Mathematics / Analysis Integral Calculus & Series 100% Free Open Access
Chapter 6 โ€ข Theory & Derivations

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:

$$\mathbf{\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} \, dt}$$

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 $dv = e^{-t} dt \implies v = -e^{-t}$:

$$\Gamma(z+1) = \int_0^\infty t^z e^{-t} \, dt = \Big[ -t^z e^{-t} \Big]_0^\infty + z \int_0^\infty t^{z-1} e^{-t} \, dt$$

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:

$$\mathbf{\Gamma(z+1) = z \Gamma(z)}$$

3. Factorial Interpolation for Non-Negative Integers

Evaluating the base case at $z = 1$:

$$\Gamma(1) = \int_0^\infty e^{-t} \, dt = \Big[ -e^{-t} \Big]_0^\infty = 0 - (-1) = 1$$

By mathematical induction, for any positive integer $n \in \mathbb{N}$:

$$\mathbf{\Gamma(n+1) = n \cdot \Gamma(n) = n(n-1)\cdots 1 \cdot \Gamma(1) = n!}$$

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:

$$\Gamma(1/2) = \int_0^\infty t^{-1/2} e^{-t} \, dt$$

Substitute $t = u^2 \implies dt = 2u \, du$:

$$\Gamma(1/2) = \int_0^\infty \frac{1}{u} e^{-u^2} (2u \, du) = 2 \int_0^\infty e^{-u^2} \, du = \int_{-\infty}^\infty e^{-u^2} \, du = \mathbf{\sqrt{\pi}}$$

Using the recurrence $\Gamma(x+1) = x\Gamma(x)$, higher half-integers are directly calculated:

$$\Gamma(3/2) = \frac{1}{2}\sqrt{\pi}, \quad \Gamma(5/2) = \frac{3}{4}\sqrt{\pi}, \quad \Gamma(7/2) = \frac{15}{8}\sqrt{\pi} = \frac{(2n-1)!!}{2^n}\sqrt{\pi}$$

2. Euler's Reflection Formula

For any non-integer $z \in \mathbb{C} \setminus \mathbb{Z}$:

$$\mathbf{\Gamma(z) \Gamma(1 - z) = \frac{\pi}{\sin(\pi z)}}$$

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

$$\mathbf{\Gamma(z) \Gamma\left(z + \frac{1}{2}\right) = 2^{1 - 2z} \sqrt{\pi} \, \Gamma(2z)}$$

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:

$$\mathbf{B(p, q) = \int_0^1 t^{p-1} (1 - t)^{q-1} \, dt}$$

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 $t = 1 \to u = 0$:

$$B(p, q) = \int_1^0 (1 - u)^{p-1} u^{q-1} (-du) = \int_0^1 u^{q-1} (1 - u)^{p-1} \, du = \mathbf{B(q, p)}$$

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

$$\mathbf{B(p, q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p + q)}}$$

2. Rigorous Multi-Dimensional Transformation Proof

Consider the product of two Gamma functions expressed under substitutions $t = x^2$ and $s = y^2$:

$$\Gamma(p) = \int_0^\infty t^{p-1} e^{-t} \, dt = 2 \int_0^\infty x^{2p-1} e^{-x^2} \, dx$$
$$\Gamma(q) = \int_0^\infty s^{q-1} e^{-s} \, ds = 2 \int_0^\infty y^{2q-1} e^{-y^2} \, dy$$

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 $\mathbb{R}^2$:

$$\Gamma(p) \Gamma(q) = 4 \int_0^\infty \int_0^\infty x^{2p-1} y^{2q-1} e^{-(x^2 + y^2)} \, dx \, dy$$

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 $\theta \in [0, \pi/2]$:

$$\begin{aligned} \Gamma(p) \Gamma(q) &= 4 \int_0^{\pi/2} \int_0^\infty (r\cos\theta)^{2p-1} (r\sin\theta)^{2q-1} e^{-r^2} r \, dr \, d\theta \ &= 4 \left( \int_0^\infty r^{2(p+q) - 1} e^{-r^2} \, dr \right) \left( \int_0^{\pi/2} \cos^{2p-1}(\theta) \sin^{2q-1}(\theta) \, d\theta \right) \end{aligned}$$

Recognize both parenthetical factors:

  1. The radial factor: substituting $u = r^2 \implies du = 2r\,dr$ gives $2 \int_0^\infty r^{2(p+q)-1} e^{-r^2} dr = \int_0^\infty u^{(p+q)-1} e^{-u} du = \mathbf{\Gamma(p + q)}$.
  2. 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:

$$\Gamma(p) \Gamma(q) = \Gamma(p + q) \cdot B(p, q) \implies \mathbf{B(p, q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p + q)}} \quad \blacksquare$$

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 $\mathbb{R}^n$:

$$\int_{\mathbb{R}^n} e^{-\|x\|^2} \, d^n x = \left( \int_{-\infty}^\infty e^{-x^2} dx \right)^n = (\sqrt{\pi})^n = \pi^{n/2}$$

Converting to spherical shell integration $d^n x = S_{n-1}(r) dr = n C_n r^{n-1} dr$:

$$\int_{\mathbb{R}^n} e^{-\|x\|^2} \, d^n x = n C_n \int_0^\infty r^{n-1} e^{-r^2} \, dr$$

With substitution $u = r^2$, the integral evaluates to $\frac{1}{2}\Gamma(n/2)$. Thus:

$$\pi^{n/2} = n C_n \cdot \frac{1}{2}\Gamma(n/2) = C_n \cdot \frac{n}{2}\Gamma(n/2) = C_n \Gamma\left(\frac{n}{2} + 1\right)$$
$$\mathbf{V_n(R) = \frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2} + 1\right)} R^n}$$

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)
TIERED UNIVERSITY HONORS PROBLEMS

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.

Tier 1: Foundational Example 6.1: Trigonometric Definite Integral via Beta Functions

Evaluate the definite integral using the Beta and Gamma functions:

$$I = \int_0^{\pi/2} \sin^6(\theta) \cos^4(\theta) \, d\theta$$

Step 1: Match with the Trigonometric Beta Representation

Recall the identity:

$$2 \int_0^{\pi/2} \sin^{2p-1}(\theta) \cos^{2q-1}(\theta) \, d\theta = B(p, q)$$

Equating exponents:

$$2p - 1 = 6 \implies 2p = 7 \implies p = \frac{7}{2}$$
$$2q - 1 = 4 \implies 2q = 5 \implies q = \frac{5}{2}$$
$$I = \frac{1}{2} B\left(\frac{7}{2}, \frac{5}{2}\right) = \frac{1}{2} \frac{\Gamma(7/2) \Gamma(5/2)}{\Gamma(7/2 + 5/2)} = \frac{1}{2} \frac{\Gamma(7/2) \Gamma(5/2)}{\Gamma(6)}$$

Step 2: Evaluate the Gamma Values

$$\Gamma(7/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi} = \frac{15}{8} \sqrt{\pi}$$
$$\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi} = \frac{3}{4} \sqrt{\pi}$$
$$\Gamma(6) = 5! = 120$$

Step 3: Combine and Simplify

$$I = \frac{1}{2} \frac{\left( \frac{15}{8}\sqrt{\pi} \right) \left( \frac{3}{4}\sqrt{\pi} \right)}{120} = \frac{1}{2} \frac{\frac{45}{32} \pi}{120} = \frac{45\pi}{64 \times 120} = \frac{3\pi}{64 \times 8} = \frac{3\pi}{512}$$
Final Answer & Physical Insight

$I = \frac{3\pi}{512}$

Tier 2: Intermediate Exam Example 6.2: Improper Integral via Algebraic Beta Transformation

Evaluate the improper integral:

$$I = \int_0^\infty \frac{x^2}{(1 + x^4)^2} \, dx$$

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 $x \to \infty \implies u \to \infty$.

$$x^2 = (u^{1/4})^2 = u^{1/2}$$
$$I = \int_0^\infty \frac{u^{1/2}}{(1 + u)^2} \left( \frac{1}{4} u^{-3/4} \, du \right) = \frac{1}{4} \int_0^\infty \frac{u^{1/2 - 3/4}}{(1 + u)^2} \, du = \frac{1}{4} \int_0^\infty \frac{u^{-1/4}}{(1 + u)^2} \, du$$

Step 2: Match with the Infinite Beta Representation

Recall the identity:

$$B(p, q) = \int_0^\infty \frac{u^{p-1}}{(1 + u)^{p+q}} \, du$$

Equating exponents:

$$p - 1 = -\frac{1}{4} \implies p = \frac{3}{4}$$
$$p + q = 2 \implies \frac{3}{4} + q = 2 \implies q = \frac{5}{4}$$
$$I = \frac{1}{4} B\left(\frac{3}{4}, \frac{5}{4}\right) = \frac{1}{4} \frac{\Gamma(3/4) \Gamma(5/4)}{\Gamma(3/4 + 5/4)} = \frac{1}{4} \frac{\Gamma(3/4) \Gamma(5/4)}{\Gamma(2)}$$

Step 3: Apply the Reflection Formula

Since $\Gamma(2) = 1! = 1$ and $\Gamma(5/4) = \frac{1}{4}\Gamma(1/4)$:

$$I = \frac{1}{4} \cdot \Gamma\left(\frac{3}{4}\right) \left( \frac{1}{4}\Gamma\left(\frac{1}{4}\right) \right) = \frac{1}{16} \Gamma\left(\frac{1}{4}\right) \Gamma\left(1 - \frac{1}{4}\right)$$

By Euler's Reflection Formula $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$ with $z = 1/4$:

$$\Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{3}{4}\right) = \frac{\pi}{\sin(\pi/4)} = \frac{\pi}{1/\sqrt{2}} = \pi \sqrt{2}$$
$$I = \frac{1}{16} \Big( \pi \sqrt{2} \Big) = \mathbf{\frac{\pi \sqrt{2}}{16}}$$
Final Answer & Physical Insight

$I = \frac{\pi\sqrt{2}}{16}$

Tier 3: Honors / Proof Challenge Example 6.3: Derivation of the Fractional Dimension Gamma Peak

Consider the volume $V_n$ of an $n$-dimensional unit ball ($R = 1$) as a continuous function of real dimension $n \in [0, \infty)$:

$$V(n) = \frac{\pi^{n/2}}{\Gamma(n/2 + 1)}$$

(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 $x = n/2$:

$$V(n) \sim \frac{\pi^{n/2}}{\sqrt{\pi n} \left( \frac{n}{2e} \right)^{n/2}} = \frac{1}{\sqrt{\pi n}} \left( \frac{2\pi e}{n} \right)^{n/2}$$

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:

$$\mathbf{\lim_{n \to \infty} V(n) = 0} \quad \blacksquare$$

Part (c): Global Maximum Dimension $n^*$

Taking the natural logarithm of $V(n)$:

$$\ln V(n) = \frac{n}{2}\ln(\pi) - \ln\Gamma\left(\frac{n}{2} + 1\right)$$

Differentiating with respect to $n$ and setting to zero:

$$\frac{d}{dn} \ln V(n) = \frac{1}{2}\ln(\pi) - \frac{1}{2} \psi\left(\frac{n}{2} + 1\right) = 0 \implies \psi\left(\frac{n}{2} + 1\right) = \ln(\pi)$$

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:

$$\frac{n^*}{2} + 1 \approx 3.627 \implies \frac{n^*}{2} \approx 2.627 \implies \mathbf{n^* \approx 5.2569}$$

Thus, the unit hypersphere volume peaks at dimension $n^* \approx 5.26$, explaining why among integers, the maximum occurs at dimension $n = 5$.

Final Answer & Physical Insight

$V(n) \to 0$ as $n \to \infty$; global maximum occurs at $n^* \approx 5.26$ with integer maximum at $n = 5$ ($V(5) \approx 5.2638$).