Compute the line integral along the curve $\gamma(l)=(l\cos(\frac{2\pi}{l}),l\sin(\frac{2\pi}{l})),\ l\in(0,1)$.
The default procedure is to calculate $\gamma'(l)$, then $\|\gamma'(l)\|$, and finally $\int_0^1\|\gamma'(l)\|\,dl$.
In this case $\|\gamma'(l)\|=\sqrt{1+\frac{4\pi^2}{l^2}}$, so we have to express the integral as an improper one as $l=0$ is problematic. For instance, we could compute $\lim_{n\to\infty}\int_{1/n}^1\sqrt{1+\frac{4\pi^2}{l^2}} \, dl$.
If we calculate the integral (which is tedious) we will see that it is not convergent, so the length of the curve is infinite. Maybe we should use some method before integrating to see whether our improper integral is convergent or not. In this case, I think we can apply limit comparison:
$$\lim_{l\to0^+}\frac{\|\gamma'(l)\|}{\frac{1}{l}} = \lim_{l\to0^+}l\sqrt{1+\frac{4\pi^2}{l^2}} = \lim_{l\to0^+}\sqrt{l^2+4\pi^2}=2\pi$$
So as $0<2\pi<+\infty,\ \int_0^1\|\gamma'(l)\|\,dl$ has the same behavior as $\int_0^1\frac{1}{l}dl$, so they are both divergent. Thus, we deduce that the length of the curve is $+\infty$ without computing the integral.
My questions are: Is this second method right? Is there any other faster method to deduce (formally, not by simple intuition) the result? In which cases is better to use convergence criteria before computing the integral (as I think this case is an example)?
$\int_0^{1}\|\gamma '(l)\|\,dl \geq \int_0^1 \frac { 2\pi} l \,dl=\infty$. So the curve has infinite length.