Problem 63 - Thomas Calculus

Question

In Exercises \(35-64\) , use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$ \int_{-\infty}^{\infty} \frac{d x}{\sqrt{x^{4}+1}} $$

Step-by-Step Solution

Answer

Integral converges using the Direct Comparison Test.

Step 1: Determine the type of Integral

This is an improper integral because it includes the limits of integration from \(-\infty\) to \(\infty\). We need to check whether the integral converges or diverges.

Step 2: Choose a Method for Testing Convergence

For this integral, we can use the Direct Comparison Test by comparing with a simpler function that we understand better. The function \(f(x) = \frac{1}{\sqrt{x^4+1}}\) seems complicated, but we can estimate its behavior by approximating \(\frac{1}{x^2}\) for \(|x|\) large.

Step 3: Define Comparison Function for the Test

For very large values of \(|x|\), the dominant term in the denominator of \( x^4 + 1 \) is \( x^4 \). Thus, \( \sqrt{x^4 + 1} \approx x^2 \). Therefore, \( \frac{1}{\sqrt{x^4 + 1}} \approx \frac{1}{x^2} \). The function \( g(x) = \frac{1}{x^2} \), which is known to converge as an improper integral over \((-\infty, \infty)\).

Step 4: Apply the Direct Comparison Test

On any symmetric interval other than zero, we can use\(\frac{1}{x^2 + 1} \leq \frac{1}{\sqrt{x^4+1}} \leq \frac{1}{x^2}\) for large values of \(|x|\). Because \(\int_{-\infty}^{\infty}\frac{1}{x^2}dx\) converges, by comparison, the integral \(\int_{-\infty}^{\infty}\frac{1}{\sqrt{x^4+1}}dx\) must also converge.

Step 5: Conclude the Analysis

Since we confirmed that the given integral is dominated by the function that converges as per the Direct Comparison Test, we conclude that \(\int_{-\infty}^{\infty}\frac{1}{\sqrt{x^4+1}}dx\) converges.

Key Concepts

Improper Integrals

Improper integrals are a type of integral where either the interval of integration is infinite, or the function to be integrated becomes infinite within the interval. This makes evaluating them a bit more complex, as traditional methods might not apply directly. To handle improper integrals, we often convert them into a limit problem.
For instance, if you have an integral from \(-\infty\) to \(\infty\), like \(\int_{-\infty}^{\infty} \frac{dx}{\sqrt{x^4+1}}\), you separate it into two parts.

• Consider splitting it at a point, for example, \(c\), giving you \(\int_{-\infty}^{c} f(x) dx + \int_{c}^{\infty} f(x) dx\).
• Then, for each part, you evaluate the limit as your lower bound approaches \(-\infty\) or your upper bound approaches \(\infty\).

The objective is to evaluate the resulting limit. If the limit exists and is finite, the improper integral converges. Otherwise, it diverges. This approach is essential in mathematical analysis for determining the behavior of functions over infinite domains.

Direct Comparison Test

The Direct Comparison Test is a useful method for determining the convergence of improper integrals. Here's how it works:

• You compare the function in your integral to another function that is known to converge or diverge.
• In your example, with the function \(f(x) = \frac{1}{\sqrt{x^4+1}}\), you estimate its behavior for large values of \(|x|\) by comparing it to \(g(x) = \frac{1}{x^2}\).

This comparison relies on the idea that:
If \(0 \leq f(x) \leq g(x)\) for all \(x\) in the domain and \(\int g(x) dx\) converges, then \(\int f(x) dx\) also converges.
In the direct comparison, because \(\int_{-\infty}^{\infty} \frac{1}{x^2} dx\) is known to converge, \(\int_{-\infty}^{\infty} \frac{1}{\sqrt{x^4+1}} dx\) must also converge. This technique is particularly handy for complex functions, where directly evaluating the integral is difficult.

Limit Comparison Test

The Limit Comparison Test is another comparative method for checking the convergence of improper integrals. This test is particularly useful when the direct comparison might be difficult to establish.
To apply this test:

• You take two functions, \(f(x)\) and \(g(x)\), and study their ratio as \(x\) approaches infinity.
• If \(\lim_{x \to \infty} \frac{f(x)}{g(x)} = L\), where \(0 < L < \infty\), it means both integrals \(\int f(x) dx\) and \(\int g(x) dx\) will either both converge or both diverge.

Imagine trying to check the convergence for another function similar in complexity to \(\int_{-\infty}^{\infty} \frac{1}{\sqrt{x^4+1}} dx\), where such direct comparisons aren't easily apparent. The Limit Comparison Test steps in beautifully by allowing you to study the behavior in the 'infinite limit', especially for functions behaving similarly as \(x\) approaches infinity. This method can robustly affirm the conclusions about convergence or divergence.