tracing a solution

[oguzkan]

Indefinite integration of equation

[1/Sqrt (a+2 Cosh[Y])] dY

in Mathematica gives a solution in terms of AppellF1 function as

AppellF1[0.5, 0.5, 0.5, 1.5, (a + 2 Cosh[Y])/(a + 2), ( a + 2 Cosh[Y])/(a - 2)]
Sqrt[(2- 2 Cosh[Y])/(2+ a)] Sqrt[( 2 + 2 Cosh[Y])/(2- a)](a + 2 Cosh[Y])^0.5 Csch[Y]

Can anybody help me how this integral leads to this solution?

I tried Trace Function in Mathematica to see the intermediate steps but with no success.

Thank you in advance

Oguz

 

[galactus]

This may not be doable by elementary means. The El at the beginning of your solution may indicate an Elliptic Integral of some kind.

EDIT: \(\displaystyle \int\frac{dy}{\sqrt{a+2cosh(y)}}\)


Just as I thought. It involves an Elliptic integral of the first kind.

Running it through Mathematica's Integrator, it gives:

\(\displaystyle \frac{-2i\cdot \sqrt{\frac{a+2cosh(y)}{a+2}}\cdot F(\frac{ix}{2}|\frac{4}{a+2})}{\sqrt{a+2cosh(y)}}\)

I ran it through Maple 10 and got

\(\displaystyle \frac{\sqrt{\frac{a+4cosh^{2}(\frac{y}{2})-2}{a-2}}\cdot\sqrt{1-cosh^{2}(\frac{y}{2})}\cdot \text{EllipticF}\left(2cosh(\frac{y}{2})\sqrt{\frac{1}{2-a}},\frac{\sqrt{2-a}}{2}\right)}{\sqrt{\frac{1}{2-a}}\cdot sinh(\frac{y}{2})\cdot \sqrt{a+4cosh^{2}(\frac{y}{2})-2}}\)

A wee bit more cumbersome than Mathematica, but I assume they are equivalent.

This would be very difficult to do by hand.

 

[oguzkan]

Dear Galactus,

First of all, Thank You for your time..

As we had talked before, integration of equation

[1/Sqrt[(a + 2 Cosh[Y])]dY

gives two identical results

AppellF1[0.5, 0.5, 0.5, 1.5, (a + 2 Cosh[Y])/(a + 2), ( a + 2 Cosh[Y])/(a - 2)]
Sqrt[(2- 2 Cosh[Y])/(2+ a)] Sqrt[( 2 + 2 Cosh[Y])/(2- a)](a + 2 Cosh[Y])^0.5 Csch[Y]

or

{-2i Sqrt[(a + 2 Cosh[Y])/(a + 2)] F[ix/2; 4/(a+2)] }/Sqrt[(a + 2 Cosh[Y])/(a + 2)]

Then, there must be a correlation or some transformation rules between Appell F1 function and Elliptic F1 Function? I searched all I can and could find none.

Take care

Oguzkan