Necessary and sufficient condition for a curve to have infinite length What is the necessary and sufficient condition for a curve to have infinite length in a compact interval? 
Say the curve is restricted to $[0, 1]$.
I vaguely remember that it is related to the boundedness of the total variation. I checked already the answers here but they are related to specific examples. 
 A: Given a function
$$f:\quad [a,b]\to{\mathbb R}^n, \qquad t\mapsto f(t)\ ,$$
the total variation of $f$ over $[a,b]$ is defined by
$$V(f):=\sup_{\cal P}\sum_{k=1}^N|f(t_k)-f(t_{k-1})|\leq\infty\ ,\tag{1}$$
whereby the $\sup$ ranges over all partitions
$${\cal P}:\qquad a=t_0<t_1<\ldots<t_N=b\ ,\qquad N=N_{\cal P}\ .$$
If $V(f)<\infty$ the function $f$ is called of bounded variation.
Consider now a curve
$$\gamma:\quad [a,b]\to {\bf z}(t):=\bigl(x(t), y(t)\bigr)\tag{2}$$
in the plane. Then the length  $L(\gamma)$ is by definition the total variation of the vector-valued function ${\bf z}(\cdot)$ used for the parametrization of $\gamma$:
$$L(\gamma):=V({\bf z}(\cdot))\leq\infty\ .$$
Note that for any two points ${\bf z}_0$, ${\bf z}_1$ one has
$$\max\{|x_1-x_0|,\>|y_1-y_0|\}\leq|{\bf z}_1-{\bf z}_0|\leq|x_1-x_0|+|y_1-y_0|\ .$$
From $(1)$ it then easily follows that the function ${\bf z}(\cdot)$ in $(2)$ is of bounded variation iff both $x(\cdot)$ and $y(\cdot)$ are of bounded variation.This allows to  conclude that a graph
$$\gamma:\quad [a,b]\to{\mathbb R}^2,\qquad x\mapsto\bigl(x,f(x)\bigr)$$
has finite length iff $V(f)<\infty$, since the total variation of the first coordinate is $=b-a<\infty$ in any case.
A: The total variation of a differentiable function on $[0,1]$ is given by
$$
\int_0^1\left|f'(x)\right|\mathrm{d}x
=\int_0^1\sqrt{f'(x)^2}\,\mathrm{d}x
$$
The length of the graph of a differentiable function on $[0,1]$ is given by
$$
\int_0^1\sqrt{1+f'(x)^2}\,\mathrm{d}x
$$
Furthermore, the triangle inequality tells us that
$$
\int_0^1\sqrt{f'(x)^2}\,\mathrm{d}x
\le\int_0^1\sqrt{1+f'(x)^2}\,\mathrm{d}x
\le1+\int_0^1\sqrt{f'(x)^2}\,\mathrm{d}x
$$
That is, the length is finite if and only if the total variation is finite.
A: A curve will have infinite length in an interval $[a, b]$ if the integral:
$$ \int_a^b \sqrt{\left(1 + (f'(x))^2\right)}\mathrm{d}x$$
evaluates to $\infty$.
Here $f'(x)$ represents the derivative of $f(x)$. This can be derived as follows.
Consider a length of the curve $\text ds$. It's position in the $XY$ plane is $(x,y)$. Then:
$$\mathrm ds = \sqrt{(\mathrm dx)^2 + (\mathrm dy)^2}$$
where $y=f(x)$. As $\displaystyle \frac {\text dy}{\text dx} = f'(x)$, so: $\text dy = f'(x)\,\text dx$. Thus,
$$\large \int_0^s\text ds = s = \int_a^b\sqrt{\left(1 + (f'(x))^2\right)}\text dx$$
where $s$ is the total length of the curve $f(x)$ in the interval $[a,b]$. If $s\to\infty$, then the curve will have infinite length in $[a,b]$.
Note: I had help from here. Also, from @Martin Sleziak's comment, this technique works only for differentiable functions.
Please correct me if I'm wrong. Hope this helps!
