# arc length of logarithmic curve

The arc length  of the graph of logarithm function (http://planetmath.org/NaturalLogarithm2) is expressible in closed form (other cases are listed in the entry arc length of parabola).  The usual arc length

 $s\;=\;\int_{a}^{b}\!\sqrt{1+(f^{\prime}(x))^{2}}\,dx$

gives, if  $0,  for  $f(x):=\ln{x}$,  $f^{\prime}(x)=\frac{1}{x}$,  the expression

 $\displaystyle s\;=\;\int_{a}^{b}\!\frac{\sqrt{1\!+\!x^{2}}}{x}\,dx.$ (1)

Here, finding a suitable substitution for integration may be a bit difficult.  E.g.  $x:=\tan{t}$ leads to

 $\int\!\frac{\sqrt{1\!+\!x^{2}}}{x}\,dx\;=\;\int\frac{dt}{\sin{t}\,\cos^{2}{t}},$

the substitution  $x:=\sinh{t}$  to

 $\int\!\frac{\sqrt{1\!+\!x^{2}}}{x}\,dx\;=\;\int\frac{\cosh^{2}{t}}{\sinh{t}}\,dt,$

which both seem to require a new substitution.  As well the Euler’s substitutions (1st and 2nd ones) lead to awkward rational functions as integrands.

But there is the straightforward substitution

 $\sqrt{1\!+\!x^{2}}\;:=\;t,\quad x\;=\;\sqrt{t^{2}\!-\!1},\quad dx\;=\;\frac{t% \,dt}{\sqrt{t^{2}\!-\!1}}$

yielding

 $\int\!\frac{\sqrt{1\!+\!x^{2}}}{x}\,dx\;=\;\int\!\frac{t^{2}\,dt}{t^{2}\!-\!1}% \;=\;t+\frac{1}{2}\ln\frac{t\!-\!1}{t\!+\!1}+C\;=\;t-\operatorname{arcoth}{t}+C$

(see area functions) and then

 $\int\!\frac{\sqrt{1\!+\!x^{2}}}{x}\,dx\;=\;\sqrt{1\!+\!x^{2}}+\frac{1}{2}\ln% \frac{\sqrt{1\!+\!x^{2}}-1}{\sqrt{1\!+\!x^{2}}+1}+C\;=\;\sqrt{1\!+\!x^{2}}+\ln% \frac{x}{1+\sqrt{1\!+\!x^{2}}}+C.$

Using this antiderivative, one can obtain the arc length (1).  For example, if  $a=\sqrt{3}$  and  $b=\sqrt{15}$,  the result is  $s=2+\ln\frac{3}{\sqrt{5}}$.

As for finding the arc length of the graph of the http://planetmath.org/node/2541exponential function    $x\mapsto e^{x}$,  which actually is the same curve as the graph of the inverse function$x\mapsto\ln{x}$,  one may write the expression

 $\displaystyle s\;=\;\int_{\alpha}^{\beta}\!\sqrt{1\!+\!e^{2x}}\,dx.$ (2)

Since here the substitution

 $e^{x}\;:=\;t,\quad x\;=\;\ln{t},\quad dx\;=\;\frac{dt}{t}$

shows that

 $\int\!\sqrt{1\!+\!e^{2x}}\,dx\;=\;\int\!\frac{\sqrt{1\!+\!t^{2}}}{t}\,dt,$

we see that it’s really a question of the same task as above.  The antiderivative is

 $\int\!\sqrt{1\!+\!e^{2x}}\,dx\;=\;\sqrt{1\!+\!e^{2x}}-\operatorname{arsinh}{e^% {-x}}+C\;=\;\sqrt{1\!+\!e^{2x}}+\ln\frac{e^{x}}{1+\sqrt{1\!+\!e^{2x}}}+C.$
Title arc length of logarithmic curve ArcLengthOfLogarithmicCurve 2013-03-22 19:01:45 2013-03-22 19:01:45 pahio (2872) pahio (2872) 11 pahio (2872) Example msc 53A04 msc 26A42 msc 26A09 msc 26A06 arc length of exponential curve