m,n)=0.$

Other Forms of the Beta Function

A substitution turns the definition into other integrals. Each of them is a tool: when an integral has one of these forms, its value is a value of the Beta function.

Trigonometric Form

Substitute $t=\sin^2\theta.$ Then $1-t=\cos^2\theta,$ $dt=2\sin\theta\cos\theta\,d\theta,$ and $\theta$ runs from $0$ to $\frac{\pi}{2}\text{:}$

${B(m,n)}={\int_0^{\pi/2}\sin^{2m-2}\theta\,\cos^{2n-2}\theta}\cdot {2\sin\theta\cos\theta\,d\theta}={2\int_0^{\pi/2}\sin^{2m-1}\theta\,\cos^{2n-1}\theta\,d\theta}$

With $p=2m-1$ and $q=2n-1,$ this form computes the integrals of powers of sine and cosine:

For example, with $p=3$ and $q=1\text{:}$ ${\int_0^{\pi/2}\sin^3\theta\,\cos\theta\,d\theta}={\frac{1}{2}\,B(2,1)}={\frac{1}{2}\cdot\frac{1}{2}}={\frac{1}{4}.}$ The substitution $u=\sin\theta$ gives the same: $\int_0^1 u^3\,du=\frac{1}{4}.$

Example 11. Find $\int_0^{\pi/2}\sin^6\theta\,\cos^4\theta\,d\theta.$

${\int_0^{\pi/2}\sin^6\theta\,\cos^4\theta\,d\theta}={\frac{1}{2}\,B\left(\frac{7}{2},\frac{5}{2}\right)}={\frac{1}{2}\cdot\frac{\Gamma\left(\frac{7}{2}\right)\Gamma\left(\frac{5}{2}\right)}{\Gamma(6)}}$

$={\frac{1}{2}\cdot\frac{\frac{15}{8}\sqrt{\pi}\cdot\frac{3}{4}\sqrt{\pi}}{120}}={\frac{1}{2}\cdot\frac{45\pi}{32\cdot 120}}={\frac{3\pi}{512}}$

Example 12. Find $\int_0^{\pi/2}\sqrt{\tan\theta}\,d\theta.$

Write $\sqrt{\tan\theta}=\sin^{1/2}\theta\,\cos^{-1/2}\theta.$ Then $p=\frac{1}{2},$ $q=-\frac{1}{2},$ and by symmetry and Example 9

${\int_0^{\pi/2}\sqrt{\tan\theta}\,d\theta}={\frac{1}{2}\,B\left(\frac{3}{4},\frac{1}{4}\right)}={\frac{1}{2}\,B\left(\frac{1}{4},\frac{3}{4}\right)}={\frac{1}{2}\cdot\frac{\pi}{\sin\frac{\pi}{4}}}={\frac{\pi}{\sqrt{2}}}={\frac{\sqrt{2}\,\pi}{2}}$

Integral from $0$ to $\infty$

In the definition, substitute $t=\frac{x}{1+x}$ (that is, $x=\frac{t}{1-t}$). Then $1-t=\frac{1}{1+x},$ $dt=\frac{dx}{(1+x)^2},$ and $x$ runs from $0$ to $\infty\text{:}$

${B(m,n)}={\int_0^\infty\frac{x^{m-1}}{(1+x)^{m-1}}}\cdot {\frac{1}{(1+x)^{n-1}}\cdot\frac{dx}{(1+x)^2}}={\int_0^\infty\frac{x^{m-1}}{(1+x)^{m+n}}\,dx}$

Example 13. ${\int_0^\infty\frac{x^2}{(1+x)^6}\,dx}={B(3,3)}={\frac{2!\cdot 2!}{5!}}={\frac{1}{30}}$ (here $m-1=2$ and $m+n=6$).

Example 14. Find $\int_0^\infty\frac{dx}{1+x^4}.$

Substitute $x^4=s,$ so $x=s^{1/4}$ and $dx=\frac{1}{4}s^{-3/4}\,ds.$ The new integral has the form above with $m-1=-\frac{3}{4}$ and $m+n=1,$ so $m=\frac{1}{4}$ and $n=\frac{3}{4}\text{:}$

${\int_0^\infty\frac{dx}{1+x^4}}={\frac{1}{4}\int_0^\infty\frac{s^{-3/4}}{1+s}\,ds}={\frac{1}{4}\,B\left(\frac{1}{4},\frac{3}{4}\right)}={\frac{1}{4}\cdot\frac{\pi}{\sin\frac{\pi}{4}}}={\frac{\sqrt{2}\,\pi}{4}}$

In the same way, ${\int_0^\infty\frac{x^{p-1}}{1+x}\,dx}={B(p,1-p)}={\frac{\pi}{\sin\pi p}}$ for $0\lt p\lt 1.$

Integral from $a$ to $b$

Substitute $x=a+(b-\nobreak a)t.$ Then $x-a=(b-\nobreak a)t,$ $b-x=(b-\nobreak a)(1-\nobreak t)$ and $dx=(b-\nobreak a)\,dt,$ so ${(x-a)^{m-1}(b-x)^{n-1}\,dx}={(b-a)^{m+n-1}\,t^{m-1}(1-t)^{n-1}\,dt,}$ and $t$ runs from $0$ to $1.$

Example 15. Find $\int_{-1}^1\left(1-x^2\right)^{3/2}dx.$

Since $1-x^2=(x+\nobreak 1)(1-\nobreak x),$ this is the form above with $a=-1,$ $b=1$ and $m=n=\frac{5}{2}\text{:}$

${\int_{-1}^1 (x+1)^{3/2}(1-x)^{3/2}\,dx}={2^4\,B\left(\frac{5}{2},\frac{5}{2}\right)}={16\cdot\frac{\Gamma\left(\frac{5}{2}\right)^2}{\Gamma(5)}}={16\cdot\frac{\frac{9\pi}{16}}{24}}={\frac{3\pi}{8}}$

A Form with a Parameter

Write the definition with the letter $s\text{:}$ ${B(m,n)}={\int_0^1 s^{m-1}(1-s)^{n-1}\,ds,}$ and substitute $s=\frac{(r+1)t}{r+t}.$ Then $1-s=\frac{r(1-t)}{r+t}$ and $ds=\frac{r(r+1)}{(r+t)^2}\,dt;$ when $t$ runs from $0$ to $1,$ so does $s.$ Multiplying gives ${s^{m-1}(1-s)^{n-1}\,ds}={\frac{r^n(r+1)^m\,t^{m-1}(1-t)^{n-1}}{(r+t)^{m+n}}\,dt.}$ For example, with $m=n=1\text{:}$

${r(r+1)\int_0^1\frac{dt}{(r+t)^2}}={r(r+1)\left(\frac{1}{r}-\frac{1}{r+1}\right)}={1}={B(1,1)}$

Table of Values

The values of $B(m,n)$ for $m,n=\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},3.$ The table is symmetric, because $B(m,n)=B(n,m).$

$m\,\backslash\,n$	$\frac{1}{2}$	$1$	$\frac{3}{2}$	$2$	$\frac{5}{2}$	$3$
$\frac{1}{2}$	$\pi$	$2$	$\frac{\pi}{2}$	$\frac{4}{3}$	$\frac{3\pi}{8}$	$\frac{16}{15}$
$1$	$2$	$1$	$\frac{2}{3}$	$\frac{1}{2}$	$\frac{2}{5}$	$\frac{1}{3}$
$\frac{3}{2}$	$\frac{\pi}{2}$	$\frac{2}{3}$	$\frac{\pi}{8}$	$\frac{4}{15}$	$\frac{\pi}{16}$	$\frac{16}{105}$
$2$	$\frac{4}{3}$	$\frac{1}{2}$	$\frac{4}{15}$	$\frac{1}{6}$	$\frac{4}{35}$	$\frac{1}{12}$
$\frac{5}{2}$	$\frac{3\pi}{8}$	$\frac{2}{5}$	$\frac{\pi}{16}$	$\frac{4}{35}$	$\frac{3\pi}{128}$	$\frac{16}{315}$
$3$	$\frac{16}{15}$	$\frac{1}{3}$	$\frac{16}{105}$	$\frac{1}{12}$	$\frac{16}{315}$	$\frac{1}{30}$

The Beta Distribution

In probability, a random variable $X$ with values between $0$ and $1$ has the Beta distribution with parameters $\alpha\gt 0$ and $\beta\gt 0$ when its probability density for $0\lt x\lt 1$ is

$f(x)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}$

Dividing by $B(\alpha,\beta)$ makes the total area under the density equal to $1.$ The Beta distribution describes proportions and probabilities, for example the share of defective items, which varies from batch to batch. The recurrence relation gives its mean value:

${E(X)}={\int_0^1 x\,f(x)\,dx}={\frac{B(\alpha+1,\beta)}{B(\alpha,\beta)}}={\frac{\alpha}{\alpha+\beta}}$

The figure shows four densities. When $\alpha\lt\beta,$ values near $0$ are more likely; when $\alpha=\beta,$ the graph is symmetric about $x=\frac{1}{2}.$

Practice Problems

Solve the problems, then open them to see the answers. You can check the values of $B(m,n)$ with the calculator at the top of this page.

1. Find $B(3,2)$ from the definition.

${B(3,2)}={\int_0^1 t^2(1-t)\,dt}={\int_0^1\left(t^2-t^3\right)dt}={\frac{1}{3}-\frac{1}{4}}={\frac{1}{12}}$ (the same as $B(2,3)\text{:}$ symmetry). Answer: $\frac{1}{12}.$

2. Find $B(4,4).$

${B(4,4)}={\frac{3!\cdot 3!}{7!}}={\frac{36}{5040}}={\frac{1}{140}.}$ Answer: $\frac{1}{140}.$

3. Find $\int_0^1 x^5(1-\nobreak x)^2\,dx.$

This is ${B(6,3)}={\frac{5!\cdot 2!}{8!}}={\frac{240}{40320}}={\frac{1}{168}.}$ Answer: $\frac{1}{168}.$

4. Find $B\left(\frac{5}{2},\frac{1}{2}\right).$

${B\left(\frac{5}{2},\frac{1}{2}\right)}={\frac{\Gamma\left(\frac{5}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(3)}}={\frac{\frac{3}{4}\sqrt{\pi}\cdot\sqrt{\pi}}{2}}={\frac{3\pi}{8}.}$ Answer: $\frac{3\pi}{8}.$

5. Find $B\left(\frac{1}{2},4\right).$

$n=4$ is a positive integer: ${B\left(\frac{1}{2},4\right)}={\frac{3!}{\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdot\frac{7}{2}}}={\frac{6}{\frac{105}{16}}}={\frac{32}{35}.}$ Answer: $\frac{32}{35}.$

6. Knowing that $B\left(\frac{1}{3},\frac{2}{3}\right)=\frac{2\pi}{\sqrt{3}},$ find $B\left(\frac{4}{3},\frac{2}{3}\right).$

By the recurrence relation with $m=\frac{1}{3}$ and $n=\frac{2}{3}\text{:}$ ${B\left(\frac{4}{3},\frac{2}{3}\right)}={\frac{\frac{1}{3}}{1}\cdot\frac{2\pi}{\sqrt{3}}}={\frac{2\pi}{3\sqrt{3}}}={\frac{2\sqrt{3}\,\pi}{9}.}$ Answer: $\frac{2\sqrt{3}\,\pi}{9}.$

7. Find $\int_0^{\pi/2}\sin^5\theta\,\cos^3\theta\,d\theta.$

$p=5,$ $q=3,$ so $\frac{p+1}{2}=3,$ $\frac{q+1}{2}=2\text{:}$ ${\frac{1}{2}\,B(3,2)}={\frac{1}{2}\cdot\frac{2!\cdot 1!}{4!}}={\frac{1}{24}.}$ Answer: $\frac{1}{24}.$

8. Find $\int_0^{\pi/2}\sin^4\theta\,d\theta.$

$p=4,$ $q=0,$ so $\frac{p+1}{2}=\frac{5}{2},$ $\frac{q+1}{2}=\frac{1}{2}\text{:}$ ${\frac{1}{2}\,B\left(\frac{5}{2},\frac{1}{2}\right)}={\frac{1}{2}\cdot\frac{3\pi}{8}}={\frac{3\pi}{16}}$ (see problem 4). Answer: $\frac{3\pi}{16}.$

9. Find $\int_0^\infty\frac{x}{(1+x)^6}\,dx.$

$m-1=1$ and $m+n=6,$ so $m=2,$ $n=4\text{:}$ ${B(2,4)}={\frac{1!\cdot 3!}{5!}}={\frac{6}{120}}={\frac{1}{20}.}$ Answer: $\frac{1}{20}.$

10. Find $\int_0^\infty\frac{dx}{1+x^3}.$

With $x^3=s\text{:}$ $\frac{1}{3}\int_0^\infty\frac{s^{-2/3}}{1+s}\,ds;$ here $m-1=-\frac{2}{3}$ and $m+n=1,$ so $m=\frac{1}{3},$ $n=\frac{2}{3}\text{:}$ ${\frac{1}{3}\,B\left(\frac{1}{3},\frac{2}{3}\right)}={\frac{1}{3}\cdot\frac{2\pi}{\sqrt{3}}}={\frac{2\pi}{3\sqrt{3}}}={\frac{2\sqrt{3}\,\pi}{9}}$ (the same number as in problem 6). Answer: $\frac{2\sqrt{3}\,\pi}{9}.$

11. Find $\int_0^1\frac{x^2}{\sqrt{1-x^2}}\,dx.$

With $x^2=t\text{:}$ $x=t^{1/2},$ $dx=\frac{1}{2}t^{-1/2}\,dt,$ and the integral is ${\frac{1}{2}\int_0^1 t^{1/2}(1-t)^{-1/2}\,dt}={\frac{1}{2}\,B\left(\frac{3}{2},\frac{1}{2}\right)}={\frac{1}{2}\cdot\frac{\frac{1}{2}\sqrt{\pi}\cdot\sqrt{\pi}}{1}}={\frac{1}{2}\cdot\frac{\pi}{2}}={\frac{\pi}{4}.}$ Answer: $\frac{\pi}{4}.$

12. Find $\int_0^2 x\sqrt{2x-x^2}\,dx.$

Since $2x-x^2=x(2-\nobreak x),$ ${x\sqrt{2x-x^2}}={x^{3/2}(2-x)^{1/2}.}$ This is the form with $a=0,$ $b=2,$ $m-1=\frac{3}{2},$ $n-1=\frac{1}{2}\text{:}$ ${2^3\,B\left(\frac{5}{2},\frac{3}{2}\right)}={8\cdot\frac{\frac{3}{4}\sqrt{\pi}\cdot\frac{1}{2}\sqrt{\pi}}{3!}}={\frac{\pi}{2}.}$ Answer: $\frac{\pi}{2}.$

13. Find $B\left(-\frac{1}{2},3\right).$

$n=3$ is a positive integer: ${\frac{2!}{\left(-\frac{1}{2}\right)\cdot\frac{1}{2}\cdot\frac{3}{2}}}={\frac{2}{-\frac{3}{8}}}={-\frac{16}{3}.}$ Answer: $-\frac{16}{3}.$

14. A random variable $X$ has the Beta distribution with $\alpha=3$ and $\beta=2.$ Find $E(X)$ and the probability $P\left(X\lt\frac{1}{2}\right).$

$E(X)=\frac{3}{3+2}=\frac{3}{5}.$ Since $B(3,2)=\frac{1}{12},$ the density is $f(x)=12x^2(1-\nobreak x),$ and ${P\left(X\lt\frac{1}{2}\right)}={12\int_0^{1/2}\left(x^2-x^3\right)dx}={12\left(\frac{1}{24}-\frac{1}{64}\right)}={\frac{5}{16}.}$ Answer: $E(X)=\frac{3}{5},$ $P=\frac{5}{16}.$

More: definite integrals · binomial coefficients