Applications of integrals (2): Gamma and Beta functions

-

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, \[ \begin{eqnarray} \Gamma(n) &=& (n-1)\Gamma(n-1)\\ &=&(n-1)(n-2)\Gamma(n-2)\\ &\vdots&\\ &=& (n-1)(n-2)\cdots 2\cdot 1\cdot\Gamma(1)\\ &=& (n-1)!. \end{eqnarray}\]
■
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, \[ \begin{eqnarray} pB(p, q+1) &= & p\int_0^1x^{p-1}(1 - x)^{q}dx\\ &=& \left[x^p(1-x)^q\right]_0^1 + q\int_0^1x^p(1-x)^{q-1}dx\\ &=& qB(p+1, q). \end{eqnarray} \]
■

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. □


Comments

Post a Comment

Popular posts from this blog

Open sets and closed sets in \(\mathbb{R}^n\)

-
Image
Open sets In \(\mathbb{R}\), we have the notion of an open interval such as \((a, b) = \{x \in \mathbb{R} | a < x < b\}\). We want to extend this idea to apply to \(\mathbb{R}^n\). We also introduce the notions of bounded sets and closed sets in \(\mathbb{R}^n\). Recall that the \(\varepsilon\)-neighbor of a point \(x\in\mathbb{R}^n\) is defined as \(N_{\varepsilon}(x) = \{y \in \mathbb{R}^n | d(x, y) < \varepsilon \}\) where \(d(x,y)\) is the distance between \(x\) and \(y\). Definition (Open set) A subset \(U\) of \(\mathbb{R}^n\) is said to be an  open set  if the following holds: \[\forall x \in U ~ \exists \delta > 0 ~ (N_{\delta}(x) \subset U).\tag{Eq:OpenSet}\] That is, for every point in an open set \(U\), we can always find an open ball centered at that point, that is included in \(U\). See the following figure. Perhaps, it is instructive to see what is  not  an open set. Negating (Eq:OpenSet), we have \[\exists x \in U ~ \forall \delta > 0 ~ (N_{\delta}(x) \not
Read more

Euclidean spaces

-
Image
We would like to study multivariate functions (i.e., functions of many variables), continuous multivariate functions in particular. To define continuity, we need a measure of "closeness" between points. One measure of closeness is the Euclidean distance. The set \(\mathbb{R}^n\) (with \(n \in \mathbb{N}\)) with the Euclidean distance function is called a Euclidean space. This is the space where our functions of interest live. The real line is a geometric representation of \(\mathbb{R}\), the set of all real numbers. That is, each \(a \in \mathbb{R}\) is represented as the point \(a\) on the real line. The coordinate plane , or the \(x\)-\(y\) plane , is a geometric representation of \(\mathbb{R}^2\), the set of all pairs of real numbers. Each pair of real numbers \((a, b)\) is visualized as the point \((a, b)\) in the plane. Remark . Recall that \(\mathbb{R}^2 = \mathbb{R}\times\mathbb{R} = \{(x, y) | x, y \in \mathbb{R}\}\) is the Cartesian product of \(\mathbb{R}\) with i
Read more

Applications of multiple integrals

-
Image
We can use multiple integrals to compute areas and volumes of various shapes. Area of a planar region Definition (Area) Let \(D\) be a bounded closed region in \(\mathbb{R}^2\). \(D\) is said to have an area if the multiple integral of the constant function 1 over \(D\), \(\iint_Ddxdy\), exists. Its value is denoted by \(\mu(D)\): \[\mu(D) = \iint_Ddxdy.\] Example . Let us calculate the area of the disk \(D = \{(x,y)\mid x^2 + y^2 \leq a^2\}\). Using the polar coordinates, \(x = r\cos\theta, y = r\sin\theta\), \(dxdy = rdrd\theta\), and the ranges of \(r\) and \(\theta\) are \([0, a]\) and \([0, 2\pi]\), respectively. Thus, \[\begin{eqnarray*} \mu(D) &=& \iint_Ddxdy\\ &=&\int_0^a\left(\int_0^{2\pi}rd\theta\right)dr\\ &=&2\pi\int_0^a rdr\\ &=&2\pi\left[\frac{r^2}{2}\right]_0^a = \pi a^2. \end{eqnarray*}\] □ Volume of a solid figure Definition (Volume) Let \(V\) be a solid figure in the \((x,y,z)\) space \(\mathbb{R}^3\). \(V\) is sai
Read more