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Chapter 6: Problem 28

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{0}^{\infty} \sin \frac{x}{2} d x $$

Short Answer

Expert verified
The improper integral diverges.

Step by step solution

01

Identify the improper integral

The given integral \(\int_{0}^{\infty} \sin \frac{x}{2} dx\) is an improper integral as it has an infinity in the interval of integration.
02

Express the integral in terms of limits

Since it is an improper integral, express it in terms of limits: \(\int_{0}^{\infty} \sin \frac{x}{2} dx = \lim_{{t \to \infty}} \int_{0}^{t} \sin \frac{x}{2} dx\)
03

Apply u-substitution

For the integral, u-substitution is easiest. Let \(u = \frac{x}{2}\), then \(du = \frac{1}{2} dx\) and \(dx = 2 du\). The limits of the integral, originally 0 to \(t\), change to 0 to \(t/2\). The integral becomes \(\lim_{{t \to \infty}} \int_{0}^{t/2} 2 \sin u du\)
04

Integrate

Now, it is a straightforward integral to compute: \(\lim_{{t \to \infty}} [-2 \cos(u)]_{0}^{t/2}\) = \(\lim_{{t \to \infty}} [2 \cos(0) - 2 \cos(t/2)]\) = \(\lim_{{t \to \infty}} [2 - 2 \cos(t/2)]\)
05

Check for convergence/divergence

Since \(\cos(t/2)\) oscillates between -1 and 1 as \(t\) approaches \(\infty\), it means the \(\lim_{{t \to \infty}} [2 - 2 \cos(t/2)]\) does not exist. Therefore, the limit does not exist, meaning the improper integral diverges.

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Most popular questions from this chapter

True or False? In Exercises 67-70, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(p(x)\) is a polynomial, then \(\lim _{x \rightarrow \infty}\left[p(x) / e^{x}\right]=0\).

Determine all values of \(p\) for which the improper integral converges. $$ \int_{1}^{\infty} \frac{1}{x^{p}} d x $$

Find the area of the region bounded by the graphs of the equations.$$ y=\sin x, \quad y=\sin ^{3} x, \quad x=0, \quad x=\pi / 2 $$

Evaluate the definite integral. $$ \int_{0}^{\pi / 2} \frac{\cos t}{1+\sin t} d t $$

Use integration by parts to verify the reduction formula. $$ \int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x $$
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