Mathematica gives a solution in terms of AppellF1 function

[oguzkan]

Indefinite integration of equation

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

Can anybody help me how this integral leads to this solution? That is I would like to know the intermediate steps or transformations between the two entries if anybody could help.

Thank you in advance

Oguz

 

[jvicens]

Dear Oguz,

Thank you for your question regarding the indefinite integration of an equation in Mathematica. I will do my best to explain the intermediate steps and transformations that lead to the solution using the AppellF1 function.

First, let's take a look at the original equation:

[1/Sqrt (a+2 CoshY)] dY

To integrate this equation, we can use the substitution method. Let u = a + 2 CoshY, then du = 2 SinhY dY. We can rewrite the equation as:

[1/Sqrt u] (du/2SinhY)

Now, we can use the AppellF1 function, which is defined as:

AppellF1[a, b, b', c, x, y] = 1/Beta[b, b'] ∫[0,1] t^(b-1) (1-t)^(b'-1) (1-xt)^(-a) (1-yt)^(-c) dt

Using this definition, we can rewrite our equation as:

(1/2) ∫[0,1] t^(-1/2) (1-t)^(-1/2) (u-2a)^(-1/2) (1-u)^(-1/2) du

Now, let's look at the parameters of the AppellF1 function in our solution:

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

We can see that the first three parameters (0.5, 0.5, 0.5) correspond to the exponents in our integral. The fourth parameter (1.5) corresponds to the exponent of u in the integral. The last two parameters ((a + 2 Cosh Y)/(a + 2), ( a + 2 Cosh Y)/(a - 2)) correspond to the values of x and y in the integral.

Using the definition of the AppellF1 function and the substitutions we made earlier, we can rewrite the integral as:

(1/2) ∫[0,1] t^(-0.5) (1-t)^(-0.5) (a+2CoshY-2a