Consider an indefinite "elliptic integralβ, $$ J_n(r):=\int^r_{r_0}\frac{x^n}{\sqrt{f(x)}}, \quad f(x):=(x-x_1)(x-x_2)(x-x_3). $$ where $r_0>0, n=1,0,β1$, and $x_{1,2,3}$ are roots of the cubic equation $π(π₯)=0$.
I am interested in a case when $π(π₯)=0$ has one real root $π₯_1>0$ and two complex roots $π₯_2$ and $π₯_3$ (since it's a cubic equation, they are also complex conjugate, $π₯_3=\bar{x_2}$).
I want to know if $π½_{β1}$ can be written without imaginary number $π$, because this integral appears in a problem of physics, $π½_π$ should be real. But, so far, I failed to make it. I write down what I faced:
I have tried an integration by substitution. In the present case, let $π₯_2=π+ππ$ and thus $π₯_3=\bar{x_2}=πβππ$. Then we have $π(π₯)=(π₯βπ₯_1)((π₯βπ)^2+π^2)$.
Now, transform the variable from $π₯$ to $π¦$ with $$ π₯=π₯_1+π \tan^2π¦, π:=\sqrt{(π₯_1βπ)^2+π^2}. $$ After some manipulation, $π½_π$ reduces to $$ \frac{\sqrt{p}}{2}π½_π(π)=\int\frac{(x_1+π\tan^2y)^n {\rm d}y}{\sqrt{1βπ΄\sin^2π¦\cos^2π¦}}, $$ where $π΄:=2(πβ(π₯_1βπ))/π$ and thus $0<π΄<2$ by definition.
For $π=0$ and $n=+1$, there is no problem: For $π=0$, a software "Mathematica" returns $$ \frac{\sqrt{p}}{2}π½_0(π)=\frac{1}{2}πΉ(2π¦,\frac{A}{4}), $$ where $πΉ(\phi,π)$ is Elliptic integral of the first kind. For $π=1$, Mathematica returns a combination of elliptic functions and elementary functions (I do not express them here, though).
Note that there is no imaginary number $i$ so far.
For $π=β1$, however, integration returns a messy form with $i$:
where $\Pi(n,\phi,m)$ is incomplete Elliptic integral of the third kind.
I guess there should be a simpler form of $π½_{β1}$ without $i$. Maybe the above transformation is not good for $π½_{β1}$? Is there some identities that eliminate $i$? I am not comfortable that in the messy form, because $\sqrt{A-4}$ appears, although $0<π΄<2$.
I am beginner of elliptic integral, any comments are welcome. Thank you.
