Beta Function Gamma Function Relationship Proof

The Intimate Relationship Between Beta and Gamma Functions: A Proof

The Beta and Gamma functions, seemingly disparate special functions, are actually deeply intertwined. Understanding their relationship is crucial for various applications in mathematics, physics, and statistics. This article will delve into the proof demonstrating this connection, highlighting the elegance and power of mathematical analysis. We'll explore how the Beta function can be elegantly expressed in terms of the Gamma function, unlocking a wealth of computational advantages.

The Beta function, denoted as B(x, y), is defined as:

B(x, y) = ∫₀¹ tˣ⁻¹(1 - t)ʸ⁻¹ dt, where Re(x) > 0 and Re(y) > 0.

The Gamma function, denoted as Γ(z), is a generalization of the factorial function to complex numbers, defined as:

Γ(z) = ∫₀^∞ t^(z-1)e⁻ᵗ dt, where Re(z) > 0.

Proving the Relationship: B(x, y) = Γ(x)Γ(y) / Γ(x+y)

The proof hinges on clever manipulation of the double integral representation. Let's break it down step-by-step:

1. Start with the Gamma function definition: We'll use two independent Gamma functions, Γ(x) and Γ(y):

   Γ(x) = ∫₀^∞ uˣ⁻¹e⁻ᵘ du Γ(y) = ∫₀^∞ vʸ⁻¹e⁻ᵛ dv

2. Formulate a double integral: Multiply the two Gamma function equations:

   Γ(x)Γ(y) = ∫₀^∞ ∫₀^∞ uˣ⁻¹vʸ⁻¹e⁻ᵘ⁻ᵛ du dv

3. Change of variables: This is the crucial step. We'll perform a change of variables to convert the double integral into a more manageable form. Let's introduce the following substitutions:

   • u = x²
   • v = y²

   This implies:

   • du = 2x dx
   • dv = 2y dy

   Substituting these into the double integral gives:

   Γ(x)Γ(y) = 4 ∫₀^∞ ∫₀^∞ x²ˣ⁻¹y²ʸ⁻¹e⁻ˣ²⁻ʸ² x y dx dy

4. Converting to polar coordinates: To simplify the integration, we switch to polar coordinates. Let:

   • x = r cos θ
   • y = r sin θ

   Then:

   • x² + y² = r²
   • dx dy = r dr dθ

   The integral becomes:

   Γ(x)Γ(y) = 4 ∫₀^π/₂ ∫₀^∞ (r cos θ)²ˣ⁻¹(r sin θ)²ʸ⁻¹e⁻ʳ² r dr dθ

   This can be simplified to:

   Γ(x)Γ(y) = 4 ∫₀^π/₂ cos²ˣ⁻¹θ sin²ʸ⁻¹θ dθ ∫₀^∞ r²ˣ⁺²ʸ⁻²e⁻ʳ² r dr

5. Solving the integrals: Let's tackle each integral separately. The first integral is related to the Beta function, and the second can be solved using the Gamma function definition through another substitution. Let's focus on the second integral, which is related to the Gamma function:

   ∫₀^∞ r²ˣ⁺²ʸ⁻¹e⁻ʳ² dr

   Using the substitution t = r², we get:

   1/2 ∫₀^∞ tˣ⁺ʸ⁻¹e⁻ᵗ dt = ½ Γ(x+y)

6. Putting it together: Substitute back into the previous equation, we have the first integral as ∫₀^π/₂ cos²ˣ⁻¹θ sin²ʸ⁻¹θ dθ = 1/4 B(x, y). Combining this result with the result from the second integral will give us the final relationship:

   Γ(x)Γ(y) = ½ Γ(x+y) B(x, y)

   Therefore, we obtain the fundamental relationship:

   B(x, y) = Γ(x)Γ(y) / Γ(x+y)

This proof showcases the deep connection between the Beta and Gamma functions. This relationship is incredibly useful for simplifying complex integrals and provides a powerful tool for various mathematical and scientific applications. Mastering this connection opens doors to a richer understanding of these important special functions.