Question

Express each of the following integrals as a \(\Gamma\) function. By computer, evaluate numerically both the \(\Gamma\) function and the original integral. \(\int_{0}^{\infty} x^{5} e^{-x^{2}} d x\) Hint: Put \(x^{2}=u\)

Short answer

The integral \(\int_{0}^{\infty} x^{5} e^{-x^{2}} dx\) evaluates to 1, expressed as \(\frac{1}{2} \Gamma(3)\).

Step-by-step solution

Substitution

Let’s use the substitution provided. Let \(u = x^2\). Then, differentiate both sides to find the differential \(du\). Since \(u = x^2\), \(du = 2x dx\), which implies \(dx = \frac{du}{2x}\).

Change of Variables

Rewrite the original integral using the new variable \(u\). Notice that when \(x = 0\), \(u = 0\), and when \(x \to \infty\), \(u \to \infty\). Thus the integral becomes: \[\int_{0}^{\infty} x^{5} e^{-x^{2}} dx = \int_{0}^{\infty} x^{5} e^{-u} \frac{du}{2x}\].

Simplify the Integral

Simplify the integral by substituting \(u\) for \(x\) entirely. Rewriting \(x^5\), we get \(x = u^{1/2}\), so \[x^5 = (u^{1/2})^5 = u^{5/2}\]. This transforms the integral into: \[\int_{0}^{\infty} u^{5/2} e^{-u} \frac{du}{2u^{1/2}} = \int_{0}^{\infty} \frac{1}{2} u^{5/2 - 1/2} e^{-u} du = \int_{0}^{\infty} \frac{1}{2} u^{2} e^{-u} du\].

Recognize the Gamma Function

The integral now matches the form of a Gamma function: \(\Gamma(n) = \int_{0}^{\infty} u^{n-1} e^{-u} du\). In this case, \(\Gamma(3) = \int_{0}^{\infty} u^{2} e^{-u} du\). We have \(\int_{0}^{\infty} x^{5} e^{-x^{2}} dx = \frac{1}{2} \Gamma(3)\).

Evaluate the Gamma Function

Using the known properties of the Gamma function, \(\Gamma(n) = (n-1)!\). Therefore, \(\Gamma(3) = 2! = 2\). Thus, \(\frac{1}{2} \Gamma(3) = \frac{1}{2} \cdot 2 = 1\).

Final Result

Therefore, we have confirmed that \(\int_{0}^{\infty} x^{5} e^{-x^{2}} dx = 1\).

Key concepts

Integral Transformation

Integral transformations are crucial techniques for solving complex integrals. They involve changing the variable of integration to simplify the integral. In the given problem, we use the substitution method, transforming the integral into a more manageable form. We start with the substitution \(u = x^2\). This change helps to simplify the integrand, making the integral easier to evaluate. The differential change is derived as well, where \(dx = \frac{du}{2x}\). This process of substituting and differentiating helps in transforming the original integral into a new form that can be recognized as a Gamma function.

Change of Variables

The change of variables technique is essential in transforming the integral to facilitate simpler computation. When we apply the substitution \(u = x^2\), we must also convert all parts of the integral accordingly. The limits of integration change: as \(x\) goes from 0 to \(\infty\), \(u\) goes from 0 to \(\infty\). Additionally, the differential \(dx\) is replaced by \(\frac{du}{2x}\).
The given problem transforms the integral \(\int_{0}^{\infty} x^{5} e^{-x^{2}} dx\) into \(\int_{0}^{\infty} x^{5} e^{-u} \frac{du}{2x}\). Substituting \(x\) as \( u^{1/2} \), the integral simplifies further. This step-by-step substitution and variable change are fundamental techniques in calculus to handle complex integrals.

Gamma Function Properties

The Gamma function, denoted by \( \Gamma(n) \), extends the concept of factorials to real and complex numbers. It is defined as \( \Gamma(n) = \int_{0}^{\infty} u^{n-1} e^{-u} du \). This function is very useful in evaluating integrals that appear in various fields of mathematics and engineering.
In our problem, after transforming the integral, we get \( \int_{0}^{\infty} \frac{1}{2} u^{2} e^{-u} du \), which matches the form \( \Gamma(3) \). According to the properties of the Gamma function, \( \Gamma(n) = (n-1)! \). Therefore, \( \Gamma(3) = 2! = 2 \). Recognizing the integral as a Gamma function allows us to use these properties to easily evaluate the integral.

Integration Techniques

Several integration techniques come into play when solving complex integral problems. For this problem, the substitution method was used to change variables, simplifying the integral into a recognizable form.
Both the differentiation of the substituted variable and the use of the Gamma function's properties were pivotal in solving the problem. After the transformation and simplification, recognizing the integral as a Gamma function made the evaluation straightforward. Finally, using the property \( \Gamma(3) = 2! = 2 \), we found our result quickly.
These techniques collectively help handle otherwise difficult integrals efficiently and accurately.