Question

Let \(I_{n}=\int x^{n} e^{-x^{2}} d x,\) where \(n\) is a nonnegative integer. a. \(I_{0}=\int e^{-x^{2}} d x\) cannot be expressed in terms of elementary functions. Evaluate \(I_{1}\). b. Use integration by parts to evaluate \(I_{3}\). c. Use integration by parts and the result of part (b) to evaluate \(I_{5}\). d. Show that, in general, if \(n\) is odd, then \(I_{n}=-\frac{1}{2} e^{-x^{2}} p_{n-1}(x)\) where \(p_{n-1}\) is a polynomial of degree \(n-1\). e. Argue that if \(n\) is even, then \(I_{n}\) cannot be expressed in terms of elementary functions.

Short answer

Question: Analyze and evaluate the integral \(I_n = \int x^n e^{-x^2} dx\) for different values of \(n\), and discuss the relationship and observations for odd and even values of \(n\). Answer: For odd values of \(n\), we have observed that \(I_n = -\frac{1}{2} e^{-x^2} p_{n-1}(x)\), where \(p_{n-1}\) is a polynomial of degree \(n-1\). However, for even values of \(n\), no pattern was observed, and the integration does not simplify into an expression involving elementary functions.

Step-by-step solution

a. Evaluate \(I_1\)

The first step is to evaluate \(I_1\) which is given by: \(I_1 = \int x e^{-x^2} dx\) To evaluate this, we use the substitution technique. We let \(u = -x^2\), so \(\frac{du}{dx} = -2x\). Now, we can rewrite the integral: \(I_1 = \dfrac{1}{2}\int -du e^{u}\) \(I_1 = -\dfrac{1}{2}\int e^{u}du\) Now we can integrate: \(I_1 = -\dfrac{1}{2} e^{u} + C\) \(I_1 = -\dfrac{1}{2} e^{-x^2} + C\)

b. Evaluate \(I_3\) using integration by parts

We will now evaluate \(I_3 = \int x^3 e^{-x^2} dx\) using integration by parts. For integration by parts, we will consider: \(u=x^2\) and \(dv=x^2e^{-x^2}dx\) Now, differentiate u with respect to x and integrate dv with respect to x: \(du=2xdx\) and \(v=I_1=-\dfrac{1}{2} e^{-x^2}\) Applying the integration by parts formula: \(I_3 = uv-\int v du\) \(I_3= (-\dfrac{1}{2} e^{-x^2}x^2) - \int (-\dfrac{1}{2} e^{-x^2})(2x)dx\) \(I_3=-\dfrac{1}{2} e^{-x^2} x^2 + \int x e^{-x^2} dx\) Using the result from part a, we have: \(I_3=-\dfrac{1}{2} e^{-x^2} x^2 - I_1\)

c. Evaluate \(I_5\) using integration by parts and the result from part b

To evaluate \(I_5 = \int x^5 e^{-x^2} dx\), we will again use integration by parts. Let's take \(u=x^4\) and \(dv=x e^{-x^2} dx\) By differentiating u with respect to x and integrating dv with respect to x, we have: \(du=4x^3dx\) and \(v=-\dfrac{1}{2} e^{-x^2}\) Now using the integration by parts formula: \(I_5 = uv-\int v du\) \(I_5= (-\dfrac{1}{2} e^{-x^2}x^4) - \int (-\dfrac{1}{2} e^{-x^2})(4x^3)dx\) \(I_5= -\dfrac{1}{2} e^{-x^2}x^4 + 2\int x^3 e^{-x^2} dx\) Now, let's substitute the expression for \(I_3\) derived in part b: \(I_5 = -\dfrac{1}{2} e^{-x^2}x^4 + 2\left(-\dfrac{1}{2} e^{-x^2} x^2 - I_1\right)\) \(I_5 = -\dfrac{1}{2} e^{-x^2}x^4 - e^{-x^2} x^2 + 2I_1\)

d. Show the relationship between \(I_n\) and \(p_{n-1}(x)\) for odd values of \(n\)

From the solutions we derived for \(I_1\), \(I_3\), and \(I_5\), we can observe the following relationship for odd integers n: \(I_1 = -\frac{1}{2} e^{-x^2} \) \(I_3 = -\frac{1}{2} e^{-x^2} x^2 - I_1\) \(I_5 = -\frac{1}{2} e^{-x^2} x^4 - e^{-x^2} x^2 + 2I_1\) Notice that for each odd integer \(n\), \(I_n = -\frac{1}{2} e^{-x^2} p_{n-1}(x)\), where \(p_{n-1}\) is a polynomial of degree \(n-1\). This pattern holds for odd values of \(n\), as shown by the examples above.

e. Argue that \(I_n\) cannot be expressed in terms of elementary functions for even values of \(n\)

Let's consider an even value of \(n\). We have the integral \(I_n = \int x^n e^{-x^2} dx\). If we use integration by parts or substitution, we can notice that, in general, at each step of the integration process for even values of \(n\), we still end up with the same function multiplied by some power of x, i.e., a function of the form \(x^k e^{-x^2}\) for some k. In the case of odd values of \(n\), the integration process led to a pattern where \(I_n\) could be expressed as \(-\frac{1}{2} e^{-x^2} p_{n-1}(x)\). However, for even values of \(n\), no such pattern is observed, and the integration does not simplify to an expression in terms of elementary functions. This indicates that it is highly likely that \(I_n\) for even values of \(n\) cannot be expressed in terms of elementary functions, i.e., in terms of common functions such as polynomials, exponentials, logarithms, and trigonometric functions.

Key concepts

Polynomial Integration

Polynomial integration is a process of finding the integral, or antiderivative, of a polynomial expression. In simple terms, this involves reversing the differentiation process. Let's consider the integral of a polynomial function like \(x^n\). Each term of the polynomial can be integrated separately. The basic rule is to increase the power of the term by one and then divide by this new power, which can be expressed as:

• \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), assuming \(n eq -1\),

where \(C\) is the constant of integration.

This makes polynomial integration relatively straightforward. However, when polynomial terms are multiplied by functions like exponentials, as in the exercise, additional integration techniques, such as substitution or integration by parts, may be necessary.

Substitution Technique

The substitution technique is a method used to evaluate integrals by simplifying the integrand into a more familiar form. This method is akin to the chain rule in reverse and is helpful when dealing with composite functions. Here's how it works:

• Identify the part of the integrand that can be substituted. Often, this is the innermost function.
• Let \(u\) be this part, and define \(du\), the differential of \(u\).

In the given problem, where \(I_1 = \int x e^{-x^2} dx\), substituting \(u = -x^2\) leads to \(du = -2x \, dx\). The integral translates to a more manageable form:

• \(I_1 = -\frac{1}{2}\int e^u \, du\).

This substitution simplifies the integration process, allowing us to apply basic integral rules more effectively.

Elementary Functions

Elementary functions are the basic building blocks of mathematical analysis, including functions like polynomials, exponentials, logarithms, and trigonometric functions. They are typically the first types of functions introduced in calculus classes due to their tractable nature in differentiation and integration.

Despite their simplicity, not all integrals involving elementary functions can be simplified back into elementary functions. This is evident in the original exercise, where \(\int e^{-x^2} \, dx\) (denoted as \(I_0\)) cannot be expressed in terms of elementary functions. The function \(e^{-x^2}\) is closely related to the Gaussian or bell curve function, which does not have an elementary antiderivative, and often requires more advanced methods, such as series expansions or special functions, for its evaluation.

Odd and Even Functions

In mathematics, functions are classified as odd or even, influencing symmetry and integrability properties. A function \(f(x)\) is:

• Even if \(f(-x) = f(x)\), symmetric about the y-axis.
• Odd if \(f(-x) = -f(x)\), symmetric about the origin.

This classification plays a role in the evaluation of integrals, particularly definite integrals over symmetric limits, where certain terms may cancel out.

In the context of the exercise, there’s a distinction between integrating odd and even powers of \(x\) multiplied by \(e^{-x^2}\). For odd \(n\), the integral \(I_n\) results in an expression with \(-\frac{1}{2} e^{-x^2} p_{n-1}(x)\), where \(p_{n-1}(x)\) is a polynomial. However, for even \(n\), such simplification is not possible, highlighting that when \(n\) is even, \(I_n\) cannot be expressed in terms of elementary functions, indicating deeper algebraic complexities.