1. Introduction: Extending the Factorial

The factorial function $n! = 1 \cdot 2 \cdot 3 \cdots n$ is only defined for non-negative integers. But what if we want to compute $(1/2)!$ or $\pi!$ ?

The Gamma function $\Gamma(\alpha)$ provides this extension, defined for all positive real numbers (and beyond, via analytic continuation).

The Key Relationship

\Gamma(n+1) = n!

For positive integers n

Famous Value

\Gamma(1/2) = \sqrt{\pi}

Related to Gaussian integral

Remark 6.20: Why is Gamma Important?

The Gamma function appears throughout mathematics and science:

Probability: Gamma, Beta, and Chi-squared distributions
Physics: Quantum mechanics, statistical mechanics
Combinatorics: Binomial coefficients for non-integer arguments
Complex analysis: Analytic continuation to ℂ \\ ℤ₋

2. Definition of the Gamma Function

Definition 6.11: The Gamma Function

For $\alpha > 0$ , the Gamma function is defined by the improper integral:

\Gamma(\alpha) = \int_0^{+\infty} t^{\alpha-1} e^{-t}\,dt

This integral combines a Type II singularity at $t = 0$ (when $\alpha < 1$ ) with a Type I infinite limit at $+\infty$ .

Theorem 6.21: Convergence of Gamma Integral

The integral $\int_0^{+\infty} t^{\alpha-1} e^{-t}\,dt$ converges for all $\alpha > 0$ .

Proof:

Split at t = 1: $\Gamma(\alpha) = \int_0^1 t^{\alpha-1} e^{-t}\,dt + \int_1^{+\infty} t^{\alpha-1} e^{-t}\,dt$

Near t = 0:

t^{\alpha-1} e^{-t} \leq t^{\alpha-1} \quad \text{(since } e^{-t} \leq 1 \text{)}

Since $\int_0^1 t^{\alpha-1}\,dt = \frac{1}{\alpha}$ converges for $\alpha > 0$ (p = 1-α < 1), the first part converges.

Near t = ∞:

For any $\alpha$ , exponential decay dominates polynomial growth:

t^{\alpha-1} e^{-t} \leq C \cdot e^{-t/2} \quad \text{for large } t

Since $\int_1^{+\infty} e^{-t/2}\,dt = 2e^{-1/2}$ converges, the second part converges. ∎

∎

Example 6.33: Computing Γ(1)

Problem: Evaluate $\Gamma(1)$

Solution:

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

Example 6.34: Computing Γ(2)

Problem: Evaluate $\Gamma(2)$

Solution:

\Gamma(2) = \int_0^{+\infty} t \cdot e^{-t}\,dt

Integration by parts: $u = t, dv = e^{-t}dt$

= [-te^{-t}]_0^{+\infty} + \int_0^{+\infty} e^{-t}\,dt = 0 + 1 = 1

So $\Gamma(2) = 1 = 1!$

Remark 6.21: The Integral Representation

The choice of variable t is conventional. Alternative forms include:

$\Gamma(\alpha) = \int_0^{+\infty} x^{\alpha-1} e^{-x}\,dx$
$\Gamma(\alpha) = 2\int_0^{+\infty} u^{2\alpha-1} e^{-u^2}\,du$ (using $t = u^2$ )

3. The Fundamental Recurrence Relation

Theorem 6.22: Gamma Recurrence Relation

For all $\alpha > 0$ :

\Gamma(\alpha + 1) = \alpha \cdot \Gamma(\alpha)

Proof:

Use integration by parts on $\Gamma(\alpha + 1) = \int_0^{+\infty} t^\alpha e^{-t}\,dt$ :

Let $u = t^\alpha$ , $dv = e^{-t}dt$ , so $du = \alpha t^{\alpha-1}dt$ , $v = -e^{-t}$ .

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

The boundary term vanishes: at $t = 0$ , $t^\alpha e^{-t} = 0$ ; at $t = +\infty$ , exponential decay dominates.

\Gamma(\alpha+1) = 0 + \alpha \cdot \Gamma(\alpha) = \alpha \cdot \Gamma(\alpha)

∎

Corollary 6.3: Gamma and Factorial

For positive integers $n$ :

\Gamma(n+1) = n!

Proof:

By induction using $\Gamma(\alpha+1) = \alpha\Gamma(\alpha)$ and $\Gamma(1) = 1$ :

\Gamma(2) = 1 \cdot \Gamma(1) = 1 = 1!

\Gamma(3) = 2 \cdot \Gamma(2) = 2 \cdot 1 = 2 = 2!

\Gamma(4) = 3 \cdot \Gamma(3) = 3 \cdot 2 = 6 = 3!

In general: $\Gamma(n+1) = n \cdot (n-1) \cdots 2 \cdot 1 \cdot \Gamma(1) = n!$ ∎

∎

Remark 6.22: Extending the Factorial

The recurrence lets us define factorial for non-integers:

$(1/2)! = \Gamma(3/2) = \frac{1}{2}\Gamma(1/2) = \frac{\sqrt{\pi}}{2}$
$(3/2)! = \Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2}\Gamma(1/2) = \frac{3\sqrt{\pi}}{4}$

Example 6.35: Computing Γ(5)

Problem: Evaluate $\Gamma(5)$

Solution:

Using the recurrence repeatedly:

\Gamma(5) = 4 \cdot \Gamma(4) = 4 \cdot 3 \cdot \Gamma(3) = 4 \cdot 3 \cdot 2 \cdot \Gamma(2) = 4 \cdot 3 \cdot 2 \cdot 1 = 24 = 4!

4. Γ(1/2) and the Gaussian Integral

Theorem 6.23: Gamma at Half-Integer

\Gamma(1/2) = \sqrt{\pi}

Proof:

By definition:

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

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

= \int_0^{+\infty} u^{-1} e^{-u^2} \cdot 2u\,du = 2\int_0^{+\infty} e^{-u^2}\,du

This equals the Gaussian integral from 0 to ∞, which is $\sqrt{\pi}/2$ :

\Gamma(1/2) = 2 \cdot \frac{\sqrt{\pi}}{2} = \sqrt{\pi}

∎

Theorem 6.24: The Gaussian Integral

\int_{-\infty}^{+\infty} e^{-x^2}\,dx = \sqrt{\pi}

Proof:

Let $I = \int_{-\infty}^{+\infty} e^{-x^2}\,dx$ . Consider $I^2$ :

I^2 = \left(\int_{-\infty}^{+\infty} e^{-x^2}\,dx\right)\left(\int_{-\infty}^{+\infty} e^{-y^2}\,dy\right) = \int\int_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dx\,dy

Convert to polar coordinates: $x^2 + y^2 = r^2$ , $dx\,dy = r\,dr\,d\theta$ :

I^2 = \int_0^{2\pi} \int_0^{+\infty} e^{-r^2} r\,dr\,d\theta = 2\pi \int_0^{+\infty} r e^{-r^2}\,dr

Let $u = r^2$ :

= 2\pi \cdot \frac{1}{2} \int_0^{+\infty} e^{-u}\,du = \pi \cdot 1 = \pi

Therefore $I = \sqrt{\pi}$ . ∎

∎

Remark 6.23: Half-Integer Values

Using the recurrence $\Gamma(\alpha+1) = \alpha\Gamma(\alpha)$ :

\Gamma(3/2) = \frac{1}{2}\sqrt{\pi}

\Gamma(5/2) = \frac{3}{4}\sqrt{\pi}

\Gamma(7/2) = \frac{15}{8}\sqrt{\pi}

\Gamma(n+1/2) = \frac{(2n-1)!!}{2^n}\sqrt{\pi}

5. The Beta Function

Definition 6.12: The Beta Function

For $p, q > 0$ , the Beta function is defined by:

B(p, q) = \int_0^1 x^{p-1}(1-x)^{q-1}\,dx

Theorem 6.25: Beta-Gamma Relationship

The Beta function is related to Gamma by:

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

Proof:

Start with the product $\Gamma(p)\Gamma(q)$ :

\Gamma(p)\Gamma(q) = \int_0^{+\infty} u^{p-1}e^{-u}\,du \int_0^{+\infty} v^{q-1}e^{-v}\,dv

Substitute $u = t\cdot s$ and $v = t(1-s)$ where $t > 0$ and $0 < s < 1$ :

= \int_0^{+\infty} \int_0^1 (ts)^{p-1}(t(1-s))^{q-1} e^{-t} \cdot t\,ds\,dt

= \int_0^{+\infty} t^{p+q-1} e^{-t}\,dt \cdot \int_0^1 s^{p-1}(1-s)^{q-1}\,ds

This equals $\Gamma(p+q) \cdot B(p,q)$ . Solving: $B(p,q) = \Gamma(p)\Gamma(q)/\Gamma(p+q)$ . ∎

∎

Remark 6.24: Symmetry

The Beta function is symmetric: $B(p, q) = B(q, p)$

This follows from the substitution $x \mapsto 1-x$ in the integral.

Example 6.36: Computing B(2,3)

Problem: Evaluate $B(2, 3)$

Solution:

B(2, 3) = \frac{\Gamma(2)\Gamma(3)}{\Gamma(5)} = \frac{1! \cdot 2!}{4!} = \frac{1 \cdot 2}{24} = \frac{1}{12}

Example 6.37: Computing B(1/2, 1/2)

Problem: Evaluate $B(1/2, 1/2)$

Solution:

B(1/2, 1/2) = \frac{\Gamma(1/2)\Gamma(1/2)}{\Gamma(1)} = \frac{\sqrt{\pi} \cdot \sqrt{\pi}}{1} = \pi

This can also be verified directly: $\int_0^1 \frac{1}{\sqrt{x(1-x)}}dx = \pi$

6. Integral Evaluation Using Gamma

Theorem 6.26: Exponential Integral Formula

For $a > 0$ and $n > -1$ :

\int_0^{+\infty} x^n e^{-ax}\,dx = \frac{\Gamma(n+1)}{a^{n+1}}

Proof:

Substitute $t = ax$ , so $x = t/a$ and $dx = dt/a$ :

\int_0^{+\infty} x^n e^{-ax}\,dx = \int_0^{+\infty} \left(\frac{t}{a}\right)^n e^{-t} \cdot \frac{dt}{a} = \frac{1}{a^{n+1}} \int_0^{+\infty} t^n e^{-t}\,dt = \frac{\Gamma(n+1)}{a^{n+1}}

∎

Example 6.38: Using the Formula

Problem: Evaluate $\int_0^{+\infty} x^3 e^{-2x}\,dx$

Solution:

Apply the formula with n = 3 and a = 2:

\int_0^{+\infty} x^3 e^{-2x}\,dx = \frac{\Gamma(4)}{2^4} = \frac{3!}{16} = \frac{6}{16} = \frac{3}{8}

Example 6.39: Non-Integer Exponent

Problem: Evaluate $\int_0^{+\infty} \sqrt{x} \cdot e^{-x}\,dx$

Solution:

Here n = 1/2 and a = 1:

\int_0^{+\infty} x^{1/2} e^{-x}\,dx = \Gamma(3/2) = \frac{1}{2}\Gamma(1/2) = \frac{\sqrt{\pi}}{2}

Theorem 6.27: Trigonometric Integral

For $m, n > -1$ :

\int_0^{\pi/2} \sin^m x \cos^n x\,dx = \frac{1}{2}B\left(\frac{m+1}{2}, \frac{n+1}{2}\right)

Example 6.40: Wallis-type Integral

Problem: Evaluate $\int_0^{\pi/2} \sin^4 x \cos^2 x\,dx$

Solution:

Here m = 4, n = 2:

= \frac{1}{2}B\left(\frac{5}{2}, \frac{3}{2}\right) = \frac{1}{2} \cdot \frac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(4)}

= \frac{1}{2} \cdot \frac{(3/4)\sqrt{\pi} \cdot (1/2)\sqrt{\pi}}{6} = \frac{1}{2} \cdot \frac{3\pi/8}{6} = \frac{\pi}{32}

7. More Gamma Function Examples

Example 6.41: Power Times Exponential

Problem: Evaluate $\int_0^{+\infty} x^4 e^{-x}\,dx$

Solution:

This is $\Gamma(5)$ by definition:

\int_0^{+\infty} x^4 e^{-x}\,dx = \Gamma(5) = 4! = 24

Example 6.42: Scaled Exponential

Problem: Evaluate $\int_0^{+\infty} x^2 e^{-3x}\,dx$

Solution:

Use the formula $\int_0^\infty x^n e^{-ax}dx = \Gamma(n+1)/a^{n+1}$ :

\int_0^{+\infty} x^2 e^{-3x}\,dx = \frac{\Gamma(3)}{3^3} = \frac{2!}{27} = \frac{2}{27}

Example 6.43: Beta Function Application

Problem: Evaluate $\int_0^1 x^2(1-x)^3\,dx$

Solution:

This is $B(3, 4)$ :

\int_0^1 x^2(1-x)^3\,dx = B(3, 4) = \frac{\Gamma(3)\Gamma(4)}{\Gamma(7)} = \frac{2! \cdot 3!}{6!} = \frac{2 \cdot 6}{720} = \frac{1}{60}

Example 6.44: Gaussian with Power

Problem: Evaluate $\int_0^{+\infty} x^2 e^{-x^2}\,dx$

Solution:

Substitute $u = x^2$ , so $x = u^{1/2}$ and $dx = \frac{1}{2}u^{-1/2}du$ :

\int_0^{+\infty} x^2 e^{-x^2}\,dx = \int_0^{+\infty} u \cdot e^{-u} \cdot \frac{1}{2}u^{-1/2}\,du = \frac{1}{2}\int_0^{+\infty} u^{1/2} e^{-u}\,du

= \frac{1}{2}\Gamma(3/2) = \frac{1}{2} \cdot \frac{\sqrt{\pi}}{2} = \frac{\sqrt{\pi}}{4}

Example 6.45: Trig Power Integral

Problem: Evaluate $\int_0^{\pi/2} \sin^3 x \cos^2 x\,dx$

Solution:

Use the formula with m = 3, n = 2:

= \frac{1}{2}B\left(2, \frac{3}{2}\right) = \frac{1}{2} \cdot \frac{\Gamma(2)\Gamma(3/2)}{\Gamma(7/2)}

= \frac{1}{2} \cdot \frac{1 \cdot \frac{\sqrt{\pi}}{2}}{\frac{15\sqrt{\pi}}{8}} = \frac{1}{2} \cdot \frac{4}{15} = \frac{2}{15}

Example 6.46: General Exponential

Problem: Evaluate $\int_0^{+\infty} e^{-x^3}\,dx$

Solution:

Substitute $u = x^3$ , so $x = u^{1/3}$ and $dx = \frac{1}{3}u^{-2/3}du$ :

\int_0^{+\infty} e^{-x^3}\,dx = \frac{1}{3}\int_0^{+\infty} u^{-2/3} e^{-u}\,du = \frac{1}{3}\Gamma(1/3)

8. The Duplication Formula

Theorem 6.28: Legendre Duplication Formula

For all $z > 0$ :

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

Remark 6.25: Alternative Form

This can also be written as:

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

Example 6.47: Verifying with z = 1

Problem: Verify the duplication formula for $z = 1$

Solution:

Left side: $\Gamma(1)\Gamma(3/2) = 1 \cdot \frac{\sqrt{\pi}}{2} = \frac{\sqrt{\pi}}{2}$

Right side: $\frac{\sqrt{\pi}}{2^1}\Gamma(2) = \frac{\sqrt{\pi}}{2} \cdot 1 = \frac{\sqrt{\pi}}{2}$

Both sides match. ✓

Example 6.48: Using Duplication to Find Γ(3/2)

Problem: Use the duplication formula with z = 1/2 to find $\Gamma(3/2)$

Solution:

With z = 1/2:

\Gamma(1/2)\Gamma(1) = \frac{\sqrt{\pi}}{2^0}\Gamma(1)

Since $\Gamma(1) = 1$ and $\Gamma(1/2) = \sqrt{\pi}$ :

The formula checks out. Now use z = 1: $\Gamma(1)\Gamma(3/2) = \frac{\sqrt{\pi}}{2}\Gamma(2)$

So $\Gamma(3/2) = \frac{\sqrt{\pi}}{2}$

Remark 6.26: Applications of Duplication

The duplication formula is useful for:

Computing Gamma at half-integers
Simplifying products of Gamma functions
Proving identities involving factorials and double factorials
Evaluating certain definite integrals

9. Advanced Gamma Function Examples

Example 6.49: Laplace Transform Connection

Problem: Find the Laplace transform of $t^n$

Solution:

The Laplace transform is:

\mathcal{L}\{t^n\} = \int_0^{+\infty} t^n e^{-st}\,dt = \frac{\Gamma(n+1)}{s^{n+1}} = \frac{n!}{s^{n+1}}

This fundamental result connects Gamma to transform methods.

Example 6.50: Gaussian Moments

Problem: Evaluate $\int_0^{+\infty} x^4 e^{-x^2}\,dx$

Solution:

Substitute $u = x^2$ :

\int_0^{+\infty} x^4 e^{-x^2}\,dx = \frac{1}{2}\int_0^{+\infty} u^{3/2} e^{-u}\,du = \frac{1}{2}\Gamma(5/2)

Using $\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3\sqrt{\pi}}{4}$ :

= \frac{1}{2} \cdot \frac{3\sqrt{\pi}}{4} = \frac{3\sqrt{\pi}}{8}

Example 6.51: Beta with Equal Arguments

Problem: Evaluate $\int_0^1 [x(1-x)]^{n-1}\,dx$ for $n > 0$

Solution:

This is $B(n, n)$ :

\int_0^1 x^{n-1}(1-x)^{n-1}\,dx = B(n, n) = \frac{\Gamma(n)^2}{\Gamma(2n)}

For example, with n = 2:

B(2, 2) = \frac{(1!)^2}{3!} = \frac{1}{6}

Example 6.52: Stirling Approximation

Problem: Estimate $\Gamma(10.5)$

Solution:

Using the recurrence from $\Gamma(1/2) = \sqrt{\pi}$ :

\Gamma(10.5) = \Gamma(9.5 + 1) = 9.5 \cdot \Gamma(9.5)

Continuing: $= 9.5 \cdot 8.5 \cdot 7.5 \cdots 0.5 \cdot \sqrt{\pi}$

This equals $\frac{(19)!!}{2^{10}} \sqrt{\pi} \approx 1133278.4$

Example 6.53: Volume of n-Sphere

Problem: Show that the volume of an n-sphere with radius R involves Gamma

Solution:

The volume formula is:

V_n(R) = \frac{\pi^{n/2}}{\Gamma(n/2 + 1)} R^n

For n = 2: $V_2 = \frac{\pi}{\Gamma(2)} R^2 = \pi R^2$ (area of circle)

For n = 3: $V_3 = \frac{\pi^{3/2}}{\Gamma(5/2)} R^3 = \frac{\pi^{3/2}}{(3/4)\sqrt{\pi}} R^3 = \frac{4\pi}{3} R^3$

Example 6.54: Generalized Gaussian

Problem: Evaluate $\int_0^{+\infty} e^{-ax^p}\,dx$ for a, p > 0

Solution:

Substitute $u = ax^p$ , so $x = (u/a)^{1/p}$ and $dx = \frac{1}{pa}(u/a)^{1/p-1}du$ :

= \frac{1}{pa^{1/p}} \int_0^{+\infty} u^{1/p - 1} e^{-u}\,du = \frac{1}{pa^{1/p}} \Gamma(1/p)

10. Study Tips for Gamma Function

1. Remember the Shift

$\Gamma(n+1) = n!$ , not $\Gamma(n) = n!$ . The exponent in the integral is $t^{\alpha-1}$ , causing the off-by-one.

2. Know Key Values

Memorize: $\Gamma(1) = 1$ , $\Gamma(1/2) = \sqrt{\pi}$ , and use recurrence for others.

3. Master Substitutions

For $\int x^n e^{-ax}dx$ , use $t = ax$ . For $\int e^{-x^p}dx$ , use $u = x^p$ .

4. Beta Simplifies [0,1] Integrals

$\int_0^1 x^{p-1}(1-x)^{q-1}dx = B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$

5. Recurrence is Your Friend

$\Gamma(\alpha+1) = \alpha \Gamma(\alpha)$ lets you reduce to known values.

6. Half-Integer Pattern

$\Gamma(n + 1/2) = \frac{(2n-1)!!}{2^n}\sqrt{\pi}$ where $(2n-1)!! = 1 \cdot 3 \cdot 5 \cdots (2n-1)$ .

7. Check Scaling Factors

When the exponent of e is not just -t, account for the scaling: $\int x^n e^{-ax}dx = \Gamma(n+1)/a^{n+1}$ .

8. Verify with Special Cases

Always check your answer with known values. For n = 0, 1, 2 you can verify directly.

11. Common Mistakes with Gamma Function

Mistake 1: Off-by-One Error

Wrong: Thinking $\Gamma(n) = n!$

Right: $\Gamma(n+1) = n!$ , so $\Gamma(n) = (n-1)!$

Mistake 2: Forgetting the Exponent Shift

Wrong: $\int_0^\infty t^n e^{-t}dt = \Gamma(n)$

Right: $\int_0^\infty t^n e^{-t}dt = \Gamma(n+1)$ since the definition uses $t^{\alpha-1}$

Mistake 3: Wrong Beta-Gamma Formula

Wrong: $B(p,q) = \Gamma(p)\Gamma(q)$

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

Mistake 4: Forgetting to Apply Substitution

Wrong: $\int_0^\infty x^n e^{-ax}dx = \Gamma(n+1)$

Right: $\int_0^\infty x^n e^{-ax}dx = \frac{\Gamma(n+1)}{a^{n+1}}$

Mistake 5: Misremembering Γ(1/2)

Wrong: $\Gamma(1/2) = \pi$ or $\Gamma(1/2) = 1/2$

Right: $\Gamma(1/2) = \sqrt{\pi} \approx 1.77$

Mistake 6: Gamma at Non-Positive Integers

Wrong: Computing $\Gamma(0) = 0! = 1$

Right: $\Gamma(0)$ is undefined (has a pole). Similarly for all non-positive integers.

12. Gamma Function Value Table

α	Γ(α)	Notes
1	1	Base case: Γ(1) = 0!
2	1	Γ(2) = 1! = 1
3	2	Γ(3) = 2! = 2
4	6	Γ(4) = 3! = 6
5	24	Γ(5) = 4! = 24
1/2	$\sqrt{\pi}$	≈ 1.7725
3/2	$\frac{1}{2}\sqrt{\pi}$	≈ 0.8862
5/2	$\frac{3}{4}\sqrt{\pi}$	≈ 1.3293
7/2	$\frac{15}{8}\sqrt{\pi}$	≈ 3.3234

13. Challenge Problems

Challenge 1

Evaluate using Gamma function:

\int_0^{+\infty} x^5 e^{-3x}\,dx

Show hint

Use Γ(n+1)/a^{n+1} with n=5, a=3.

Challenge 2

Evaluate using Beta function:

\int_0^1 x^3(1-x)^4\,dx

Show hint

This is B(4,5) = Γ(4)Γ(5)/Γ(9).

Challenge 3

Prove using Gamma functions:

\int_0^{+\infty} e^{-x^4}\,dx = \Gamma(5/4)

Show hint

Substitute u = x⁴.

Challenge 4

Evaluate the integral:

\int_0^{\pi/2} \sqrt{\sin x}\,dx

Show hint

Use the trigonometric Beta formula with m=1/2, n=0.

14. Applications in Probability

Gamma Distribution

A random variable X has a Gamma distribution with parameters α > 0 and β > 0 if:

f(x) = \frac{1}{\beta^\alpha \Gamma(\alpha)} x^{\alpha-1} e^{-x/\beta}, \quad x > 0

The Gamma function in the denominator normalizes the distribution.

Beta Distribution

A random variable X has a Beta distribution with parameters α, β > 0 if:

f(x) = \frac{1}{B(\alpha, \beta)} x^{\alpha-1} (1-x)^{\beta-1}, \quad 0 < x < 1

The Beta function normalizes this probability density on [0,1].

Chi-Squared Distribution

The chi-squared distribution with k degrees of freedom is Gamma(k/2, 2):

f(x) = \frac{1}{2^{k/2} \Gamma(k/2)} x^{k/2-1} e^{-x/2}, \quad x > 0

This is fundamental in statistical hypothesis testing.

15. Beta Function Value Table

p	q	B(p,q)	Notes
1	1	1	Γ(1)²/Γ(2) = 1
2	2	1/6	Γ(2)²/Γ(4) = 1/6
3	3	1/30	Γ(3)²/Γ(6) = 4/120
1/2	1/2	π	Γ(1/2)²/Γ(1) = π
1	2	1/2	Γ(1)Γ(2)/Γ(3) = 1/2
2	3	1/12	Γ(2)Γ(3)/Γ(5) = 2/24
3	4	1/60	Γ(3)Γ(4)/Γ(7) = 12/720
1/2	3/2	π/2	Used in arc length integrals

16. Formula Reference Card

Gamma Formulas

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

$\Gamma(\alpha+1) = \alpha\Gamma(\alpha)$

$\Gamma(n+1) = n!$

$\Gamma(1/2) = \sqrt{\pi}$

Beta Formulas

$B(p,q) = \int_0^1 x^{p-1}(1-x)^{q-1}dx$

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

$B(p,q) = B(q,p)$

Integral Formulas

$\int_0^\infty x^n e^{-ax}dx = \frac{\Gamma(n+1)}{a^{n+1}}$

$\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}$

$\int_0^{\pi/2} \sin^m x \cos^n x\,dx = \frac{1}{2}B(\frac{m+1}{2}, \frac{n+1}{2})$

Duplication Formula

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

$\Gamma(n+\frac{1}{2}) = \frac{(2n-1)!!}{2^n}\sqrt{\pi}$

17. Historical Note

The Gamma function was introduced by Leonhard Euler in 1729 as an extension of the factorial to non-integer arguments. He discovered the integral representation while trying to find a smooth curve passing through the points (n, n!) for positive integers n.

The notation Γ(z) was introduced by Adrien-Marie Legendre in the early 19th century. The duplication formula is also named after Legendre.

Today, the Gamma function is one of the most important special functions in mathematics, appearing in number theory, complex analysis, probability, physics, and engineering applications.

Gamma Function Quiz

12

Questions

0

Correct

0%

Accuracy

1

The Gamma function

\Gamma(n+1)

for positive integer

n

equals:

Easy

Not attempted

2

The value of

\Gamma(1/2)

is:

Medium

Not attempted

3

The recurrence relation for Gamma is:

Easy

Not attempted

4

The Beta function

B(p,q)

equals:

Medium

Not attempted

5

The Gaussian integral

\int_{-\infty}^{+\infty} e^{-x^2}dx

equals:

Medium

Not attempted

6

The value of

\Gamma(3/2)

is:

Medium

Not attempted

7

The integral

\int_0^{+\infty} x^2 e^{-x} dx

equals:

Easy

Not attempted

8

\int_0^{+\infty} x^4 e^{-2x} dx

equals:

Hard

Not attempted

9

The Beta function

B(1/2, 1/2)

equals:

Medium

Not attempted

10

For the Gamma integral to converge at

t=0

, we need:

Hard

Not attempted

11

The value of

\Gamma(1)

is:

Easy

Not attempted

12

Which identity is TRUE?

Hard

Not attempted

Frequently Asked Questions

What is the Gamma function?

Γ(α) = ∫₀^∞ t^{α-1}e^{-t}dt for α > 0. It extends the factorial to non-integers: Γ(n+1) = n! for positive integers n.

Why is Γ(1/2) = √π important?

It connects the Gamma function to the Gaussian integral ∫e^{-x²}dx = √π, which appears in probability (normal distribution), physics, and many other areas.

What is the Beta function used for?

B(p,q) = ∫₀¹ x^{p-1}(1-x)^{q-1}dx appears in probability (Beta distribution), combinatorics, and simplifies many integrals via B(p,q) = Γ(p)Γ(q)/Γ(p+q).

How do I evaluate integrals using Gamma?

Transform to the form ∫₀^∞ t^{α-1}e^{-t}dt using substitution. For example, ∫₀^∞ x^n e^{-ax}dx = Γ(n+1)/a^{n+1} = n!/a^{n+1}.

What is the relationship between Gamma and factorial?

Γ(n+1) = n! for positive integers. This means Γ(1) = 0! = 1, Γ(2) = 1! = 1, Γ(3) = 2! = 2, Γ(4) = 3! = 6, etc.

How do I compute Γ at half-integer values?

Use Γ(1/2) = √π and the recurrence. For example: Γ(3/2) = (1/2)Γ(1/2) = √π/2, Γ(5/2) = (3/2)(1/2)Γ(1/2) = 3√π/4.

Why does Gamma converge for α > 0?

Near t=0, t^{α-1} is integrable for α > 0 (p-test). Near t=∞, e^{-t} dominates any polynomial growth, ensuring convergence.

What is the duplication formula?

Γ(2z) = (2^{2z-1}/√π) Γ(z)Γ(z+1/2). This relates Gamma at 2z to its values at z and z+1/2.

Can Gamma be extended to negative arguments?

Yes, via Γ(α) = Γ(α+1)/α. This gives poles at 0, -1, -2, ... but defines Gamma for all other real (and complex) numbers.

Where does the Gamma function appear in probability?

The Gamma distribution has PDF f(x) = x^{α-1}e^{-x/β}/(β^α Γ(α)). The Chi-squared, Exponential, and Erlang distributions are special cases.

Gamma Function and Applications

1. Introduction: Extending the Factorial

2. Definition of the Gamma Function

3. The Fundamental Recurrence Relation

4. Γ(1/2) and the Gaussian Integral

5. The Beta Function

6. Integral Evaluation Using Gamma

7. More Gamma Function Examples

8. The Duplication Formula

9. Advanced Gamma Function Examples

10. Study Tips for Gamma Function

1. Remember the Shift

2. Know Key Values

3. Master Substitutions

4. Beta Simplifies [0,1] Integrals

5. Recurrence is Your Friend

6. Half-Integer Pattern

7. Check Scaling Factors

8. Verify with Special Cases

11. Common Mistakes with Gamma Function

Mistake 1: Off-by-One Error

Mistake 2: Forgetting the Exponent Shift

Mistake 3: Wrong Beta-Gamma Formula

Mistake 4: Forgetting to Apply Substitution

Mistake 5: Misremembering Γ(1/2)

Mistake 6: Gamma at Non-Positive Integers

12. Gamma Function Value Table

13. Challenge Problems

14. Applications in Probability

Gamma Distribution

Beta Distribution

Chi-Squared Distribution

15. Beta Function Value Table

16. Formula Reference Card

Gamma Formulas

Beta Formulas

Integral Formulas

Duplication Formula

17. Historical Note

Frequently Asked Questions

What is the Gamma function?

Why is Γ(1/2) = √π important?

What is the Beta function used for?

How do I evaluate integrals using Gamma?

What is the relationship between Gamma and factorial?

How do I compute Γ at half-integer values?

Why does Gamma converge for α > 0?

What is the duplication formula?

Can Gamma be extended to negative arguments?

Where does the Gamma function appear in probability?

Gamma Definition

Recurrence Relation

Factorial Connection

Half-Integer

Beta-Gamma Relation

Remember