arc length of logarithmic curve
ArcLengthOfLogarithmicCurve
2009-09-06 10:45:58
2009-09-13 12:41:44
Example
arc length
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\DeclareMathOperator{\arsinh}{arsinh}
\DeclareMathOperator{\arcosh}{arcosh}
\DeclareMathOperator{\artanh}{artanh}
\DeclareMathOperator{\arcoth}{arcoth}
\PMlinkescapeword{substitution}
The arc length of the graph of \PMlinkname{logarithm function}{NaturalLogarithm2} is expressible in closed form (other cases are listed in the entry arc length of parabola).\, The usual arc length \PMlinkescapetext{formula}
$$s \;=\; \int_a^b\!\sqrt{1+(f'(x))^2}\,dx$$
gives, if\, $0 < a < b$,\, for\, $f(x) := \ln{x}$,\, $f'(x) = \frac{1}{x}$,\, the expression
\begin{align}
s \;=\; \int_a^b\!\frac{\sqrt{1\!+\!x^2}}{x}\,dx.
\end{align}
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 \PMlinkescapetext{simple} substitution
$$\sqrt{1\!+\!x^2} \;:=\; t, \quad x \;=\; \sqrt{t^2\!-\!1}, \quad dx \;=\; \frac{t\,dt}{\sqrt{t^2\!-\!1}}$$
yielding
$$\int\!\frac{\sqrt{1\!+\!x^2}}{x}\,dx \;=\; \int\!\frac{t^2\,dt}{t^2\!-\!1}
\;=\; t+\frac{1}{2}\ln\frac{t\!-\!1}{t\!+\!1}+C \;=\; t-\arcoth{t}+C$$
(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 \PMlinkid{exponential function}{2541} \,$x \mapsto e^x$,\, which actually is the same curve as the graph of the inverse function \,$x \mapsto \ln{x}$,\, one may write the expression
\begin{align}
s \;=\; \int_\alpha^\beta\!\sqrt{1\!+\!e^{2x}}\,dx.
\end{align}
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
$$\int\!\sqrt{1\!+\!e^{2x}}\,dx \;=\; \sqrt{1\!+\!e^{2x}}-\arsinh{e^{-x}}+C
\;=\; \sqrt{1\!+\!e^{2x}}+\ln\frac{e^x}{1+\sqrt{1\!+\!e^{2x}}}+C.$$