For the case $s\>1$:

${\int }_1^{\infty }\frac{1}{x^s} \mathit{\ⅆ}x\=\underset{t\to \infty }{lim}{\int }_1^t{x}^{-s} \mathit{\ⅆ}x$ = $\underset{t\to \infty }{lim}\left({\left[\frac{x^{-s\+1}}{-s\+1}\right]}_1^t\right)$ = $\frac{1}{1-s}$ $\underset{t\to \infty }{lim}\left(t^{1-s}-1\right)$ = $\frac{-1}{1-s}\=\frac{1}{s-1}$ 

Since $s\>1$, the exponent $1-s$ is negative, so $t^{1-s}\to 0$ as $t\to \infty$.

In Example 4.5.2 where $s\=1$, the integral has been shown to diverge.

For the case $s<1$:

${\int }_1^{\infty }\frac{1}{x^s} \mathit{\&DifferentialD;}x\&equals;\underset{t\to \infty }{lim}{\int }_1^t{x}^{-s} \mathit{\&DifferentialD;}x$ = $\underset{t\to \infty }{lim}\left({\left[\frac{x^{-s\&plus;1}}{-s\&plus;1}\right]}_1^t\right)$ = $\frac{1}{1-s}$ $\underset{t\to \infty }{lim}\left(t^{1-s}-1\right)\&equals; \infty$ 

Since $s<1$, the exponent $1-s$ is positive, so $t^{1-s}\to \infty \mathrm{as} t\to \infty$.

Unfortunately, it is no longer possible to evaluate this integral in a syntax-free way via the Context Panel. One must use the "assuming" option explicitly.