Question

Why does it make sense that ${\int }_1^{\infty }\frac{1}{x^p}dx$ diverges when $0<p\le 1$ ? Consider how $\frac{1}{x^p}$ compares with $\frac{1}{x}$ in this case.

Short answer

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

Step-by-step solution

Step 1. Given information.   

We are given,

${\int }_1^{\infty }\frac{1}{x^p}dx$

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

Step 2. Comparing the integral

When the integral ${\int }_1^{\infty }\frac{1}{x^p}dx$is solved in the interval $0<p\le 1$, 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,

$$\begin{gathered} {\int }_1^{\infty }\frac{1}{x}dx=[\ln x{]}_1^x \\ =\ln \infty -\ln 1 \\ =\infty \end{gathered}$$