Question

The gamma function is defined by \(\Gamma(p)=\int_{0}^{\infty} x^{p-1} e^{-x} d x,\) for \(p\) not equal to zero or a negative integer. a. Use the reduction formula $$\int_{0}^{\infty} x^{p} e^{-x} d x=p \int_{0}^{\infty} x^{p-1} e^{-x} d x \quad \text { for } p=1,2,3, \ldots$$ to show that \(\Gamma(p+1)=p !(p\) factorial). b. Use the substitution \(x=u^{2}\) and the fact that \(\int_{0}^{\infty} e^{-u^{2}} d u=\frac{\sqrt{\pi}}{2}\) to show that \(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\).

Short answer

Question: Prove that \(\Gamma(p+1)=p!\) and find the value of \(\Gamma\left(\frac{1}{2}\right)\). Solution: (a) We have shown that \(\Gamma(p+1)=p!\) using the reduction formula for \(p=1,2,3,\ldots\). (b) We have found the value of \(\Gamma\left(\frac{1}{2}\right)\) using the substitution method, and the result is \(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\).

Step-by-step solution

Part (a): Using the reduction formula to show that \(\Gamma(p+1)=p!\)

Use the reduction formula provided, for the case when \(p=1\), we can find the integral as follows: $$\begin{aligned} \int_{0}^{\infty} x^{(1+1)-1} e^{-x} dx &= 1 \int_{0}^{\infty} x^{1-1} e^{-x} dx \\ &= 1 \int_{0}^{\infty} e^{-x} dx \\ &= 1(-e^{-x}\Big|_0^\infty) \\ &= 1(0-(-1)) \end{aligned} $$ In general we have, $$\begin{aligned} \Gamma(p+1) &= \int_{0}^{\infty} x^{(p+1)-1} e^{-x} dx \\ &= p \int_{0}^{\infty} x^{p-1} e^{-x} dx \\ &= p \Gamma(p) \end{aligned} $$ Applying this reduction iteratively, we find: $$\begin{aligned} \Gamma(p+1) &= p \Gamma(p) \\ &= p (p-1) \Gamma(p-1) \\ &= p(p-1)(p-2) \Gamma(p-2) \\ &= \cdots \\ &= p! \Gamma(1) \end{aligned} $$ Since we found that \(\Gamma(1)=1\), we can conclude that \(\Gamma(p+1)=p!\) for \(p=1,2,3,\ldots\).

Part (b): Using the substitution method to find \(\Gamma\left(\frac{1}{2}\right)\)

Let us first use the substitution \(x=u^{2}\) in the gamma function definition for \(\Gamma\left(\frac{1}{2}\right)\). We have: $$dx = 2u du$$ and $$x^{\frac{1}{2} - 1} = u^{-1}.$$ Now, we can rewrite \(\Gamma\left(\frac{1}{2}\right)\) as: $$\begin{aligned} \Gamma\left(\frac{1}{2}\right) &= \int_{0}^{\infty} (u^2)^{\frac{1}{2} - 1} e^{-u^{2}} (2u) du \\ &= 2 \int_{0}^{\infty} u^{-1} e^{-u^{2}} u du \\ &= 2 \int_{0}^{\infty} e^{-u^{2}} du \end{aligned} $$ Using the provided information, we have: $$ \int_{0}^{\infty} e^{-u^{2}} du = \frac{\sqrt{\pi}}{2} $$ Now, substituting it into the equation for \(\Gamma\left(\frac{1}{2}\right)\): $$\begin{aligned} \Gamma\left(\frac{1}{2}\right) &= 2 \left(\frac{\sqrt{\pi}}{2}\right) \\ &= \sqrt{\pi} \end{aligned} $$ So, we have shown that \(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\).

Key concepts

Integration Techniques

In calculus, there are several integration techniques that help us find the antiderivatives of functions. One of these techniques is the use of a reduction formula, which simplifies a given integral into a form that is easier to evaluate. The reduction formula is a powerful tool especially when dealing with complicated integrals that involve exponentials and factorials, such as those found in the gamma function.

Another technique, the substitution method, involves changing the variable of integration to transform the integral into a simpler form. This method can also be used in tandem with improper integrals, where the limits of integration may extend to infinity. The key to mastering these methods is practice and a clear understanding of the rules and properties of integrals.

Reduction Formula

A reduction formula in calculus is a kind of recursive relation that relates an integral to another integral with a lower power. This formula reduces a complicated integral into a simpler one until it reaches a base case that can be easily solved. In the context of the gamma function, the reduction formula allows us to find \(\Gamma(p+1)\) iteratively by reducing the power of the function step by step until we reach \(\Gamma(1)\), which is known to be 1. This technique simplifies the process of finding factorials of positive integers as it relates directly to the gamma function.

Factorial

The factorial of a non-negative integer \(n\), denoted as \(n!\), is the product of all positive integers less than or equal to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials play a crucial role in various branches of mathematics, including calculus, where they are often encountered in series expansions, probability, and permutations. The gamma function provides a way to generalize the factorial operation to non-integer values. The connection between the gamma function and factorials is vital in many areas of advanced mathematics and physics.

Substitution Method

The substitution method is an integral part of calculus which simplifies the integration process. By substituting a section of the integrand with a new variable, we can often convert a difficult integral into a simpler one. This method is similar to using a substitution in algebra to solve equations. When using the substitution method, it's essential to replace both the chosen part of the integrand and the differential accordingly. This technique is particularly useful in the case of the gamma function, where an appropriate substitution turns the integral into a standard form that is already known or easier to integrate.

Improper Integrals

Improper integrals occur when the integrand becomes infinite or when the limits of integration contain infinity. In the context of the gamma function, we deal with an integral that extends from 0 to infinity, making it an improper integral.

Handling Improper Integrals

Key to evaluating an improper integral is understanding how to take limits of the integral as one of its bounds goes to infinity. Calculating improper integrals often involves using comparison tests to determine convergence or divergence. In the exercise related to the gamma function, the evaluation of such an integral provides a key step in deriving properties of the gamma function, including its specific values for fractional inputs.