Applications of integrals (2): Gamma and Beta functions

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The use of integrals is not limited to computing areas and lengths. Integrals are also helpful for defining new functions. Here, we study two special functions: the Gamma and Beta functions. These functions are widely used in various fields of science and engineering, as well as statistics.



Gamma function

Lemma

For any \(s > 0\), the improper integral \(\int_0^{\infty}e^{-x}x^{s-1}dx\) converges.
Proof. Let \(f(x) = e^{-x}x^{s-1}\). We decompose the given integral into \(\int_{0}^{1}f(x)dx\) and \(\int_1^{\infty}f(x)dx\) and show that both of them converge.

First, consider \(f(x)e^{\frac{x}{2}} = \frac{x^{s-1}}{e^{\frac{x}{2}}}\) on \([1,\infty)\). If we take \(n\in\mathbb{N}\) such that \(n \geq s-1\), then
\[f(x)e^{\frac{x}{2}} = \frac{x^{s-1}}{e^{\frac{x}{2}}} \leq \frac{x^{n}}{e^{\frac{x}{2}}}.\]
Applying L'Hôpital's rule \(n\) times, we can see that \(\lim_{x\to\infty}\frac{x^n}{e^{\frac{x}{2}}} = 0\). Hence \(\lim_{x\to\infty}f(x)e^{\frac{x}{2}} = 0\). In particular, \(f(x)e^{\frac{x}{2}}\) is bounded on \([1, \infty)\) (why?). This means there exists some constant \(c > 0\) such that
\[f(x) < ce^{-\frac{x}{2}}.\]
Integrating both sides gives
\[\int_1^{\infty}f(x)\,dx < \int_1^{\infty}ce^{-\frac{x}{2}}\,dx = 2ce^{-\frac{1}{2}} < \infty.\]
Thus, the improper integral \(\int_1^{\infty}f(x)dx\) converges.

Next, on \((0,1]\), \(f(x)x^{1-s} = e^{-x} < 1\). This means
\[f(x) < x^{s-1}.\]
Integrating both sides gives
\[\int_0^1f(x)\,dx < \int_0^1x^{s-1}\,dx = \frac{1}{s} < \infty.\]
Thus, \(\int_0^1f(x)dx\) converges. ■

Definition (Gamma function)

The gamma function is defined by
\[\Gamma(s) = \int_0^{\infty}e^{-x}x^{s-1}dx\]
where \(s\in (0, \infty)\).
By the above lemma, the Gamma function is well-defined.

Theorem (Properties of the gamma function)

  1. For any \(s > 0\), \(\Gamma(s) > 0\).
  2. For any \(s > 0\), \(\Gamma(s+1) = s\Gamma(s)\).
  3. For any \(n\in\mathbb{N}\), \(\Gamma(n) = (n-1)!\).
Proof. 
  1. Trivial.
  2. By integration by parts, \[\begin{eqnarray} \Gamma(s+1) &=& \int_0^{\infty}e^{-x}x^{s}dx\\ &=&\left[-e^{-x}x^s\right]_{0}^{\infty} - \int_0^{\infty}(-e^{-x})sx^{s-1}dx\\ &=&s\int_0^{\infty}e^{-x}x^{s-1}dx = s\Gamma(s). \end{eqnarray}\]
  3. From part 2, we have for any \(n\in\mathbb{N}\) \[ \Gamma(n+1) = n\Gamma(n)\] and \[ \Gamma(1) = \int_0^{\infty}e^{-x}dx = \left[-e^{-x}\right]_0^{\infty} = 1.\] Therefore, \[ \begin{eqnarray} \Gamma(n) &=& (n-1)\Gamma(n-1)\\ &=&(n-1)(n-2)\Gamma(n-2)\\ &\vdots&\\ &=& (n-1)(n-2)\cdots 2\cdot 1\cdot\Gamma(1)\\ &=& (n-1)!. \end{eqnarray}\]
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Remark. Note that the Gamma function extends factorial \(n!\) (defined for natural numbers) to real numbers. □

Beta function

The Beta function is also of practical importance.

Lemma

For any \(p > 0\) and \(q > 0\), the improper integral \(\int_0^1x^{p-1}(1-x)^{q-1}dx\) converges.
Proof. Let \(f(x) = x^{p-1}(1-x)^{q-1}\). We decompose the given integral into \(\int_0^{\frac{1}{2}}f(x)dx\) and \(\int_{\frac{1}{2}}^{1}f(x)dx\) and show that both of these converge.

For \(0 \leq x \leq \frac{1}{2}\), \(f(x)x^{1-p} = (1-x)^{q-1}\) is bounded. Thus \(\int_0^{\frac{1}{2}}f(x)dx\) converges.

For \(\frac{1}{2}\leq x < 1\), \(f(x)(1-x)^{1-q} = x^{p-1}\) is bounded. Thus \(\int_{\frac{1}{2}}^{1}f(x)dx\) converges (why?). ■

Exercise. Fill in the details of the above proof. □

Definition (Beta function)

The beta function is a function of \(p\) and \(q\) defined by
\[B(p, q) = \int_0^1x^{p-1}(1 - x)^{q-1}dx\]
where \(p > 0\) and \(q > 0\).
By the above lemma, the Beta function is well-defined.

Theorem (Properties of the beta function)

  1. For any \(p, q > 0\), \(B(p,q) > 0\).
  2. \(B(p, q) = B(q, p)\).
  3. \(B(p, q + 1) = \frac{q}{p}B(p+1, q)\).
Proof. 
  1. Trivial. 
  2. Substitute \(t = 1 - x\), and we have \[ \begin{eqnarray} B(p,q) &=& \int_0^1x^{p-1}(1 - x)^{q-1}dx\\ &=& \int_1^0(1-t)^{p-1}t^{q-1}(-dt)\\ &=& \int_0^1t^{q-1}(1-t)^{p-1}dt = B(q,p). \end{eqnarray} \]
  3. By integration by parts, \[ \begin{eqnarray} pB(p, q+1) &= & p\int_0^1x^{p-1}(1 - x)^{q}dx\\ &=& \left[x^p(1-x)^q\right]_0^1 + q\int_0^1x^p(1-x)^{q-1}dx\\ &=& qB(p+1, q). \end{eqnarray} \]
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Example. For \(a, b > -1\), the following holds:
\[\int_0^{\frac{\pi}{2}}\sin^a\theta\cos^b\theta~ d\theta = \frac{1}{2}B\left(\frac{a+1}{2}, \frac{b+1}{2}\right).\]
The proof is the following.
In \(B(p,q) = \int_0^1x^{p-1}(1-x)^{q-1}dx\), substitute \(x = \sin^2 \theta\). This is an injection and, as \(x\) moves from 0 to 1, \(\theta\) moves from \(0\) to \(\frac{\pi}{2}\).  We have \(dx = 2\sin\theta\cos\theta ~ d\theta\). Thus,
\[ \begin{eqnarray} B(p,q) &=& \int_0^{\frac{\pi}{2}}\sin^{2(p-1)}\theta\cos^{2(q-1)}\theta(2\sin\theta\cos\theta)d\theta\\ &=&2\int_0^{\frac{\pi}{2}}\sin^{2p-1}\theta\cos^{2q-1}\theta d\theta. \end{eqnarray} \]
Now let \(2p - 1 = a\) and \(2q - 1 = b\) and we are done. □


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