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Chapter 7: Problem 12

Evaluate the following integrals or state that they diverge. $$\int_{4 / \pi}^{\infty} \frac{1}{x^{2}} \sec ^{2}\left(\frac{1}{x}\right) d x$$

Short Answer

Expert verified
Question: Determine whether the improper integral converges or diverges: $$\int_{4 / \pi}^{\infty} \frac{1}{x^{2}} \sec ^{2}\left(\frac{1}{x}\right) dx$$ Answer: The given improper integral diverges.

Step by step solution

01

Rewriting the integral as a limit

We can rewrite the given improper integral as a limit by extending the upper limit to infinity: $$ \int_{4 / \pi}^{\infty} \frac{1}{x^{2}} \sec ^{2}\left(\frac{1}{x}\right) dx = \lim_{b\to\infty}\int_{4 / \pi}^{b} \frac{1}{x^{2}} \sec ^{2}\left(\frac{1}{x}\right) dx $$
02

Substitution

We can perform a substitution to simplify the integral. Let \(u = \frac{1}{x}\). Then, \(du = -\frac{1}{x^2}dx\). $$ \begin{aligned} \lim_{b\to\infty}\int_{4 / \pi}^{b} \frac{1}{x^{2}} \sec ^{2}\left(\frac{1}{x}\right) dx &=-\lim_{\substack{a\to \frac{\pi}{4}\\b\to\infty}}\int_{a}^{\infty} \sec ^{2}(u) du \end{aligned} $$ To find the new limits of integration, observe that $$ u(4 / \pi)=\frac{1}{(4 / \pi)}=\frac{\pi}{4} \quad \text{and} \quad u\left(\frac{1}{b}\right)=\frac{1}{(1/b)}=b. $$
03

Antiderivative of the integrand

Now, we need to find the antiderivative of \(-\sec^2(u)\): $$ \begin{aligned} \int -\sec^{2}(u) du &= -\int \sec^{2}(u) du \\ &= -\tan(u) + C \\ \end{aligned} $$ The antiderivative of \(\sec^2(u)\) is \(\tan(u)\), which we found by applying the basic rules of integration, and C is the constant of integration.
04

Applying the limits to find the value of the integral

Now, we will apply the limits of the integral and take the limit as \(b\to \infty\): $$ \begin{aligned} -\lim_{\substack{a\to \frac{\pi}{4}\\b\to\infty}}\int_{a}^{\infty} \sec^2(u) du &= -\lim_{\substack{a\to \frac{\pi}{4}\\b\to\infty}}[\tan(u)]^{\infty}_{\frac{\pi}{4}} \\ &= -\lim_{\substack{a\to \frac{\pi}{4}\\b\to\infty}} [\tan(\infty) - \tan(\frac{\pi}{4})] \\ \end{aligned} $$
05

Evaluating the limit

Evaluating the limit, we can see that it diverges because \(\tan(\infty)\) does not exist: $$ \begin{aligned} -\lim_{\substack{a\to \frac{\pi}{4}\\b\to\infty}} [\tan(\infty) - \tan(\frac{\pi}{4})] = \text{Diverge} \end{aligned} $$ Therefore, the given improper integral diverges.

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