Improper Integral $\int_0^1\frac{dx}{x^p}$

Question

Is this integral convergent only for $p<1$?

$$\int_0^1\frac{dx}{x^p}$$

Answer 1

Yes. \begin{align} \int \dfrac{dx}{x^p} = \begin{cases} \dfrac{x^{1-p}}{1-p} + \text{constant} & \text{ for } p \neq 1\\ \log(x) & \text{ for }p=1\end{cases} \end{align} From the above, note that for $p\geq1$ as $x \to 0^+$, we have $x^{1-p} \to \infty$ and for $p<1$ as $x \to 0^+$, we have $x^{1-p} \to 0$. Hence, \begin{align} \int_0^1 \dfrac{dx}{x^p} = \begin{cases} \dfrac{1}{1-p} & \text{ for } p < 1\\ \text{does not exist} & \text{ for }p\geq1\end{cases} \end{align}

Answer 2

If $p\le 0$, there is no problem of any kind: the function is continuous in the interval $[0,1]$, so the integral exists.

But if $p\gt 0$, there is a potential problem, since the function blows up as $x$ approaches $0$ from the right. So to find out whether the improper integral converges or not, for any $\epsilon \gt 0$, we let $$I_\epsilon=\int_\epsilon^1 \frac{1}{x^p}\,dx,$$ and examine the behaviour of $I_\epsilon$ as $\epsilon$ approaches $0$ from the right.

If $\lim_{\epsilon\to 0+} I_\epsilon$ exists, then the improper integral converges, otherwise it diverges. (Note that $\lim_{\epsilon\to 0^+}I_\epsilon$ means the limit as $\epsilon$ approaches $0$ from the right, that is, through positive values.) If $p\ne 1$, then $$I_\epsilon=\left. \frac{1}{1-p}x^{-p+1}\right|_\epsilon^1=\frac{1}{1-p}(1-\epsilon^{1-p}),$$ while if $p=1$ then $$I_\epsilon =-\ln(\epsilon).$$ First deal with the case $p=1$. As $\epsilon$ approaches $0$ from the right, $\ln(\epsilon)\to-\infty$, so $\lim_{\epsilon\to 0^+}I_\epsilon$ does not exist. If $p\gt 1$, then $1-p$ is negative, and again $I_\epsilon$ blows up as $\epsilon$ approaches $0$ from the right.

Finally, if $p\lt 1$, then $1-p$ is positive, so $\epsilon^{1-p}$ approaches $0$ as $\epsilon$ approaches $0$ from the right, so the improper integral converges.