Which plane curve parametrized by arc-length has curvature $\kappa(s) = \frac{1}{s}$ Here's how far I have gotten:
$$\kappa(s) = 1/s \Rightarrow \theta(s) = \int \frac{ds}{s} = \ln(s)$$
Then 
$$T(s) = e^{i \cdot \ln(s)} = s^i$$
Then the (non-arc-length parametrized) curve $\gamma(s)$ we're looking for is
$$\gamma(s) = \int T(s) ds = \int s^{i} ds = \frac{1}{i+1} s^{i+1} = (\frac{1}{2} - \frac{i}{2}) s^{1+i}$$
I solved that integral, but now I have to parameterized by arc-length. This is where I got stuck. Any hints?
EDIT: I now realize that my result is already parameterized by arc-length, since:
$$\gamma(s) = (\frac{1}{2} - \frac{i}{2}) s^{1+i} = \frac{1}{2}s \cdot (1-i) s^i = \frac{1}{2}s \cdot (1-i) e^{\ln(s) i}$$
$$=\frac{1}{2}s (1-i) \cdot (\cos(\ln s) + i \sin(\ln s))$$ 
$$= \frac{1}{2} s \cdot (\cos(\ln s) + \sin(\ln s) + i\cdot(\sin(\ln s) - \cos(\ln s)))$$
$$ = \frac{1}{2}s \begin{pmatrix}
    \cos(\ln s) + \sin(\ln s)\\
    \sin(\ln s) - \cos(\ln s)
\end{pmatrix}
$$
Whereas $$|\gamma'(s)| = | \begin{pmatrix}
    \cos(\ln s)\\
    \sin(\ln s)
\end{pmatrix} | = \sqrt{\cos^2(\ln s) + \sin^2(\ln s)} = \sqrt{1} = 1$$
 A: By Fundamental Theorem of Plane Curves we have that for given continuous $\kappa: I \to \mathbb{R}, p_0 = (x_0,y_0)$ and $v_0=(a,b)$ there's a unique unit-speed plane curve $\alpha: I \to \mathbb{R}^2$, with curvature $\kappa$ starting from $p_0$ and with initial velocity $v_0$.
To find it explicitly take angle $\theta_0$ s.t. $\cos \theta_0 = a$ and $\sin \theta_0 = b$. Then define angle function $\theta(s) = \theta_0 + \int_0^s \kappa(t)dt$. Then let $x(s) = x_0 + \int_0^s \cos(\theta(t))dt$ and $y(s) = y_0 + \int_0^s \sin(\theta(t))dt$. Finally $\alpha(s) = (x(s),y(s))$ is the wanted curve.
In your case finding $x$ and $y$ explicitly isn't straightforward, but it's doable with enough time, as you need to evaluate $\int_0^s \sin(\ln(t))dt$ and $\int_0^s \cos(\ln(t))dt$
A: The curve is a logarithmic spiral. Let a $c>0$ and an $a>0$ be given, and consider the curve $\gamma$ with the polar representation
$$\gamma:\quad r(\phi):= a e^{c\,\phi}\quad(-\infty<\phi<\infty)\ .$$
For this curve one computes
$$\kappa(\phi)={e^{-c\,\phi}\over a\sqrt{1+c^2}}\ .\tag{1}$$
On the other hand we have
$$s'(\phi)=\sqrt{r^2+r'^2}=a\sqrt{1+c^2}\>e^{c\,\phi}$$
and therefore, if we let $s=0$ at the inner end of the spiral,
$$s(\phi)=\int_{-\infty}^\phi s'(\phi)\>d\phi=a\sqrt{1+c^2}\int_{-\infty}^\phi e^{c\,t}\>dt={a\sqrt{1+c^2}\over c} e^{c\,\phi}\ .$$
Comparing this with $(1)$ we see that
$$\kappa(s)={1\over c\,s}\ .$$
