Question

The integral $$ \int_{-\infty}^{\infty} e^{-x x^{2}} d x $$ has the value \((\pi / \alpha)^{1 / 2}\). Use this result to evaluate $$ J(n)=\int_{-\infty}^{\infty} x^{2 n} e^{-x^{2}} d x $$ where \(n\) is a positive integer. Express your answer in terms of factorials.

Short answer

\( J(n) = \sqrt{\pi} \frac{(2n)!}{2^n n!} \)

Step-by-step solution

Recognize the given standard integral

The integral \[ \int_{-\infty}^{\infty} e^{-x^2} \, dx \]is given to be \[ \sqrt{\pi} \]. This will be useful for solving the given problem.

Define the integral J(n)

The integral we need to solve is \[ J(n) = \int_{-\infty}^{\infty} x^{2n} e^{-x^2} \, dx. \]

Change of variables

Use the substitution \[ x = \sqrt{u} \]and rewrite the integral in terms of u.

Simplify the integrand

Using the substitution, the integral becomes \[ J(n) = \int_{0}^{\infty} (\sqrt{u})^{2n} e^{-u} \, \frac{1}{2\sqrt{u}} \, du \],\( which simplifies to \) \[ \frac{1}{2} \int_{0}^{\infty} u^{n-1/2} e^{-u} \, du. \]

Identify the resulting integrand

Recognize that the resulting integral is a Gamma function \( \Gamma(t) = \int_{0}^{\infty} u^{t-1} e^{-u} du \). In this case, \( t = n + \frac{1}{2}. \)

Evaluate the Gamma function

The Gamma function can be evaluated to \[ (n - 1/2)! = \Gamma(n + 1/2) \]. Using the identity \( \Gamma(z) \Gamma(z + 1/2) = \sqrt{\pi} \quad \text{for}\quad z = n+1 \), we get the final result:

Final expression for J(n)

Combine all the results: \( J(n) = \frac{\sqrt{\pi}}{2^n} \frac{(2n)!}{n!} \).

Key concepts

Integral Evaluation

To understand integral evaluation, we begin with identifying the integral we need to solve. In this case, it is \ J(n) = \ \(\int_{-\infty}^{\infty} x^{2n} e^{-x^{2}} \ dx\).
Recognizing the form of the integral and its similarities with known integrals helps simplify the problem. For example, the integral \ \( \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi} \) is a standard Gaussian integral. This forms the basis for evaluating more complex integrals by connecting them to simpler known results.
Pursuing step-by-step reduction and transformation of the integral from its original form to a more recognizable one makes the evaluation process manageable. Using substitution and understanding patterns play a crucial role in integral evaluation.

Substitution Method

To simplify complex integrals, one useful technique is the substitution method. This involves changing the variable of integration to transform the integrand into a simpler form.
For the integral \ \(J(n) = \int_{-\infty}^{\infty} x^{2n} e^{-x^{2}} \ dx\), we use the substitution \ \(u = x^2\). This changes the original integral to: \ \( \int_{0}^{\infty} u^{n-1/2} e^{-u} \ du \).
After substitution, always remember to adjust the differential and the limits of integration accordingly. This can result in an integrand that is equivalent to a known form, such as the Gamma function. The substitution method simplifies the integrand, making it more tractable for further evaluation.

Factorials

Factorials are essential in dealing with integrals involving the Gamma function, particularly when the Gamma function involves half-integer values. The Gamma function \ \(\Gamma(t) = \int _{0}^{\infty} u^{t-1} e^{-u} du\) generalizes the factorial function for non-integer arguments. For integer \(n\), \ \(\Gamma(n + 1) = n!\). Similarly, we use identities involving the Gamma function to simplify expressions.
Using the identity \ \(\Gamma(z)\Gamma(z + \frac{1}{2}) = \sqrt{\pi}\) for any non-negative integer \(z\), enables the evaluation of integrals into terms of factorials.
Thus, we can express our final evaluated integral as: \ \ \(J(n) = \frac{\sqrt{\pi}}{2^n} \frac{(2n)!}{n!}\).
Understanding the relationship between the Gamma function and factorials helps in transitioning from the integral's form to a more familiar combinatorial form involving factorials.