The p-integral test

Theorem. The improper integral ${\int }_1^{\infty }\frac{1}{x^p}dx$ converges if $p>1$ and diverges if $p\le 1$.

Proof. By definition of a Type I improper integral, ${\int }_1^{\infty }{x}^{-p}dx={\lim }_{t\to \infty }{\int }_1^t{x}^{-p}dx,$ so the integral converges precisely when this limit exists and is finite. The antiderivative of $x^{-p}$ takes a different form when $p=1$, so we split into cases.

Case 1: $p=1$. Here ${\int }_1^t\frac{1}{x}dx=[\ln x{]}_1^t=\ln t-\ln 1=\ln t.$ As $t\to \infty$, $\ln t\to \infty$, so the limit does not exist (as a finite number). The integral diverges.

Case 2: $p\ne 1$. Here ${\int }_1^t{x}^{-p}dx={\left[\frac{x^{1-p}}{1-p}\right]}_1^t=\frac{t^{1-p}-1}{1-p}.$ Everything now depends on the sign of the exponent $1-p$.

If $p>1$, then $1-p<0$, so we may write $t^{1-p}=\frac{1}{t^{p-1}}$ with $p-1>0$. As $t\to \infty$, the denominator grows without bound, so $t^{1-p}\to 0$. Therefore ${\lim }_{t\to \infty }\frac{t^{1-p}-1}{1-p}=\frac{0-1}{1-p}=\frac{-1}{1-p}=\frac{1}{p-1},$ a finite value. The integral converges, to $\frac{1}{p-1}$.

If $p<1$, then $1-p>0$, so $t^{1-p}\to \infty$ as $t\to \infty$. The expression grows without bound, so the limit does not exist. The integral diverges.

Combining all cases: the integral converges only when $p>1$, and diverges for every $p\le 1$. $■$

$${\int }_1^{\infty }\frac{1}{x^p}dx=\{\begin{array}{ll}\frac{1}{p-1}& \text{if }p>1\\ \text{divergent}& \text{if }p\le 1\end{array}$$