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how can I please calculate an arc length of $\dfrac{e^x-e^{-x}}{2}$. I tried to substitute $\dfrac{e^x-e^{-x}}{2}=\sinh x$, which leads to $\int\sqrt{1+\cosh^2x}dx$, which unfortunately I can't solve.

Thank you very much.

C.

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    It was not a homework, I just found it somewhere on the web and tried to solve it. I know that $\frac{e^x+e^{-x}}{2}$ is easy to solve (as it leads to $\sqrt{\cosh^2 x}$), but thank you anyway.2011-04-18

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I find Wolfram Alpha's solution a bit ugly, in that it returns complex results for a real entity, so I'll roll my own solution here.

We start with

$\int\sqrt{1+\cosh^2 x}\mathrm dx=\frac1{\sqrt{2}}\int\sqrt{3+\cosh\;2x}\;\mathrm dx$

which can be turned into

$\frac1{\sqrt{2}}\int\sqrt{3+\frac{1+\tanh^2 x}{1-\tanh^2 x}}\;\mathrm dx$

(recognize Weierstrass? ;) )

Let

$\tanh\;x=\mathrm{sn}\left(v|\frac12\right)$

where $\mathrm{sn}(v|m)$ is a Jacobian elliptic function, resulting in

$\frac1{\sqrt{2}}\int\sqrt{3+\frac{1+\mathrm{sn}^2\left(v|\frac12\right)}{1-\mathrm{sn}^2\left(v|\frac12\right)}}\mathrm{dc}\left(v|\frac12\right)\;\mathrm dv=\sqrt{2}\int\mathrm{dc}^2\left(v|\frac12\right)\;\mathrm dv$

where $\mathrm{dc}(v|m)=\frac{\mathrm{dn}(v|m)}{\mathrm{cn}(v|m)}$ is a Jacobian elliptic function.

From this formula, we obtain

$\sqrt{2}\left(v-E\left(\mathrm{am}\left(v|\frac12\right)|\frac12\right)+\mathrm{sn}\left(v|\frac12\right)\mathrm{dc}\left(v|\frac12\right)\right)$

or, by undoing the transformation with $v=F\left(\arcsin\left(\tanh\;x\right)|\frac12\right)$,

$\sqrt{2}\left(F\left(\arcsin\left(\tanh\;x\right)|\frac12\right)-E\left(\arcsin\left(\tanh\;x\right)|\frac12\right)+\tan\left(\arcsin\left(\tanh\;x\right)\right)\sqrt{1-\frac12\sin^2\left(\arcsin\left(\tanh\;x\right)\right)}\right)$

which simplifies to

$\sqrt{2}\left(F\left(\arcsin\left(\tanh\;x\right)|\frac12\right)-E\left(\arcsin\left(\tanh\;x\right)|\frac12\right)\right)+\tanh\;x\sqrt{1+\cosh^2 x}$

to which an arbitrary constant can be added.


As an alternative, one can start with the Mathematica result

$\frac{\sqrt{2}}{i}E\left(ix|\frac12\right)$

and simplify (get rid of the complex stuff) accordingly, using the second relation in formula 19.7.7 in the DLMF. Note that $\arctan\sinh\;x=\arcsin\tanh\;x$.

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    @claudia: It admittedly is a bit "highbrow", but there really is no way to express the arclength function elementarily. (Also, I was looking for an excuse to practice my elliptic integral manipulations.)2011-04-18
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There is no solution for arbitrary integration limits in elementary terms: we face an elliptic integral:

Wolfram alpha output

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    It depends what you call simple: a single elliptic function is quite a simple solution (from one point of view). And yes, you can be sure that it cannot be reduced to $\sin, \cos, \exp$ and their inverses.2011-02-27