Definition
Before you start: read the Gamma function lesson first. The two functions work as a pair.
For \(m > 0\), \(n > 0\), \[B(m, n) = \int_0^1 x^{m-1}(1 - x)^{n-1}\,dx\]
The integrand is a power of \(x\) times a power of \((1 - x)\) on \([0, 1]\). Whenever you see that shape, or one that can be turned into it, think Beta.
Three forms of the same function
| Form | Expression | Use it for |
|---|---|---|
| Algebraic | \(\displaystyle\int_0^1 x^{m-1}(1 - x)^{n-1}\,dx\) | polynomials on \([0, 1]\) |
| Trigonometric | \(\displaystyle 2\int_0^{\pi/2}\sin^{2m-1}\theta\,\cos^{2n-1}\theta\,d\theta\) | powers of \(\sin\) and \(\cos\) on \(\left[0, \tfrac{\pi}{2}\right]\) |
| Infinite range | \(\displaystyle\int_0^{\infty}\frac{x^{m-1}}{(1 + x)^{m+n}}\,dx\) | rational functions on \([0, \infty)\) |
The trigonometric form comes from substituting \(x = \sin^2\theta\); the infinite form from \(x = \dfrac{t}{1 + t}\).
Properties
- Symmetry: \(B(m, n) = B(n, m)\). Substitute \(x \to 1 - x\).
- Splitting: \(B(m, n) = B(m + 1, n) + B(m, n + 1)\). Multiply the integrand by \(x + (1 - x) = 1\).
- Whole numbers: \[B(m, n) = \frac{(m - 1)!\,(n - 1)!}{(m + n - 1)!}\]
The last result is a special case of the relation \(B(m, n) = \dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m + n)}\), proved in the next lesson, which also handles fractional values.
Solved examples
Example 1: A polynomial integral
EasyEvaluate \(\displaystyle\int_0^1 x^3(1 - x)^2\,dx\).
- This is \(B(m, n)\) with \(m - 1 = 3\), \(n - 1 = 2\): \(B(4, 3)\).
- \(B(4, 3) = \dfrac{3!\,2!}{6!} = \dfrac{12}{720}\).
- Check by expanding: \(\displaystyle\int_0^1(x^3 - 2x^4 + x^5)\,dx = \frac{1}{4} - \frac{2}{5} + \frac{1}{6} = \frac{1}{60}\). ✓
Answer\(\dfrac{1}{60}\)
Example 2: A trigonometric integral
MediumEvaluate \(\displaystyle\int_0^{\pi/2}\sin^5\theta\cos^3\theta\,d\theta\).
- Match \(2m - 1 = 5\) and \(2n - 1 = 3\): \(m = 3\), \(n = 2\). The integral is \(\tfrac{1}{2}B(3, 2)\).
- \(B(3, 2) = \dfrac{2!\,1!}{4!} = \dfrac{1}{12}\).
Answer\(\dfrac{1}{24}\)
Example 3: Infinite range
MediumEvaluate \(\displaystyle\int_0^\infty\frac{x^2}{(1 + x)^5}\,dx\).
- Match \(m - 1 = 2\), so \(m = 3\); and \(m + n = 5\), so \(n = 2\).
- \(B(3, 2) = \dfrac{1}{12}\).
Answer\(\dfrac{1}{12}\)
Example 4: Changing the interval
MediumEvaluate \(\displaystyle\int_0^2 x^2(2 - x)^3\,dx\).
- Substitute \(x = 2t\) to bring the interval to \([0, 1]\): \(dx = 2\,dt\).
- \(\displaystyle\int_0^1 4t^2\cdot 8(1 - t)^3\cdot 2\,dt = 64\,B(3, 4) = 64\cdot\frac{2!\,3!}{6!}\).
Answer\(\dfrac{64\cdot 12}{720} = \dfrac{16}{15}\)
Example 5: A substitution to reach Beta form
Exam levelEvaluate \(\displaystyle\int_0^1 x^5(1 - x^3)^{10}\,dx\).
- Put \(t = x^3\), so \(dt = 3x^2\,dx\). Write \(x^5\,dx = x^3\cdot x^2\,dx = t\cdot\tfrac{1}{3}\,dt\).
- The integral becomes \(\dfrac{1}{3}\displaystyle\int_0^1 t(1 - t)^{10}\,dt = \frac{1}{3}B(2, 11)\).
- \(B(2, 11) = \dfrac{1!\,10!}{12!} = \dfrac{1}{132}\).
Answer\(\dfrac{1}{396}\)
Read off \(m\) and \(n\) from the form, not from the powers. In \(x^{m-1}\) the power is one less than \(m\), and in the trigonometric form the power is \(2m - 1\). Writing the matching equations down every time avoids off-by-one errors.
Common mistakes
1. Taking the powers as \(m\) and \(n\)
\(\int_0^1 x^3(1 - x)^2\,dx\) is \(B(4, 3)\), not \(B(3, 2)\).
2. Forgetting the \(\tfrac{1}{2}\) in the trigonometric form
\(\displaystyle\int_0^{\pi/2}\sin^{2m-1}\theta\cos^{2n-1}\theta\,d\theta = \tfrac{1}{2}B(m, n)\).
3. Using Beta on the wrong interval
The algebraic form needs \([0, 1]\). For \([0, a]\), substitute \(x = at\) first, as in Example 4.
Practice questions
Try each one on paper before you open the answer.
Q1\(B(3, 3)\)
\(\dfrac{2!\,2!}{5!} = \dfrac{4}{120} = \dfrac{1}{30}\).
Q2\(\displaystyle\int_0^1 x^4(1 - x)^3\,dx\)
\(B(5, 4) = \dfrac{4!\,3!}{8!} = \dfrac{1}{280}\).
Q3\(\displaystyle\int_0^{\pi/2}\sin^3\theta\cos^5\theta\,d\theta\)
\(m = 2\), \(n = 3\): \(\tfrac{1}{2}B(2, 3) = \tfrac{1}{2}\cdot\tfrac{1}{12} = \dfrac{1}{24}\).
Q4\(\displaystyle\int_0^\infty\frac{x^3}{(1 + x)^7}\,dx\)
\(m = 4\), \(n = 3\): \(B(4, 3) = \dfrac{1}{60}\).
Q5Prove that \(B(m, n) = B(m + 1, n) + B(m, n + 1)\).
\(B(m + 1, n) + B(m, n + 1) = \displaystyle\int_0^1 x^{m-1}(1 - x)^{n-1}\big[x + (1 - x)\big]\,dx = B(m, n)\).
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Frequently asked questions
What is the Beta function?
\(B(m, n) = \displaystyle\int_0^1 x^{m-1}(1 - x)^{n-1}\,dx\) for \(m, n > 0\).
Is the Beta function symmetric?
Yes, \(B(m, n) = B(n, m)\).
How is the Beta function related to the Gamma function?
\(B(m, n) = \dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m + n)}\). See Beta and Gamma functions together.
Where the Beta function leads next
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Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.