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 ${\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{s-1}dx$ converges.

Proof. Let $f(x)=e^{-x}{x}^{s-1}$. We decompose the given integral into ${\int }_0^{1}f(x)dx$ and ${\int }_1^{\mathrm{\infty }}f(x)dx$ and show that both of them converge.

First, consider $f(x)e^{\frac{x}{2}}=\frac{x^{s-1}}{e^{\frac{x}{2}}}$ on $[1,\mathrm{\infty })$. If we take $n\in \mathbb{N}$ such that $n\ge s-1$, then

$$f(x)e^{\frac{x}{2}}=\frac{x^{s-1}}{e^{\frac{x}{2}}}\le \frac{x^n}{e^{\frac{x}{2}}}.$$

Applying L'Hôpital's rule $n$ times, we can see that $\underset{x\to \mathrm{\infty }}{lim}\frac{x^n}{e^{\frac{x}{2}}}=0$. Hence $\underset{x\to \mathrm{\infty }}{lim}f(x)e^{\frac{x}{2}}=0$. In particular, $f(x)e^{\frac{x}{2}}$ is bounded on $[1,\mathrm{\infty })$ (why?). This means there exists some constant $c>0$ such that

$$f(x)<c{e}^{-\frac{x}{2}}.$$

Integrating both sides gives

$${\int }_1^{\mathrm{\infty }}f(x)dx<{\int }_1^{\mathrm{\infty }}c{e}^{-\frac{x}{2}}dx=2c{e}^{-\frac{1}{2}}<\mathrm{\infty }.$$

Thus, the improper integral ${\int }_1^{\mathrm{\infty }}f(x)dx$ converges.

Next, on $(0,1]$, $f(x)x^{1-s}=e^{-x}<1$. This means

$$f(x)<x^{s-1}.$$

Integrating both sides gives

$${\int }_0^{1}f(x)dx<{\int }_0^1{x}^{s-1}dx=\frac{1}{s}<\mathrm{\infty }.$$

Thus, ${\int }_0^{1}f(x)dx$ converges. ■

Definition (Gamma function)

The gamma function is defined by

$$\mathrm{\Gamma }(s)={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{s-1}dx$$

where $s\in (0,\mathrm{\infty })$.

By the above lemma, the Gamma function is well-defined.

Theorem (Properties of the gamma function)

1. For any $s>0$, $\mathrm{\Gamma }(s)>0$.
2. For any $s>0$, $\mathrm{\Gamma }(s+1)=s\mathrm{\Gamma }(s)$.
3. For any $n\in \mathbb{N}$, $\mathrm{\Gamma }(n)=(n-1)!$.

Proof. 

1. Trivial.
2. By integration by parts, $$\begin{array}{rcl}\mathrm{\Gamma }(s+1)& =& {\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{s}dx\\ & =& {[-e^{-x}{x}^s]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}(-e^{-x})s{x}^{s-1}dx\\ & =& s{\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{s-1}dx=s\mathrm{\Gamma }(s).\end{array}$$
3. From part 2, we have for any $n\in \mathbb{N}$ $$\mathrm{\Gamma }(n+1)=n\mathrm{\Gamma }(n)$$ and $$\mathrm{\Gamma }(1)={\int }_0^{\mathrm{\infty }}{e}^{-x}dx={[-e^{-x}]}_0^{\mathrm{\infty }}=1.$$ Therefore, $$\begin{array}{rcl}\mathrm{\Gamma }(n)& =& (n-1)\mathrm{\Gamma }(n-1)\\ & =& (n-1)(n-2)\mathrm{\Gamma }(n-2)\\ & ⋮& \\ & =& (n-1)(n-2)\cdots 2\cdot 1\cdot \mathrm{\Gamma }(1)\\ & =& (n-1)!.\end{array}$$

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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 ${\int }_0^1{x}^{p-1}(1-x{)}^{q-1}dx$ 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 $0\le x\le \frac{1}{2}$, $f(x)x^{1-p}=(1-x{)}^{q-1}$ is bounded. Thus ${\int }_0^{\frac{1}{2}}f(x)dx$ converges.

For $\frac{1}{2}\le x<1$, $f(x)(1-x{)}^{1-q}=x^{p-1}$ 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

$$B(p,q)={\int }_0^1{x}^{p-1}(1-x{)}^{q-1}dx$$

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 $$\begin{array}{rcl}B(p,q)& =& {\int }_0^1{x}^{p-1}(1-x{)}^{q-1}dx\\ & =& {\int }_1^0(1-t{)}^{p-1}{t}^{q-1}(-dt)\\ & =& {\int }_0^1{t}^{q-1}(1-t{)}^{p-1}dt=B(q,p).\end{array}$$
3. By integration by parts, $$\begin{array}{rcl}pB(p,q+1)& =& p{\int }_0^1{x}^{p-1}(1-x{)}^{q}dx\\ & =& {[x^p(1-x{)}^q]}_0^1+q{\int }_0^1{x}^p(1-x{)}^{q-1}dx\\ & =& qB(p+1,q).\end{array}$$

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Example. For $a,b>-1$, the following holds:

$${\int }_0^{\frac{\pi }{2}}{\sin }^a\theta {\cos }^b\theta d\theta =\frac{1}{2}B(\frac{a+1}{2},\frac{b+1}{2}).$$

The proof is the following.

In $B(p,q)={\int }_0^1{x}^{p-1}(1-x{)}^{q-1}dx$, 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,

$$\begin{array}{rcl}B(p,q)& =& {\int }_0^{\frac{\pi }{2}}{\sin }^{2(p-1)}\theta {\cos }^{2(q-1)}\theta (2\sin \theta \cos \theta )d\theta \\ & =& 2{\int }_0^{\frac{\pi }{2}}{\sin }^{2p-1}\theta {\cos }^{2q-1}\theta d\theta .\end{array}$$

Now let $2p-1=a$ and $2q-1=b$ and we are done. □