Gamma function The gamma function is defined by Γ(p) = ∫ from 0 to ∞ of x^(p-1) e^(-x) dx, for p not equal to zero or a negative integer.
a. Use the reduction formula ∫ from 0 to ∞ of x^p e^(-x) dx = p ∫ from 0 to ∞ of x^(p-1) e^(-x) dx for p = 1, 2, 3, ...
to show that Γ(p + 1) = p! (p factorial).

검증된 단계별 안내

Recall the definition of the Gamma function: \(\Gamma(p) = \int_0^{\infty} x^{p-1} e^{-x} \, dx\), where \(p\) is not zero or a negative integer.

Use the given reduction formula: \(\int_0^{\infty} x^p e^{-x} \, dx = p \int_0^{\infty} x^{p-1} e^{-x} \, dx\). Notice that the left integral is \(\Gamma(p+1)\) and the right integral is \(\Gamma(p)\), so rewrite it as \(\Gamma(p+1) = p \Gamma(p)\).

Apply this recursive relation repeatedly for positive integers \(p = 1, 2, 3, \ldots\) to express \(\Gamma(p+1)\) in terms of \(\Gamma(1)\): \(\Gamma(p+1) = p \times (p-1) \times (p-2) \times \cdots \times 1 \times \Gamma(1)\).

Evaluate \(\Gamma(1)\) by substituting \(p=1\) into the Gamma function definition: \(\Gamma(1) = \int_0^{\infty} x^{0} e^{-x} \, dx = \int_0^{\infty} e^{-x} \, dx\), which is a standard integral.

Since \(\Gamma(1) = 1\), conclude that \(\Gamma(p+1) = p!\), where \(p!\) is the factorial of \(p\), completing the proof.

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Gamma Function Definition

The gamma function Γ(p) generalizes the factorial function to real and complex numbers. It is defined as the improper integral Γ(p) = ∫₀^∞ x^(p-1) e^(-x) dx for p > 0 and p not a negative integer. Understanding this integral form is essential to relate the gamma function to factorials.

Reduction Formula for the Gamma Function

The reduction formula ∫₀^∞ x^p e^(-x) dx = p ∫₀^∞ x^(p-1) e^(-x) dx expresses the integral with power p in terms of the integral with power p-1. This recursive relationship is key to proving properties of the gamma function, such as connecting Γ(p+1) to pΓ(p).

Factorial and Its Relation to the Gamma Function

The factorial of a positive integer p, denoted p!, is the product of all positive integers up to p. The gamma function satisfies Γ(p+1) = p!, linking continuous and discrete mathematics. Demonstrating this equality involves using the reduction formula and the base case Γ(1) = 1.