Derivation of the Gamma Function 📂Functions

Derivation of the Gamma Function

Non-negative Integers and the Gamma Function

For $\alpha >0$, $$ \int_{0}^{\infty} e^{-\alpha x} dx=\left[-\frac{1}{\alpha}e^{-\alpha x}\right]_{0}^{\infty}=\frac{1}{\alpha} $$ Differentiating both sides with respect to $\alpha$, according to the Leibniz integral rule, allows the differentiation to move under the integration sign, thus giving $$ \begin{align*} &&\int_{0}^\infty -xe^{-\alpha x}dx&=-\frac{1}{\alpha^2} \\ \implies && \int_{0}^\infty xe^{-\alpha x}dx &= \frac{1}{\alpha ^2} \end{align*} $$ Continuing to differentiate gives $$ \begin{align*} \int_{0}^\infty x^2e^{-\alpha x}dx&=\frac{2}{\alpha^3} \\ \int_{0}^\infty x^3e^{-\alpha x}dx&=\frac{3\cdot 2}{\alpha^4} \\ \int_{0}^\infty x^4e^{-\alpha x}dx &=\frac{4\cdot 3\cdot 2}{\alpha^5} \\ &\vdots \\ \int_{0}^\infty x^ne^{-\alpha x}dx&=\frac{n!}{\alpha^{n+1}} \end{align*} $$ If we set this as $\alpha =1$, then $$ \int_{0}^\infty x^n e^{-x}dx=n! \quad n=1,2,3,\cdots $$ From the above equation, the reason why $0!=1$ is naturally explained. If we say $n=0$, then $$ 0!=\int_{0}^\infty e^{-x}dx=\left[-e^{-x}\right]_{0}^\infty=1 $$ Hence, $0!:=1$ can be naturally defined.

Recurrence Relation of Gamma Function

$n$ does not have to be an integer to define the function using the above integral value. This is called the Gamma function. Usually, when it is an integer, it is written as $n$, and if not, it is written as $p$. $$ \Gamma (p)=\int_{0}^\infty x^{p-1}e^{-x}dx,\quad p>0 \tag{1} $$ The reason why the range is $p>0$ is that the improper integral converges only in this range. When $p\le 0$, the above integral diverges, so it cannot be used to define $\Gamma (p)$. When $p\le 0$, the method of defining the Gamma function is introduced further below. Also note that it is $\Gamma (p)=(p-1)!$ not $\Gamma (p)= p!$. If $p$ is an integer, then the Gamma function becomes the same as the factorial, hence it is obvious that $\Gamma (n+1)=n \Gamma (n)$ holds. However, this also holds when $p$ is not an integer. First, substituting $p+1$ for $p$ in $(1)$ yields $$ \Gamma (p+1)=\int_{0}^\infty x^pe^{-x}dx,\quad p>-1, \tag{2} $$ Partially integrating the right side of $(2)$ gives $$ \begin{align*} \int_{0}^{\infty} x^{p}e^{-x}dx&= \int_{0}^{\infty} (-x^{p})(-e^{-x})dx \\ &= \left[-x^{p}e^{-x}\right]_{0}^{\infty}+\int_{0}^{\infty} px^{p-1}e^{-x}dx \\ &= p\int_{0}^{\infty} x^{p-1}e^{-x}dx \\ &=p\Gamma (p) \end{align*} $$ Therefore, combining the above result and $(2)$ yields $$ \Gamma (p+1)=p\Gamma (p),\quad p>-1 \tag{3} $$ $(3)$ is called the recurrence relation of the Gamma function. This allows expressions involving the Gamma function to be simplified. For example, $$ \frac{\Gamma (1/4)}{\Gamma (9/4)}=\frac{\Gamma (1/4)}{\frac{5}{4}\Gamma (5/4)}=\frac{\Gamma (1/4)}{\frac{5}{4}\frac{1}{4}\Gamma (1/4)}=\frac{16}{5} $$

Extending the Gamma Function to Negative Numbers

The recurrence relation allows for defining the Gamma function for negative numbers. Examining $(3)$ reveals that $-1<p<0$ can be inserted into the Gamma function on the right side. Therefore, the Gamma function for $p<0$ is defined as follows. $$ \Gamma (p)=\frac{1}{p}\Gamma (p+1),\quad p<0 $$ For example, $\Gamma (-3/5)=-\frac{5}{3}\Gamma (2/5 )$ and $\Gamma (-8/5)=-\frac{5}{8}\Gamma (-3/5)=\frac{25}{24}\Gamma (2/5)$. When $p=0$, it can be shown to diverge as follows. Since $\Gamma (1)=0!=1$, $$ \lim \limits_{p \rightarrow 0} \Gamma (p)=\lim \limits_{p \rightarrow 0} \frac{\Gamma (p+1)}{p}=\infty $$