10.3 The Gamma Function
Recall the definition of an improper integral specialized to $[0, \infty) . \int_{0}^{\infty} f(t) d t$ is the number, if it exists which equals $\lim {\delta \rightarrow 0, R \rightarrow \infty} \int{\delta}^{R} f(t) d t$. Equivalently, if $\delta_{n} \rightarrow 0$ and $R_{n} \rightarrow \infty$ are any two sequences converging to 0 then $\int_{\delta_{n}}^{R_{n}} f(t) d t \rightarrow \int_{0}^{\infty} f(t) d t$

Definition 10.3.1 The gamma function is defined by $\Gamma(\alpha) \equiv \int_{0}^{\infty} e^{-t} t^{\alpha-1} d t$ whenever $\alpha>0$.
Lemma 10.3.2 The improper integral $\int_{0}^{\infty} e^{-t} t^{\alpha-1} d t$ exists for each $\alpha>0$.
Proof: Say $\sigma, \delta$ are small and positive and $R, M$ are large. Then say $\sigma<\delta$ and $R<M$. Then
$$
\begin{aligned}
\left|\int_{\delta}^{R} e^{-t} t^{\alpha-1} d t-\int_{\sigma}^{M} e^{-t} t^{\alpha-1}\right| & \leq \int_{R}^{M} e^{-t} t^{\alpha-1} d t+\int_{\sigma}^{\delta} e^{-t} t^{\alpha-1} d t \
& \leq e^{-R}+\frac{1}{\alpha}\left(\delta^{\alpha}-\sigma^{\alpha}\right)
\end{aligned}
$$
which is small if $R$ is large and $\delta$ is small. Similar reasoning shows that
$$
\left|\int_{\sigma}^{R} e^{-t} t^{\alpha-1} d t-\int_{\delta}^{M} e^{-t} t^{\alpha-1}\right|,\left|\int_{\delta}^{M} e^{-t} t^{\alpha-1} d t-\int_{\sigma}^{R} e^{-t} t^{\alpha-1}\right|
$$
is small. Thus if $\delta_{n} \rightarrow 0+$ and $R_{n} \rightarrow \infty,\left{\int_{\delta_{n}}^{R_{n}} e^{-t} t^{\alpha-1} d t\right}$ is a Cauchy sequence and that which it converges to is not dependent on the choice of sequence $\delta_{n} \rightarrow 0$ and $R_{n} \rightarrow \infty$.

This gamma function has some fundamental properties described in the following proposition. In case the improper integral exists, we can obviously compute it in the form $\lim {\delta \rightarrow 0+} \int{\delta}^{1 / \delta} f(t) d t$ which is used in what follows. Thus also the usual algebraic properties of the Riemann integral are inherited by the improper integral.

Proposition 10.3.3 For $n$ a positive integer, $n !=\Gamma(n+1) .$ In general, $\Gamma(1)=$ $1, \Gamma(\alpha+1)=\alpha \Gamma(\alpha)$

Proof: First of all, $\Gamma(1)=\lim {\delta \rightarrow 0} \int{\delta}^{\delta^{-1}} e^{-t} d t=\lim {\delta \rightarrow 0}\left(e^{-\delta}-e^{-\left(\delta^{-1}\right)}\right)=1 .$ Next, for $\alpha>0$, $$ \begin{gathered} \Gamma(\alpha+1)=\lim {\delta \rightarrow 0} \int_{\delta}^{\delta^{-1}} e^{-t} t^{\alpha} d t=\lim {\delta \rightarrow 0}\left[-\left.e^{-t} t^{\alpha}\right|{\delta} ^{\delta^{-1}}+\alpha \int_{\delta}^{\delta^{-1}} e^{-t} t^{\alpha-1} d t\right] \
=\lim {\delta \rightarrow 0}\left(e^{-\delta} \delta^{\alpha}-e^{-\left(\delta^{-1}\right)} \delta^{-\alpha}+\alpha \int{\delta}^{\delta^{-1}} e^{-t} t^{\alpha-1} d t\right)=\alpha \Gamma(\alpha)
\end{gathered}
$$
Now it is defined that $0 !=1$ and so $\Gamma(1)=0 !$. Suppose that $\Gamma(n+1)=n !$, what of $\Gamma(n+2) ?$ Is it $(n+1) ! ?$ if so, then by induction, the proposition is established. From what was just shown, $\Gamma(n+2)=\Gamma(n+1)(n+1)=n !(n+1)=(n+1) !$ and so this proves the proposition.

The properties of the gamma function also allow for a fairly easy proof about differentiating under the integral in a Laplace transform. First is a definition.

Definition 10.3.4 A function $\phi$ has exponential growth on $[0, \infty)$ if there are positive constants $\lambda, C$ such that $|\phi(t)| \leq C e^{\lambda t}$ for all $t \geq 0$.
10.3 The Gamma Function
Recall the definition of an improper integral specialized to $[0, \infty) . \int_{0}^{\infty} f(t) d t$ is the number, if it exists which equals $\lim {\delta \rightarrow 0, R \rightarrow \infty} \int{\delta}^{R} f(t) d t$. Equivalently, if $\delta_{n} \rightarrow 0$ and $R_{n} \rightarrow \infty$ are any two sequences converging to 0 then $\int_{\delta_{n}}^{R_{n}} f(t) d t \rightarrow \int_{0}^{\infty} f(t) d t$

Definition 10.3.1 The gamma function is defined by $\Gamma(\alpha) \equiv \int_{0}^{\infty} e^{-t} t^{\alpha-1} d t$ whenever $\alpha>0$.
Lemma 10.3.2 The improper integral $\int_{0}^{\infty} e^{-t} t^{\alpha-1} d t$ exists for each $\alpha>0$.
Proof: Say $\sigma, \delta$ are small and positive and $R, M$ are large. Then say $\sigma<\delta$ and $R<M$. Then
$$
\begin{aligned}
\left|\int_{\delta}^{R} e^{-t} t^{\alpha-1} d t-\int_{\sigma}^{M} e^{-t} t^{\alpha-1}\right| & \leq \int_{R}^{M} e^{-t} t^{\alpha-1} d t+\int_{\sigma}^{\delta} e^{-t} t^{\alpha-1} d t \
& \leq e^{-R}+\frac{1}{\alpha}\left(\delta^{\alpha}-\sigma^{\alpha}\right)
\end{aligned}
$$
which is small if $R$ is large and $\delta$ is small. Similar reasoning shows that
$$
\left|\int_{\sigma}^{R} e^{-t} t^{\alpha-1} d t-\int_{\delta}^{M} e^{-t} t^{\alpha-1}\right|,\left|\int_{\delta}^{M} e^{-t} t^{\alpha-1} d t-\int_{\sigma}^{R} e^{-t} t^{\alpha-1}\right|
$$
is small. Thus if $\delta_{n} \rightarrow 0+$ and $R_{n} \rightarrow \infty,\left{\int_{\delta_{n}}^{R_{n}} e^{-t} t^{\alpha-1} d t\right}$ is a Cauchy sequence and that which it converges to is not dependent on the choice of sequence $\delta_{n} \rightarrow 0$ and $R_{n} \rightarrow \infty$.

This gamma function has some fundamental properties described in the following proposition. In case the improper integral exists, we can obviously compute it in the form $\lim {\delta \rightarrow 0+} \int{\delta}^{1 / \delta} f(t) d t$ which is used in what follows. Thus also the usual algebraic properties of the Riemann integral are inherited by the improper integral.

Proposition 10.3.3 For $n$ a positive integer, $n !=\Gamma(n+1) .$ In general, $\Gamma(1)=$ $1, \Gamma(\alpha+1)=\alpha \Gamma(\alpha)$

Proof: First of all, $\Gamma(1)=\lim {\delta \rightarrow 0} \int{\delta}^{\delta^{-1}} e^{-t} d t=\lim {\delta \rightarrow 0}\left(e^{-\delta}-e^{-\left(\delta^{-1}\right)}\right)=1 .$ Next, for $\alpha>0$, $$ \begin{gathered} \Gamma(\alpha+1)=\lim {\delta \rightarrow 0} \int_{\delta}^{\delta^{-1}} e^{-t} t^{\alpha} d t=\lim {\delta \rightarrow 0}\left[-\left.e^{-t} t^{\alpha}\right|{\delta} ^{\delta^{-1}}+\alpha \int_{\delta}^{\delta^{-1}} e^{-t} t^{\alpha-1} d t\right] \
=\lim {\delta \rightarrow 0}\left(e^{-\delta} \delta^{\alpha}-e^{-\left(\delta^{-1}\right)} \delta^{-\alpha}+\alpha \int{\delta}^{\delta^{-1}} e^{-t} t^{\alpha-1} d t\right)=\alpha \Gamma(\alpha)
\end{gathered}
$$
Now it is defined that $0 !=1$ and so $\Gamma(1)=0 !$. Suppose that $\Gamma(n+1)=n !$, what of $\Gamma(n+2) ?$ Is it $(n+1) ! ?$ if so, then by induction, the proposition is established. From what was just shown, $\Gamma(n+2)=\Gamma(n+1)(n+1)=n !(n+1)=(n+1) !$ and so this proves the proposition.

The properties of the gamma function also allow for a fairly easy proof about differentiating under the integral in a Laplace transform. First is a definition.

Definition 10.3.4 A function $\phi$ has exponential growth on $[0, \infty)$ if there are positive constants $\lambda, C$ such that $|\phi(t)| \leq C e^{\lambda t}$ for all $t \geq 0$.

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