Determine if the improper integral converges or diverges. If it converges, find the limit.
$$\int_0^3 \frac{dx}{(x-1)^{2/3}}$$ $$=\int_0^1 \frac{1}{(x-1)^{2/3}}dx + \int_1^3 \frac{1}{(x-1)^{2/3}}dx$$ $$=\lim_{x \to 1^-} \int_0^1 \frac{1}{(x-1)^{2/3}}dx + \lim_{x\to1^+}\int_1^3 \frac{1}{(x-1)^{2/3}}dx$$
I'm confused here. Does it diverge because $\lim_{x \to 1^-} \int_0^1 \frac{1}{(x-1)^{2/3}}dx$ has an imaginary part to the left of 1?
There is a potential problem at $x=1$. So break up the integral into two parts, from $0$ to $1$ and from $1$ to $3$. If both integrals converge, your original integral converges. If either integral diverges, your integral diverges.
So let us consider $$\int_0^1 \frac{dx}{(x-1)^{2/3}}.$$ There is a matter of interpretation involved here. To some people, $u^{2/3}$ does not make sense for negative $u$. To others, it makes sense, because the denominator $3$ in the exponent is odd. To answer the question, we take the second view.
What we need to do is to let $$I(b)=\int_0^b \frac{dx}{(x-1)^{2/3}},$$ and find the limit (if it exists) of $I(b)$ as $b$ approaches $1$ from the left.
An antiderivative of $(x-1)^{-2/3}$ is $3(x-1)^{1/3}$. So $I(b)=3(b-1)^{1/3}-3(-1)^{1/3}$. As $b$ approaches $1$ from the left, this approaches $3$.
Do a similar calculation for the interval from $1$ to $3$.
Remark: In the reals, a standard definition of $a^b$ is $a^b=\exp(b\ln a)$. Under that definition, $a^b$ does not exist if $a$ is negative.
However, this conflicts with the usual lower level understanding that, for example, $(-8)^{1/3}=-2$, since $(-2)^3=-8$.
When $q$ is an odd integer, we can define $a^{1/q}$ as the real number whose $q$-th power is $a$. This makes sense, there is a unique such number. And then for any integer $p$, we can define $a^{p/q}$ as the $p$-th power of $a^{1/q}$. This works well for negative $a$.
We cannot make a similar definition of $a^{p/q}$ when $a$ is negative and $q$ is even.