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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
Question: Determine whether the following improper integral converges, and if so, find its value: $$\int_{e^{2}}^{\infty} \frac{d x}{x \ln ^{p} x}, p>1$$ Answer: The improper integral converges, and its value is \(\frac{-2^{1-p}}{1-p}\).

Step by step solution

01

Changing the variable and using the Comparison Test

We can change the variable by setting \(u = \ln x\) and \(du = \frac{dx}{x}\). Then, the integral becomes: $$\int_{\ln e^{2}}^{\infty} \frac{d u}{u^{p}}$$ We will consider the integral comparison test. We notice that our original integral is: $$\int_{2}^{\infty} \frac{d u}{u^p}$$ We will now compare our function with $$\int_{2}^{\infty} \frac{d u}{u^2}$$ As \(p>1\), we know that \(u^p > u^2\) for all \(u > 0\). Thus, \(\frac{1}{u^p} < \frac{1}{u^2}\) for all \(u > 0\). Since the integral $$\int_{2}^{\infty} \frac{d u}{u^2}$$ converges (it is a well-known convergent p-series for p=2), our given integral $$\int_{2}^{\infty} \frac{d u}{u^p}$$ also converges by the comparison test.
02

Calculate the antiderivative of the new integral

Now that we know the integral converges, we will find its antiderivative: $$\int \frac{d u}{u^{p}} = \frac{u^{1-p}}{1-p}$$
03

Evaluate the limit as the upper bound approaches infinity

Now that we have the antiderivative, we will evaluate the limit: $$\lim_{b\to\infty} \left(\frac{u^{1-p}}{1-p}\right|^b_2 = \frac{1}{1-p} \lim_{b \to \infty} (b^{1-p}-2^{1-p})$$ Since \(p>1\), the exponent \((1-p)\) is negative, which means \(b^{1-p}\) approaches zero as \(b\) approaches infinity: $$\lim_{b \to \infty} b^{1-p} = 0$$ So the limit is $$\frac{1}{1-p} (0 - 2^{1-p}) = \frac{-2^{1-p}}{1-p}$$
04

Convert back to x

Now, let's change back to the variable \(x\) by substituting \(u = \ln x\): $$\int_{e^{2}}^{\infty} \frac{d x}{x \ln ^{p} x} = \frac{-2^{1-p}}{1-p}$$ Thus, the integral converges to the value: \(\frac{-2^{1-p}}{1-p}\).

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Most popular questions from this chapter

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