Generalising the Factorial

Recall that for any $n \in \mathbb N_0$ and $s \in \mathbb R^+$ , $F_n(s) := \mathcal L\{t^n\} = n!/s^{n+1}$ . Setting $s = 1$ , we obtain $n! = F_n(1)$ , and consequently, $F_n(s) = F_n(1) \cdot s^{-(n+1)}$ . In particular, $F_0(1) = 1$ .

Problem 1. For $\alpha \in \mathbb N_0$ , prove that $F(s):= \mathcal L\{t^\alpha\}$ is well-defined for $\mathrm{Re}(s) > 0$ .

(Click for Solution)

Solution. Write $s = \sigma + i\tau$ with $\sigma > 0$ . Then

$\begin{aligned} \mathcal L\{t^{\alpha}\} &= \int_0^{\infty} t^{\alpha} e^{-st}\, \mathrm dt \\ &= \int_0^{\infty} t^{\alpha} e^{-(\sigma + i \tau)t}\, \mathrm dt \\ &= \int_0^{\infty} t^{\alpha} e^{-\sigma t} e^{-i (\tau t)}\, \mathrm dt. \end{aligned}$

Taking absolute values and applying the triangle inequality,

$\begin{aligned}\left| \int_0^{\infty} t^{\alpha} e^{-st}\, \mathrm dt \right| &\leq \int_0^{\infty} t^{\alpha} e^{-\sigma t} |e^{-i (\tau t)}| \, \mathrm dt \\ &=\int_0^{\infty} t^{\alpha} e^{-\sigma t} \, \mathrm dt = F(\sigma) < \infty.\end{aligned}$

For $\alpha \in \mathbb C$ and $s \in \mathbb C$ , define $F_\alpha(s) := \mathcal L\{t^\alpha\} (s)$ whenever the right-hand side is well-defined. For example, by Problem 1, $F_\alpha(s)$ is well-defined for $\alpha \in \mathbb N_0$ and $s \in \mathbb C$ such that $\mathrm{Re}(s) > 0$ .

Problem 2. Prove that for $\mathrm{Re}(\alpha) > -1$ and $\mathrm{Re}(s) > 0$ , $F_\alpha(s)$ is well-defined. Furthermore, prove that

$\displaystyle F_{\alpha+1}(s) = \frac{\alpha+1}{s} \cdot F_\alpha(s).$

Deduce that $F_\alpha(s) = F_\alpha(1) \cdot s^{-(n+1)}$ .

(Click for Solution)

Solution. Write $\alpha = \beta + i \gamma$ with $\beta > -1$ . For $t \in (0, \infty)$ ,

$t^{\alpha+1} = t^{(\beta+1) + i \gamma} = t^{\beta + 1} \cdot e^{i (\gamma \ln(t))}$

implies that $|t^{\alpha+1}| \leq |t^{\beta+1}|$ , so that $|\mathcal L\{t^{\alpha + 1}\}| \leq \mathcal L\{t^{\beta + 1}\}$ . Since $\beta+1 > 0$ , the latter converges for $\mathrm{Re}(s) > 0$ , so does $\mathcal L\{t^\alpha\} = F_\alpha(s)$ . Integrating by parts,

$\begin{aligned}F_{\alpha+1}(s) &= \int_0^\infty t^{\alpha+1}e^{-st}\, \mathrm dt \\ &= \left[t^{\alpha+1} \cdot \frac{e^{-st}}{-s} \right]_0^\infty - \frac{\alpha+1}{-s} \int_0^\infty t^\alpha e^{-st}\, \mathrm dt \\ &= (0 - 0) + \frac{\alpha + 1}{s} F_\alpha(s) \\ &= \frac{\alpha + 1}{s} \cdot F_\alpha(s). \end{aligned}$

On the other hand, by properties of Laplace transforms,

$\displaystyle F_{\alpha+1}(s) = \mathcal L\{t \cdot t^{\alpha}\} = -\frac{\mathrm d}{\mathrm ds} F_\alpha(s).$

Hence,

$\displaystyle \frac{\mathrm d}{\mathrm ds} F_\alpha(s) = - \frac{\alpha + 1}{s} \cdot F_\alpha(s).$

We can verify that $F_\alpha(s) = F_\alpha(1) \cdot s^{-(n+1)}$ satisfies the initial value problem

$\displaystyle \frac{\mathrm dy}{\mathrm ds} = -\frac{\alpha+1}{s} \cdot y,\quad y(1) = F_\alpha(1).$

By the existence and uniqueness theorem, this solution is unique, as required.

Definition 1. For $\mathrm{Re}(\alpha) > 0$ , define $\Gamma_+(\alpha) := F_{\alpha - 1}(1)$ . We remark that $\Gamma_+(\alpha) \in \mathbb R^+$ if $\alpha \in \mathbb R^+$ and $\Gamma_+(\alpha) \in \mathbb C$ if $\alpha \notin \mathbb R^+$ .

Problem 3. Prove that $\Gamma_+(1) = 1$ and that for any $\alpha \in \mathbb R^+$ ,

$\Gamma_+(\alpha+1) = \alpha \cdot \Gamma_+(\alpha).$

In particular, $\Gamma_+(n) = (n-1)!$ for $n \in \mathbb N^+$ .

(Click for Solution)

Solution. By definition,

$\displaystyle \Gamma_+(1) = F_0(1) = (\mathcal L\{1\})(1) = \frac 11 = 1.$

Setting $s = 1$ in Problem 2,

$\Gamma_+(\alpha + 1) = F_{\alpha}(1) = \alpha \cdot F_{\alpha - 1}(1) = \alpha \cdot \Gamma_+(\alpha).$

The final identity follows inductively from $\Gamma_+(1) = (1-1)! = 0! = 1$ and

$\Gamma_+(n) = (n-1) \cdot \Gamma_+(n-1) = (n-1) \cdot (n-2)! = (n-1)!.$

Problem 4. Construct an extension of $\Gamma_+$ to the gamma function $\Gamma : \mathbb C \backslash \mathbb Z_{\leq 0} \to \mathbb C$ such that $\Gamma|_{\mathbb R^+} = \Gamma_+$ and for any $\alpha \in \mathbb C \backslash \mathbb Z_{\leq 0}$ , $\Gamma(\alpha + 1) = \alpha \cdot \Gamma(\alpha)$ .

(Click for Solution)

Solution. We have defined $\Gamma(\alpha) := \Gamma_+(\alpha)$ whenever $\mathrm{Re}(\alpha) > 0$ . Suppose $\mathrm{Re}(\alpha) < 0$ . We leave it as an exercise in induction to verify that if $-n < \mathrm{Re}(\alpha) \leq -(n-1)$ , the extension yields

$\displaystyle \Gamma(\alpha) = \frac{\Gamma(\alpha+n)}{\alpha \cdot (\alpha+1) \cdot \cdots \cdot (\alpha+n-1)},$

which is well-defined so long as $\alpha \notin \mathbb Z$ . The crucial multiplicativity property is preserved, since

$\begin{aligned} \Gamma(\alpha+1) &= \frac{\Gamma((\alpha+1)+n-1)}{ (\alpha+1) \cdot \cdots \cdot ((\alpha+1)+n-2)} \\ &= \frac{\Gamma(\alpha+n)}{(\alpha+1) \cdot \cdots \cdot (\alpha+n-1)} \\ &= \alpha \cdot \frac{\Gamma(\alpha+n)}{\alpha \cdot (\alpha+1) \cdot \cdots \cdot (\alpha+n-1)} \\ &= \alpha \cdot \Gamma(\alpha). \end{aligned}$

Problem 5. Evaluate $\Gamma(3/2)$ .

(Click for Solution)

Solution. Since $\Gamma(3/2) = 1/2 \cdot \Gamma(1/2)$ , it suffices to evaluate the latter. Here,

$\begin{aligned} \Gamma(1/2) &= \Gamma_+(1/2) \\ &= F_{-1/2}(1) \\ &= (\mathcal L\{t^{-1/2}\})(1) = \int_0^\infty t^{-1/2} e^{-t}\, \mathrm dt. \end{aligned}$

For the latter, make the substitution $u = t^{1/2} \Rightarrow \mathrm dt = 2u\, \mathrm du$ so that

$\begin{aligned} \int_0^\infty t^{-1/2} e^{-t}\, \mathrm dt &= \int_0^\infty u^{-1} e^{-u^2} \cdot 2u\, \mathrm du \\ &= 2\int_0^{\infty}e^{-u^2}\, \mathrm du \\ &= \int_{-\infty}^{\infty} e^{-u^2}\, \mathrm du = \sqrt{\pi}. \end{aligned}$

Therefore, $\Gamma(3/2) = \frac 12 \sqrt{\pi}$ .

—Joel Kindiak, 7 Jun 25, 1243H

Response

A Proposed Definition of Grace – KindiakWrites

August 20, 2025 at 9:37 pm

[…] Christian who loves to talk about Jesus more so than even Cauchy’s integral formula or the Gamma function, I have been remarkably silent about grace. I am bold in defending the logical validity of […]

LikeLike

KindiakMath