Solid of revolution of $\frac{1}{x^a}$ Consider the solid of revolution $L_a$ that arises with $$f(x)= \frac{1}{x^a}$$ with $0< x\leq 1$ about the $x$-axis.
I need to find all $a > 0$ where the volume $L_a$ would be finite and infinite.
I tried the following:
$$\pi \int_0^1\frac{1}{x^{2a}} dx$$
$$= \pi (\frac{1^{-2a+1}}{-2a+1})$$
The only solution that is not possible in this case, is when $a = 1/2$. But in this case we wouldn't get infinite, we would just divide by $0$, which is not possible. So the volume would be finite for all $\mathbb{R}\setminus \{1/2\}$. Which steps did I do wrong in this process?
 A: There's a mistake on the step
$$
\int_0^1\frac{1}{x^{2a}} dx = \frac{1^{-2a+1}}{-2a+1}
$$
which is not always true. To see why we'll split the problem in cases.

*

*If $\color{blue}{a<\frac{1}{2}}$ we have that $2a<1 \implies 1-2a>0$, and since $1-2a$ is positive we can correctly assert that $\color{purple}{0^{1-2a} = 0}$. Notice that this is not always the case, as if we were to have a negative exponent we would have a $0$ in the denominator. Because of the previous discussion, we get that

\begin{align*}
\int_0^1\frac{1}{x^{2a}} dx &= \frac{x^{1-2a}}{1-2a}\Biggr|_{0}^{1}\\
&= \frac{1^{1-2a}}{1-2a}-\frac{\color{purple}{0^{1-2a}}}{1-2a} \\
& = \frac{1}{1-2a}
\end{align*}
which is correctly defined since $1-2a>0 \implies 1-2a \neq 0$. Hence, for $a < \frac{1}{2}$ the volume is finite.

*

*If $\color{blue}{a=\frac{1}{2}}$ we see that
\begin{align*}
\int_0^1\frac{1}{x^{2a}} dx &= \lim_{\varepsilon \to 0^+}\int_\varepsilon^1\frac{1}{x^{2a}} dx\\ 
&= \lim_{\varepsilon \to 0^+}\int_\varepsilon^1\frac{1}{x} dx\\
& = \ln(1) - \lim_{\varepsilon \to 0^+}\ln(\varepsilon)\\
& = - \lim_{\varepsilon \to 0^+}\ln(\varepsilon)
\end{align*}
but since the natural log diverges at $0$, the integral diverges, and hence, for $a=\frac{1}{2}$ the volume is infinte.


*If $\color{blue}{a>\frac{1}{2}}$ we have the following. Recall that $x^n  \le x$ for $x\in(0,1]$ and $n>1$ since
$$ x \le1 \overset{\color{purple}{n-1 >0}}{\implies} x^{n-1} -1  \le 0 \overset{\color{purple}{x  > 0}}{\implies}  x \left(x^{n-1} -1\right) \le 0 \implies x^n  \le x$$
Since in our integral we're interested in values of $x$ precisely on the interval from $0$ to $1$ we can say
$$
\underbrace{x^{2a} \le x}_{\color{blue}{2a>1}} \implies \frac{1}{x} \le \frac{1}{x^{2a}} \implies\underbrace{\int_0^1 \frac{1}{x} \ dx}_{\color{red}{\infty}}\le \int_0^1\frac{1}{x^{2a}} \ dx
$$
recalling that we previously showed that the case of $a = \frac{1}{2}$ is divergent. Since in this case every volume has to be greater than a divergent integral, all these volumes are also infinite.

In conclusion, the possible values of $a>0$ for which the volume would be finite would correspond to the values of $a$ in the interval
$$
a \in \left(0, \frac{1}{2}\right)
$$
and the volume would be infinite for every other positive value.
