Question

Why does it make sense that ${\int }_0^1\frac{1}{x^p}dx$diverges when $p\ge 1$? Consider how $\frac{1}{x^p}$compares with $\frac{1}{x}$in this case.

Short answer

For $p\ge 1$,$\frac{1}{x^p}$is greater than $\frac{1}{x}$in the interval $[0,1]$, whose improper integral on$[0,1]$ is known to diverge.

Step-by-step solution

Step 1. Given information.    

We are given,

${\int }_0^1\frac{1}{x^p}dx$

Step 2. Comparing the integral

When the integral ${\int }_0^1\frac{1}{x^p}dx$is solved in the interval $p\ge 1$, it gives an result of $\infty$. Consider $\frac{1}{x}$, since $p=1$.

Then,

$$\begin{gathered} {\int }_0^1\frac{1}{x}dx=[\ln x{]}_0^1 \\ =\ln 1-\ln 0 \\ =\infty \end{gathered}$$

So, the integral is diverges.