Determining the divergence of an integral. $1/x^p$

Question

I am trying to determine for which p > 0 the improper integral $\int_{0}^{1}dx/x^p$ converges. The first step I have taken for these improper integrals was to set them in a form of a limit, but I am having a hard time doing this... Can anybody help?

Answer 1

The problem lies at $x=0$, so by definition we have that $$\int_0^1\frac{1}{x^p}\,dx = \lim_{a\to 0^+}\int_a^1\frac{1}{x^p}\,dx.$$

Now, if $p\neq 1$, then $$\int\frac{1}{x^p}\,dx = \int x^{-p}\,dx = \frac{1}{1-p}x^{1-p} + C,$$ so $$\begin{align*} \int_0^1\frac{1}{x^p}\,dx &= \lim_{a\to 0^+}\int_a^1\frac{1}{x^p}\,dx\\ &= \lim_{a\to 0^+}\left( \frac{1}{1-p}x^{1-p}\Bigm|_{a}^1\right)\\ &= \lim_{a\to 0^+}\left( \frac{1}{1-p} - \frac{a^{1-p}}{1-p}\right). \end{align*}$$ Now you have to be careful: if $1-p\gt 0$, then something happens; if $1-p\lt 0$, then something else happens.

Of course, you should deal with the case of $p=1$ separately. I'll leave that to you.