Question

Use symmetry to evaluate the following integrals. a. \(\int_{-\infty}^{\infty} e^{|x|} d x \quad\) b. \(\int_{-\infty}^{\infty} \frac{x^{3}}{1+x^{8}} d x\)

Short answer

Answer: a) The integral \(\int_{-\infty}^{\infty} e^{|x|} dx\) diverges, as it doesn't converge to a finite value. b) The integral \(\int_{-\infty}^{\infty}\frac{x^3}{1+x^8} dx = 0\), since the function is odd, and the integral of an odd function over a symmetric interval is zero.

Step-by-step solution

Evaluate the integral of the even function

The function we need to integrate in part a is \(e^{|x|}\) which is even. For an even function, we can use the property \(\int_{-\infty}^{\infty} f(x) dx = 2\int_{0}^{\infty} f(x) dx\). So, we have: \(\int_{-\infty}^{\infty} e^{|x|} dx = 2\int_{0}^{\infty} e^{|x|} dx\)

Evaluate the integral between 0 and \(\infty\)

Now, we have to evaluate the integral over the interval [0, \(\infty\)]. For \(x \ge 0\), we can rewrite the function as \(e^x\): \(2\int_{0}^{\infty} e^{|x|} dx = 2\int_{0}^{\infty} e^x dx\) To solve this integral, we can use the fundamental theorem of calculus with \(u = x\), then \(du = dx\): \(2\int_{0}^{\infty} e^{u} du = 2[e^u]_{0}^{\infty}\)

Apply the limits of integration

Now, let's apply the limits of integration by subtracting the values when substituting \(u=\infty\) and \(u=0\): \(2[e^u]_{0}^{\infty} = 2(e^{\infty} - e^0) = 2(\infty - 1)\) Since the result is not convergent, the integral \(\int_{-\infty}^{\infty} e^{|x|} dx\) diverges.

Evaluate the integral of the odd function

The function we need to integrate in part b is \(\frac{x^3}{1+x^8}\) which is odd. For an odd function, we can use the property \(\int_{-\infty}^{\infty} f(x) dx = 0\). So, we have: \(\int_{-\infty}^{\infty}\frac{x^3}{1+x^8} dx = 0\) This concludes the evaluation of both integrals using symmetry properties.

Key concepts

Even Functions

Even functions have a special kind of symmetry. They are symmetrical about the y-axis. This means if you fold the graph along the y-axis, the two halves will match perfectly.
In mathematical terms, a function is even if it satisfies the condition \(f(-x) = f(x)\) for all values in its domain. This symmetry property can simplify evaluating definite integrals enormously because it reduces the domain of integration.
When dealing with an integral over the entire real line from \(-\infty\) to \(\infty\), we can exploit the even function property by using the formula:

• \(\int_{-\infty}^{\infty} f(x) \, dx = 2\int_{0}^{\infty} f(x) \, dx\)

This means we only need to integrate from \(0\) to \(\infty\) and then multiply by 2, which makes our calculations easier and faster. Remember, this shortcut is only valid if the function is even. In the original exercise, the function \(e^{|x|}\) is even, allowing us to use this property directly.

Odd Functions

Odd functions have their own unique symmetry. They are symmetric about the origin, which means if you rotate the graph 180 degrees around the origin, it would look the same.
Mathematically, a function is odd if \(f(-x) = -f(x)\) for all values in its domain. A key characteristic of odd functions is that when you integrate them over a symmetric interval around zero, the positive and negative areas cancel each other out.
This leads to a useful property for definite integrals:

• \(\int_{-\infty}^{\infty} f(x) \, dx = 0\)

Thus, if you need to evaluate an integral of an odd function over \(-\infty\) to \(\infty\), it is simply zero, provided the integral is defined. This property was used in the exercise for the function \(\frac{x^3}{1+x^8}\), which is an odd function.

Definite Integrals

Definite integrals form a significant part of calculus. They are used to calculate the area under a curve over a specific interval on the x-axis. For instance, when we write an integral such as \(\int_{a}^{b} f(x) \, dx\), we are finding the total area between the x-axis and the curve \(f(x)\) from \(x = a\) to \(x = b\).
The concept becomes even more fascinating when combined with the properties of even and odd functions, as seen in symmetry in integration examples.
With definite integrals, even and odd functions help reduce the complexity of calculations:

• For even functions, you can integrate from 0 to infinity and double the result.
• For odd functions, over symmetric limits, the integral is zero, assuming the integral exists.

These properties are foundational in calculus and can greatly simplify the work involved in evaluating integrals, making problems more approachable and solutions more elegant.