Chapter 5: Q. 8 (page 477)

Why does it make sense that $\int_{1}^{\infty} \frac{1}{x^{p}} d x$ diverges when $0 < p \leq 1$ ? Consider how $\frac{1}{x^{p}}$ compares with $\frac{1}{x}$ in this case.

Short Answer

Expert verified

The integral gives the indefinite result for $p = 1$ .

Step by step solution

Step 1. Given information.

We are given,

$\int_{1}^{\infty} \frac{1}{x^{p}} d x$

The integral diverges when $0 < p \leq 1$ .

Step 2. Comparing the integral

When the integral $\int_{1}^{\infty} \frac{1}{x^{p}} d x$ is solved in the interval $0 < p \leq 1$ , it gives an result of $\infty$ .

As when $p = 1$ is considered in the interval.

Then the resulting integral is $\ln x$ where the interval $[\infty, 1]$ is applied which gives the indefinite result.

That's why the result is diverges.

In this case, $\frac{1}{x^{p}}$ is equal to $\frac{1}{x}$ when $p = 1$ .

So,

$\int_{1}^{\infty} \frac{1}{x} d x = [\ln x]_{1}^{x} = \ln \infty - \ln 1 = \infty$

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Why does it make sense that $\int_{1}^{\infty} \frac{1}{x^{p}} d x$ diverges when $0 < p \leq 1$ ? Consider how $\frac{1}{x^{p}}$ compares with $\frac{1}{x}$ in this case.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Comparing the integral

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Why does it make sense that ∫1∞1xpdx diverges when 0<p≤1 ? Consider how 1xp compares with 1x in this case.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Comparing the integral

One App. One Place for Learning.

Most popular questions from this chapter

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Why does it make sense that $\int_{1}^{\infty} \frac{1}{x^{p}} d x$ diverges when $0 < p \leq 1$ ? Consider how $\frac{1}{x^{p}}$ compares with $\frac{1}{x}$ in this case.