English is not my first language so please excuse me for possibly not expressing myself clearly.
I am a computer science student and I've just finished calculus this semester.
I was having fun with parametric curves and their lengths when I stumbled upon this one:
$$ x = x(t)= e^t\\y=y(t)=\cos(e^{-t}) $$ graph picture
I wanted to find the arc length near zero so I got down to this integral: $$ \int_{0}^{1} \sqrt{{(e^t)}^2 + ({e^{-t}})^2 \cdot \sin^2(e^{-t})} \,dt $$
Wolfram computes it to be about 1.79 That's what I don't understand. Doesn't the arc oscilate an infinite amount of times before it gets to zero? I know that the function doesn't have a limit when $t\rightarrow-\infty$, so I guess I can't integrate it? Or am I just overthinking everything?
Your computation does not compute the arc length of the part of the curve with $x \in (0, 1]$ but rather the length of the curve defined by the parameter range $t \in [0, 1]$, which corresponds to the part of the curve with $x \in [x(0), x(1)] = [e^0, e^1] = [1, e]$.
To compute the arc length of the part of the curve with $x \in (0, 1]$, we can integrate over $t \in (-\infty, 0]$, since $\lim_{t \to -\infty} x(t) = 0$ and $x(0) = 1$: Substituting $u = \frac{1}{x} = e^{-t}$ transforms the arc length integral to $$ \int_1^\infty \sqrt{\frac{1}{u^2} + \sin^2 u} \,du,$$ but $$\int_1^\infty \sqrt{\frac{1}{u^2} + \sin^2 u} \,du > \int_1^M |\sin u| \,du,$$ and the latter quantity goes to $\infty$ as $M \to \infty$, hence the arc length is infinite as expected.
Cf. https://en.wikipedia.org/wiki/Topologist%27s_sine_curve.