# Could someone explain how to evaluate this horrible integral?

1. Feb 18, 2009

### AxiomOfChoice

The integral is

$$\int_0^{f(E)} \frac{\cosh \alpha x}{\sqrt{1-\beta \cosh^2 \alpha x}} dx,$$

where

$$f(E) = \frac{1}{\alpha}\cosh^{-1}(\sqrt{\frac{1}{\beta}}),$$

$$\alpha > 0$$, and $$0<\beta<1$$. The integral has proven very difficult to evaluate. Every time I plug it into Mathematica with values for $$\alpha$$ and $$\beta$$ that satisfy the conditions listed above, I get an imaginary number! But that can't happen because the integral is involved in computing the period of a particle undergoing periodic motion, which is necessarily real.

It is possible that the integral will not evaluate to a real number, in which case I've made a mistake in deriving its form. If that's the case, please let me know. Thanks!

2. Feb 18, 2009

### HallsofIvy

I wouldn't dare. Looks like an elliptic integral to me.

3. Feb 20, 2009

### maze

Seems like everything inside the integral is real and measurable, and its being integrated over a real interval, so the answer must be real. I know that there are commands to force Maple to assume certain variables are real, between an interval, etc. Maybe try to find the equivalent commands on Mathematical and try it again?

4. Feb 20, 2009

### maze

Ran this through Maple, the answer is (pi/2)(alpha*sqrt(beta))^-1.

Here is an image:
http://img132.imageshack.us/img132/5376/horribleintegralpd1.png [Broken]

If you want to actually calculate it, my best guess is maybe to try complex contour integration? That's probably not going to work since the upper limit is finite, but worth a try.

Last edited by a moderator: May 4, 2017
5. Feb 21, 2009

### coomast

A simple substitution does the trick. Setting
$$cosh(\alpha x)=t$$
gives:
$$\frac{1}{\alpha \sqrt{\beta}} \int_{1}^{\frac{1}{\sqrt{\beta}}} \frac{t dt} {\sqrt{t^2-1} \sqrt{\frac{1}{\beta}-t^2}}$$
Setting now:
$$t^2-1=u^2$$
gives:
$$\frac{1}{\alpha \sqrt{\beta}} \int_{0}^{\sqrt{\frac{1}{\beta}-1}} \frac{du} {\sqrt{\frac{1}{\beta}-1-u^2} }$$
giving arcsin as solution and after filling in the limits you get the result:
$$\frac{\pi}{2\alpha \sqrt{\beta}}$$

So, no elliptic integral here, HallsofIvy :-)

coomast

6. Feb 21, 2009

### Unknot

That's some deadly integration skill there.