Special Functions: Euler's Gamma & Beta Functions
ยง6.1 Euler's Gamma Function: Axiomatic Integral, Recurrence & Factorial Interpolation
1. Definition of the Gamma Function
The Euler Gamma Function $\Gamma(z)$ is defined for all complex numbers with positive real part $\operatorname{Re}(z) > 0$ (and for all real $x > 0$) by the improper integral:
2. The Fundamental Recurrence Relation: $\Gamma(z+1) = z\Gamma(z)$
Applying integration by parts with $u = t^z \implies du = z t^{z-1} dt$ and $dv = e^{-t} dt \implies v = -e^{-t}$:
For $\operatorname{Re}(z) > 0$, $\lim_{t \to \infty} t^z e^{-t} = 0$ and $\lim_{t \to 0^+} t^z e^{-t} = 0$. Thus:
3. Factorial Interpolation for Non-Negative Integers
Evaluating the base case at $z = 1$:
By mathematical induction, for any positive integer $n \in \mathbb{N}$:
Thus, $\Gamma(z)$ extends the discrete factorial function continuously across the real continuum: $\Gamma(n) = (n - 1)!$.
ยง6.2 Half-Integer Values, The Reflection Formula & Legendre Duplication
1. The Fundamental Half-Integer Value: $\Gamma(1/2) = \sqrt{\pi}$
Setting $z = 1/2$ in the definition:
Substitute $t = u^2 \implies dt = 2u \, du$:
Using the recurrence $\Gamma(x+1) = x\Gamma(x)$, higher half-integers are directly calculated:
2. Euler's Reflection Formula
For any non-integer $z \in \mathbb{C} \setminus \mathbb{Z}$:
This remarkable identity links Gamma functions to circular trigonometry. Notice that for $z = 1/2$, $\Gamma(1/2)^2 = \frac{\pi}{\sin(\pi/2)} = \pi \implies \Gamma(1/2) = \sqrt{\pi}$.
3. Legendre's Duplication Formula
ยง6.3 Euler's Beta Function: Standard, Trigonometric & Infinite Representations
1. Definition of the Beta Function
The Euler Beta Function $B(p, q)$ is defined for $p, q > 0$ by the compact interval integral:
2. Fundamental Symmetry: $B(p, q) = B(q, p)$
Substituting $u = 1 - t \implies dt = -du$ swaps the endpoints $t = 0 \to u = 1$ and $t = 1 \to u = 0$:
3. Alternative Representations
- Trigonometric Form: Substitute $t = \sin^2\theta \implies dt = 2\sin\theta\cos\theta \, d\theta$: $$\mathbf{B(p, q) = 2 \int_0^{\pi/2} \sin^{2p-1}(\theta) \cos^{2q-1}(\theta) \, d\theta}$$
- Infinite Domain Form: Substitute $t = \frac{u}{1 + u} \implies 1 - t = \frac{1}{1 + u}, dt = \frac{du}{(1+u)^2}$: $$\mathbf{B(p, q) = \int_0^\infty \frac{u^{p-1}}{(1 + u)^{p + q}} \, du}$$
ยง6.4 The Fundamental Bridge: Line-by-Line Proof of $B(p, q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$
1. The Bridge Theorem
2. Rigorous Multi-Dimensional Transformation Proof
Consider the product of two Gamma functions expressed under substitutions $t = x^2$ and $s = y^2$:
Multiplying these two independent single integrals produces a double integral over the first quadrant $Q_1 = \{ (x, y) : x \ge 0, y \ge 0 \}$ of $\mathbb{R}^2$:
Transform into 2D polar coordinates: $x = r\cos\theta, y = r\sin\theta, dx\,dy = r\,dr\,d\theta$ with $r \in [0, \infty)$ and $\theta \in [0, \pi/2]$:
Recognize both parenthetical factors:
- The radial factor: substituting $u = r^2 \implies du = 2r\,dr$ gives $2 \int_0^\infty r^{2(p+q)-1} e^{-r^2} dr = \int_0^\infty u^{(p+q)-1} e^{-u} du = \mathbf{\Gamma(p + q)}$.
- The angular factor: by the trigonometric form of the Beta function from Section 6.3: $$2 \int_0^{\pi/2} \sin^{2q-1}(\theta) \cos^{2p-1}(\theta) \, d\theta = \mathbf{B(p, q)}$$
Therefore:
ยง6.5 Applications of Gamma & Beta Functions: Hypersphere Volumes & Asymptotics
1. Volume of the $n$-Dimensional Euclidean Ball $V_n(R)$
Let $B_n(R) = \{ x \in \mathbb{R}^n : \sum_{i=1}^n x_i^2 \le R^2 \}$ be the $n$-dimensional Euclidean ball of radius $R$. Its volume scales as $V_n(R) = C_n R^n$.
Integrating the multi-variable Gaussian over $\mathbb{R}^n$:
Converting to spherical shell integration $d^n x = S_{n-1}(r) dr = n C_n r^{n-1} dr$:
With substitution $u = r^2$, the integral evaluates to $\frac{1}{2}\Gamma(n/2)$. Thus:
2. Verification in Low Dimensions
- $n = 1$: $V_1(R) = \frac{\pi^{1/2}}{\Gamma(3/2)} R = \frac{\sqrt{\pi}}{\frac{1}{2}\sqrt{\pi}} R = 2R$ (Line segment length)
- $n = 2$: $V_2(R) = \frac{\pi^1}{\Gamma(2)} R^2 = \pi R^2$ (Circular disk area)
- $n = 3$: $V_3(R) = \frac{\pi^{3/2}}{\Gamma(5/2)} R^3 = \frac{\pi^{3/2}}{\frac{3}{4}\sqrt{\pi}} R^3 = \frac{4}{3}\pi R^3$ (3D sphere volume)
- $n = 4$: $V_4(R) = \frac{\pi^2}{\Gamma(3)} R^4 = \frac{\pi^2}{2} R^4$ (4D hypersphere volume)
Step-by-Step Solved Examination Problems
Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.
Evaluate the definite integral using the Beta and Gamma functions:
Step 1: Match with the Trigonometric Beta Representation
Recall the identity:
Equating exponents:
Step 2: Evaluate the Gamma Values
Step 3: Combine and Simplify
$I = \frac{3\pi}{512}$
Evaluate the improper integral:
by mapping to the Beta function.
Step 1: Perform the Variable Transformation
Substitute $u = x^4 \implies x = u^{1/4}$, with differential $dx = \frac{1}{4} u^{-3/4} du$.
When $x = 0 \implies u = 0$, and when $x \to \infty \implies u \to \infty$.
Step 2: Match with the Infinite Beta Representation
Recall the identity:
Equating exponents:
Step 3: Apply the Reflection Formula
Since $\Gamma(2) = 1! = 1$ and $\Gamma(5/4) = \frac{1}{4}\Gamma(1/4)$:
By Euler's Reflection Formula $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$ with $z = 1/4$:
$I = \frac{\pi\sqrt{2}}{16}$
Consider the volume $V_n$ of an $n$-dimensional unit ball ($R = 1$) as a continuous function of real dimension $n \in [0, \infty)$:
(a) Evaluate $V(n)$ for $n = 0, 1, 2, 3, 4, 5, 6$. (b) Prove that as $n \to \infty$, $V(n) \to 0$ (the volume of a unit hypersphere vanishes in infinite dimensions). (c) Determine the exact dimension $n^*$ at which the unit ball volume reaches its global maximum.
Part (a): Values in Dimensions $n = 0$ to $6$
- $n = 0$: $V(0) = \frac{1}{\Gamma(1)} = 1$
- $n = 1$: $V(1) = \frac{\sqrt{\pi}}{\Gamma(3/2)} = 2$
- $n = 2$: $V(2) = \frac{\pi}{\Gamma(2)} = \pi \approx 3.1416$
- $n = 3$: $V(3) = \frac{\pi^{3/2}}{\Gamma(5/2)} = \frac{4}{3}\pi \approx 4.1888$
- $n = 4$: $V(4) = \frac{\pi^2}{\Gamma(3)} = \frac{\pi^2}{2} \approx 4.9348$
- $n = 5$: $V(5) = \frac{\pi^{5/2}}{\Gamma(7/2)} = \frac{8\pi^2}{15} \approx 5.2638$
- $n = 6$: $V(6) = \frac{\pi^3}{\Gamma(4)} = \frac{\pi^3}{6} \approx 5.1677$
Notice that the volume increases up to $n = 5$ ($V(5) \approx 5.2638$) and then drops at $n = 6$ ($V(6) \approx 5.1677$)!
Part (b): Asymptotic Limit as $n \to \infty$
By Stirling's approximation, $\Gamma(x+1) \sim \sqrt{2\pi x}\left(\frac{x}{e}\right)^x$. With $x = n/2$:
For $n > 2\pi e \approx 17.079$, the base $\frac{2\pi e}{n} < 1$, driving the power to zero at a super-exponential rate:
Part (c): Global Maximum Dimension $n^*$
Taking the natural logarithm of $V(n)$:
Differentiating with respect to $n$ and setting to zero:
where $\psi(z) = \frac{\Gamma'(z)}{\Gamma(z)}$ is the Digamma function. Solving $\psi(z) = \ln(\pi) \approx 1.14473$ numerically yields $z \approx 3.627$, which corresponds to:
Thus, the unit hypersphere volume peaks at dimension $n^* \approx 5.26$, explaining why among integers, the maximum occurs at dimension $n = 5$.
$V(n) \to 0$ as $n \to \infty$; global maximum occurs at $n^* \approx 5.26$ with integer maximum at $n = 5$ ($V(5) \approx 5.2638$).