Example 6. Investigate for which values of $\alpha>0$ the improper integral $\int_{a}^{b} \frac{d x}{(x-a)^{\alpha}}(b>a)$ converges.
Solution. If $\alpha \neq 1$, then the integral

$$
\int_{a+\varepsilon}^{b} \frac{d x}{(x-a)^{\alpha}}=\frac{1}{1-\alpha}\left[(b-a)^{1-\alpha}-\varepsilon^{1-\alpha}\right]
$$

as $\varepsilon \rightarrow 0$ has a limit of $\infty$ or a finite number $\frac{1}{1-\alpha}(b-a)^{1-\alpha}$ depending on whether $\alpha>1$ or $\alpha<1$.

If $\alpha=1$, then

$$
\int_{a+\varepsilon}^{b} \frac{d x}{x-a}=[\ln (b-a)-\ln \varepsilon] \rightarrow \infty(\text { as } \varepsilon \rightarrow 0)
$$

Therefore, the given integral converges when $\alpha<1$.