This article delves into the elegant mathematical concept of extending the factorial function to non-integer values, specifically how to prove that the factorial of 1/2 is 0.5√π. The proof utilizes the Gamma function, which generalizes the concept of factorials to the realm of real and complex numbers. We discuss the properties of the Pi function and the Gamma function, and apply integral calculus to solve the problem. Read on for a comprehensive understanding of this intriguing mathematical challenge.

Understanding the Gamma Function

The factorial function is traditionally defined for natural numbers, but the Gamma function extends this concept to all positive real numbers. For any positive integer n, the Gamma function Γ(n 1) n!. However, for non-integer values, the Gamma function is defined by the integral:

Γ(z) ∫_0^∞ t^{z-1}e^{-t} dt

For z 1/2, we can find the value of the Gamma function, which is critical for solving our problem. The integral becomes:

Γ(1/2) ∫_0^∞ t^{-1/2}e^{-t} dt

This integral can be evaluated using a substitution or by recognizing it as a standard result. It is known that:

Γ(1/2) √π

From Gamma Function to 1/2!

Using the property of the Gamma function that Γ(z 1) zΓ(z), we can express 1/2! in terms of Γ:

1/2! Γ(3/2) (1/2)Γ(1/2)

Substituting Γ(1/2) √π, we get:

1/2! (1/2)√π

This shows that the factorial of 1/2 is indeed 0.5√π, proving the statement.

Exploring the Pi Function

The Pi function, denoted as Π(z), is another approach to defining the factorial for non-integer values. The functional equation for the Pi function is:

Π(x 1) xΠ(x) with Π(0) 1

Which makes Π(z) Γ(z 1) for non-negative integers z. The integral representation of the Pi function is:

Π(z) ∫_0^∞ t^ze^{-t} dt

For z 1/2, we find:

Π(1/2) ∫_0^∞ t^{1/2-1}e^{-t} dt ∫_0^∞ √t e^{-t} dt Γ(1/2) √π

Thus, using the Pi function, we have:

1/2! Π(3/2) (1/2)Π(1/2) (1/2)√π

Integration by Parts and Fourier Transforms

Another approach involves integrating the function √t e^{-t} from 0 to ∞. This integral can be solved using integration by parts or by transforming it into a form solvable through the Fourier transform. Consider:

1/2! ∫_0^∞ √{t} e^{-t} dt

By making the substitution u e^{-t}, we get:

1/2! ∫_0^1 (-ln u)^{1/2} du

This integral is the area under the curve of e^{-x^2} from 0 to ∞, which can be found using the Fourier and inverse Fourier transforms. After a series of integrals and applying Cauchy’s integral theorem, we find:

1/2! ∫_0^∞ e^{-x^2} dx √π / 2