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Chapter 6: Problem 21

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{4}^{\infty} \frac{1}{x(\ln x)^{3}} d x $$

Short Answer

Expert verified
The improper integral \( \int_{4}^{\infty} \frac{1}{x(\ln x)^{3}} d x \) converges and its value is \( \frac{1}{8} \).

Step by step solution

01

Identify The Improper Integral

An integral is improper if it has one or more infinite limits or discontinuous integrands. Therefore, the given integral \( \int_{4}^{\infty} \frac{1}{x(\ln x)^{3}} d x \) is an improper integral because upper limit is infinity.
02

Transform The Improper Integral Into Limit Form

To evaluate this integral, we must transform it into a limit form. Then, the improper integral is recast as a limit: \( \lim_{b \to \infty} \int_{4}^{b} \frac{1}{x(\ln x)^{3}} dx \). This is to ensure that the integral becomes a 'normal' one where we can apply traditional calculus rules, which cannot handle an infinite limit.
03

Evaluate The Integral

Let \(u=\ln x\), then \(du = \frac{1}{x}dx\). Thus the limit becomes \( \lim_{b \to \infty} \int_{\ln 4}^{\ln b} \frac{1}{u^{3}} du = \lim_{b \to \infty} [-\frac{1}{2u^2}]_{\ln 4}^{\ln b} \). Here, first, apply Fundamental of calculus to find the integral of the function.
04

Apply The Limit

Iterate the limit as \(b\) goes to infinity for the evaluated limit, we get \( \lim_{b \to \infty} [-\frac{1}{2(\ln b)^{2}} + \frac{1}{8}] = -\frac{1}{2(\ln \infty)^{2}} + \frac{1}{8} = \frac{1}{8} \). The limit is finite. Here, we applied limit properties for finding limits at infinity.
05

Final Conclusion

If the limit exists and is finite, the improper integral converges and is equal to the computed limit. If the limit does not exist or is infinite, the improper integral diverges. In our case, the limit is finite.

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