integral representation of length of smooth curve

Suppose $\gamma :[0,1]\to {ℝ}^m$is a continuously differentiablecurve. Then the definition of its lengthas a rectifiable curve

is equal to its length as computed in differential geometry:

$${\int }_0^1\parallel {\gamma }^{\prime }(t)\parallel 𝑑t.$$

Proof.

Let the partition $\{t_i\}$ of $[0,1]$ be arbitrary. Then

${\displaystyle \sum _{i=1}^n}\parallel \gamma (t_i)-\gamma (t_{i-1})\parallel$$={\displaystyle \sum _{i=1}^n}\parallel {\displaystyle {\int }_{t_{i-1}}^{t_i}}{\gamma }^{\prime }(t)𝑑t\parallel$(fundamental theorem of calculus)

$\le {\displaystyle \sum _{i=1}^n}{\displaystyle {\int }_{t_{i-1}}^{t_i}}\parallel {\gamma }^{\prime }(t)\parallel 𝑑t$(triangle inequality for integrals)

$={\displaystyle {\int }_0^1}\parallel {\gamma }^{\prime }(t)\parallel 𝑑t.$

Hence $L\le {\int }_0^1\parallel {\gamma }^{\prime }(t)\parallel 𝑑t$.(By the way, this also shows that $\gamma$ is rectifiable in the first place.)

The inequality in the other direction is more tricky.Given $ϵ>0$, we know that${\int }_0^1\parallel {\gamma }^{\prime }(t)\parallel 𝑑t$ can be approximated up to $ϵ$by a Riemann sum of the form

$$\sum _{i=1}^n\parallel {\gamma }^{\prime }(t_{i-1})\parallel (t_i-t_{i-1})$$

provided the partition $\{t_i\}$ is fine enough,i.e. has mesh width $\le \mathrm{\Delta }$ for some small $\mathrm{\Delta }>0$.We want to approximate ${\gamma }^{\prime }(t_{i-1})$ with$[\gamma (t_i)-\gamma (t_{i-1})]/(t_i-t_{i-1})$, but this only worksif $t_i-t_{i-1}$ is small.

To get the precise estimates, use uniform continuityof ${\gamma }^{\prime }$ on $[0,1]$ to obtain a $\delta >0$ such that$\parallel {\gamma }^{\prime }(\tau )-{\gamma }^{\prime }(t)\parallel \le ϵ$whenever $|\tau -t|\le \delta$.Then for all and $t\in [0,1]$,

$$\parallel \frac{\gamma (t+h)-\gamma (t)}{h}-{\gamma }^{\prime }(t)\parallel \le \frac{1}{h}{\int }_t^{t+h}\parallel {\gamma }^{\prime }(\tau )-{\gamma }^{\prime }(t)\parallel 𝑑\tau \le \frac{h}{h}ϵ=ϵ.$$

Let the partition $\{t_i\}$ have a mesh width less than both $\delta$and $\mathrm{\Delta }$. Thensetting $h=t_i-t_{i-1}$ successively in each summand,we have

${\displaystyle {\int }_0^1}\parallel {\gamma }^{\prime }(t)\parallel 𝑑t$$\le {\displaystyle \sum _{i=1}^n}\parallel {\gamma }^{\prime }(t_{i-1})\parallel (t_i-t_{i-1})+ϵ$

$\le {\displaystyle \sum _{i=1}^n}{\displaystyle \frac{\parallel \gamma (t_i)-\gamma (t_{i-1})\parallel }{t_i-t_{i-1}}}(t_i-t_{i-1})+{\displaystyle \sum _{i=1}^n}ϵ(t_i-t_{i-1})+ϵ$

$={\displaystyle \sum _{i=1}^n}\parallel \gamma (t_i)-\gamma (t_{i-1})\parallel +2ϵ$

$\le L+2ϵ.$

Taking $ϵ\to 0$ yields ${\int }_0^1\parallel {\gamma }^{\prime }(t)\parallel 𝑑t\le L$.∎

We remark that $L={\int }_0^1\parallel {\gamma }^{\prime }(t)\parallel 𝑑t$ is true for piecewise smooth curves $\gamma$also, simply by adding together the results for each smooth segment of $\gamma$.