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.


  • 1
    $\begingroup$ 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

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? ;) )



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


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


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


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

  • $\begingroup$ 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

There is no solution for arbitrary integration limits in elementary terms: we face an elliptic integral:

Wolfram alpha output

  • $\begingroup$ How can I fix my link? $\endgroup$ – Rasmus Feb 27 '11 at 13:49
  • 2
    $\begingroup$ 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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