Question

Express the following integrals as \(\Gamma\) functions and evaluate them using a table of \(\Gamma\) functions. $$ \int_{0}^{\alpha} x^{2} e^{-x^{2}} d x \quad \text { Hint: Put } x^{2}=t $$

Short answer

The integral evaluates to \( \frac{1}{2} \).

Step-by-step solution

- Substitution

Set the substitution as given in the hint: let \( x^2 = t \). Then, differentiate both sides to find \( dx \). We get \( 2x dx = dt \) or \( dx = \frac{dt}{2x} \). Since \( x = \sqrt{t} \), this becomes \( dx = \frac{dt}{2 \sqrt{t}} \).

- Change of Variables

Substitute back into the integral: \[ \int_{0}^{\beta} x^{2} e^{-x^{2}} d x = \int_{0}^{\alpha^2} \sqrt{t} \cdot e^{-t} \cdot \frac{dt}{2 \sqrt{t}} \]This simplifies to: \[ \int_{0}^{\alpha^2} \frac{t^{1/2} \cdot e^{-t}}{2 \sqrt{t}} dt = \frac{1}{2} \int_{0}^{\alpha^2} e^{-t} dt \]

- Express as a Gamma Function

Rewrite the integral to reflect the form of the Gamma function. Recall that the Gamma function is defined as: \[ \Gamma(n) = \int_{0}^{\infty} e^{-t} t^{n-1} dt \]In this case, \( n-1 = 0 \) or \( n = 1 \), so:\[ \frac{1}{2} \Gamma(1)= \frac{1}{2} \int_{0}^{\infty} e^{-t} dt \]

- Evaluate the Gamma Function

The Gamma function \( \Gamma(1) \) is equal to 1. Thus,\[ \frac{1}{2} \Gamma(1) = \frac{1}{2} \cdot 1 = \frac{1}{2} \]

Key concepts

Substitution in Integrals

Substitution is a powerful technique used to simplify integrals. By changing variables, we can transform a complex integral into a more manageable one. In this exercise, we start with the integral: \[ \int_{0}^{\beta} x^{2} e^{-x^{2}} dx \] \To simplify this, we use the substitution \( x^2 = t \). This helps to change the variable of integration and simplifies the integral. Differentiating both sides, we get \( 2x dx = dt \), which we can rearrange to find \( dx = \frac{dt}{2x} \). Since \( x = \sqrt{t} \), \( dx \) becomes \( \frac{dt}{2 \sqrt{t}} \). \Substitution helps to reduce the integral into a form where standard techniques or known results (like the Gamma function) can be applied, making it easier to solve.

Gamma Function

The Gamma function, denoted \( \Gamma(n) \), is a generalization of the factorial function to real and complex numbers. It is defined as: \[\Gamma(n) = \int_{0}^{\infty} t^{n-1} e^{-t} dt \]For positive integers, \( \Gamma(n) = (n-1)! \). The Gamma function is essential in various fields, including probability and statistics, where it appears in distributions like the Gamma and Chi-squared distributions. \When solving integrals, recognizing parts of the integrand that match the Gamma function's form can simplify the evaluation. In this exercise, we rewrite the integral to match the Gamma function's definition, allowing us to leverage its known values.

Integral Evaluation

Evaluating integrals often involves rewriting them in a recognizable form. In this case, after substitution and simplification, we have: \[ \int_{0}^{\alpha^2} \frac{e^{-t}}{2} dt \] By representing this in terms of the Gamma function, we get: \[ \frac{1}{2} \int_{0}^{\infty} e^{-t} dt \] Using the fact that \( \int_{0}^{\infty} e^{-t} dt = \Gamma(1) \), and knowing \( \Gamma(1) = 1 \), we find: \[ \frac{1}{2} \Gamma(1) = \frac{1}{2} \] This process shows how substitution, understanding the Gamma function, and recognizing integrals' patterns are crucial skills for solving complex integrals efficiently and accurately.