Let $b<0<a$, and consider the function $\alpha:(0,+\infty) \to \mathbb R^2$ defined as $$\alpha(t)=(ae^{bt}\cos(t),ae^{bt}\sin(t))$$
Show that $\lim_{t \to +\infty} \alpha'(t)=(0,0)$ and $\lim_{t \to \infty} \int_0^t |\alpha'(t)|dt$ is finite. Conclude that the arc lenght of the curve is finite.
I've already shown that $\lim_{t \to +\infty} \alpha'(t)=(0,0)$, which implies $\lim_{t \to +\infty} |\alpha'(t)|=(0,0)$.
Since the arclenght of $\alpha$ is defined as $\Gamma(t)=\int_0^t |\alpha'(t)|dt$, then to show it is finite, it is sufficient to show that $\lim_{t \to \infty} \int_0^t |\alpha'(t)|dt$ is finite.
My question is, if a positive function $f(t) \to 0$ when $t \to +\infty$, is it true that $\int_0^t f(t)dt is finite? If I could show that, then this exercise is a particular case of the statement.
I've directly calculated $\int_0^t |\alpha'(t)|dt=\sqrt{a^2b^2+a^2}\int_o^t e^{bt}dt$ but I don't know if it is useful to know the exact expression of the integral.
You have $\alpha'(t) = a e^{bt} (b(\cos t, \sin t) + (-\sin t, \cos t))$, hence $|\alpha'(t)| \le a e^{bt}(|b|+1)$.
Since $\int_0^\infty e^{bt} dt = {1 \over |b|}$, we see that $t \mapsto |\alpha'(t)|$ is integrable.
In fact, since you have computed $\int_0^t |\alpha'(x) | dx = a \sqrt{b^2+1} \int_0^t e^{bx} dx $, and $t \mapsto e^{bt}$ is integrable, this is enough to conclude that $\int_0^\infty |\alpha'(x) | dx$ exists. Indeed, it gives the value $\int_0^\infty |\alpha'(x) | dx = a \sqrt{1 + {1 \over b^2}}$.
Note: It is not sufficient or necessary that $f(t) \to 0$ in order that $f$ be integrable.
For example, $f_1(t) = {1 \over t+1}$ converges to zero, but is not integrable.
Let $\tau(t) = \max(0,1-|t|)$, and consider $f_2(t) = \sum_{n=1}^\infty \tau(2^n t-2^n)$. Then $\int_0^\infty f_2(t) dt = \sum_{n=1}^\infty {1 \over 2^n} = 1$, hence $f_2$ is integrable, but $f_2(2^n) = 1$ for all $n$.