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Chapter 7: Problem 15

Evaluate the following integrals or state that they diverge. $$\int_{e^{2}}^{\infty} \frac{d x}{x \ln ^{p} x}, p>1$$

Short Answer

Expert verified
In conclusion, the integral $\int_{e^2}^{\infty} \frac{dx}{x \ln^p x}$ converges for $p > 1$, and its value is $\frac{1}{1-p} 2^{1-p}$.

Step by step solution

01

Introduce substitution

In order to simplify the integral, let us introduce a new variable \(u = \ln x\). Then, we have \(du = \frac{dx}{x}\), and also, the limits of integration transform as follows: - When \(x = e^2\), we have \(u = \ln{e^2} = 2\). - When \(x \rightarrow \infty\), we have \(u \rightarrow \infty\). Now, the integral becomes $$\int_2^\infty \frac{du}{u^p}.$$
02

Evaluate the resulting integral

Next, let's evaluate this new integral $$\int_2^\infty \frac{du}{u^p} = \int_2^\infty u^{-p} du.$$ This is a basic power rule integral. The antiderivative of \(u^{-p}\) is \(\frac{u^{-p+1}}{-p+1}\), but we have to remember that \(p>1\), so \(-p+1 < 0\). Now, we need to evaluate the integral at the limits: $$\int_2^\infty \frac{du}{u^p} = \left[\frac{u^{-p+1}}{-p+1}\right]_2^\infty.$$
03

Check the convergence of the integral

Let's evaluate the limit of the antiderivative as \(u \rightarrow \infty\): $$\lim_{u \rightarrow \infty} \frac{u^{-p+1}}{-p+1} = \lim_{u \rightarrow \infty} \frac{1}{-p+1} u^{-p+1}.$$ Since \(-p+1 < 0\), this limit converges to 0 as \(u \rightarrow \infty\). Thus, the integral converges.
04

Return to the original variable and find the result if the integral converges

Since the integral converges, we can find the resulting value: $$\int_{e^2}^{\infty} \frac{dx}{x \ln^p x} = \int_2^\infty \frac{du}{u^p} = \left[ \frac{u^{-p+1}}{-p+1} \right]_2^\infty.$$ Now, we can evaluate the antiderivative at the limits: $$\int_{e^2}^{\infty} \frac{dx}{x \ln^p x} = \frac{(2)^{-p+1}}{-p+1} - \frac{(\infty)^{-p+1}}{-p+1} = \frac{1}{1-p} 2^{1-p}.$$ So the final answer is $$\int_{e^2}^{\infty} \frac{dx}{x \ln^p x} = \frac{1}{1-p} 2^{1-p}.$$

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