When $σ(t)$ is the curve $σ(t) = cos(t) i + sin (t) j$ and set $r(t) =e^{-t} σ(t)$
I gave it a shot. I might be a little rusty at this:
$L[0, ∞)=\int_0^∞\sqrt{x_1'(t) + x_2'(t)+ ...+x_n'(t)} dt$
\begin{equation}L= [0, ∞)=\int_0^∞\sqrt{-e^{-2t}cos^2(t)-e^{-2t}sin^2(t)-e^{-2t}sin^2(t)+e^{-2t}cos^2(t)} dt \end{equation}
And in a few steps
\begin{equation}L= [0, ∞)=\int_0^∞\sqrt{e^{-2t}}\sqrt{-1+cos(2t)} dt \end{equation}
Is this correct? Can someone help me with what to do next
If a curve $\gamma$ is given in the so-called polar form $\phi\mapsto r(\phi)$, which is the same as the parametric representation $$\gamma:\quad t\mapsto r(t)(\cos t,\sin t)\qquad(a\leq t\leq b)\ ,$$ then from $$x(t)=r(t)\cos t,\quad y(t)=r(t)\sin t$$ we get $$x'(t)=r'(t)\cos t-r(t)\sin t,\quad y'(t)=r'(t)\sin t+r(t)\cos t\ .$$ In the case $r(t)=e^{-t}$ we therefore have $$x'^2(t)+y'^2(t)=r'^2(t)+r^2(t)=2e^{-2t}\ .$$ It follows that $$L(\gamma)=\int_0^\infty\sqrt{x'^2(t)+y'^2(t)}\>dt=\sqrt{2}\int_0^\infty e^{-t}\>dt=\sqrt{2}\ .$$