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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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    $\begingroup$ Assuming this is homework, are you sure the question isn't about $\frac{e^x + e^{-x}}{2}$? That is one of the few examples where the arc-length integral can be done in elementary functions. $\endgroup$ – Robert Israel Apr 17 '11 at 18:25
  • $\begingroup$ 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. $\endgroup$ – user7557 Apr 18 '11 at 13:04
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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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  • $\begingroup$ Thank you. It is quite too advanced for me, but I'll try to get it anyway. $\endgroup$ – user7557 Apr 18 '11 at 13:08
  • $\begingroup$ @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.) $\endgroup$ – J. M. is a poor mathematician Apr 18 '11 at 13:10
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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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  • $\begingroup$ How can I fix my link? $\endgroup$ – Rasmus Feb 27 '11 at 13:49
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    $\begingroup$ I fixed the link. It's not necessary to introduce a html-tag, but it looks nicer. Some characters are not compatible with the markdown engine, in your case it was the offending ^ (this is where the link started to be broken). I replaced it by its percent equivalent %5E, see en.wikipedia.org/wiki/Percent-encoding $\endgroup$ – t.b. Feb 27 '11 at 14:24
  • $\begingroup$ @Theo, thanks for your help. $\endgroup$ – Rasmus Feb 27 '11 at 14:57
  • $\begingroup$ Thank you. I've tried Wolfram alpha, but does it prove that simple solution does not exist? The plot of the integral seems simple enough. $\endgroup$ – user7557 Feb 27 '11 at 15:23
  • $\begingroup$ 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. $\endgroup$ – Fabian Feb 27 '11 at 16:08

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