Consider an indefinite "elliptic integral”, $$ J_n(r):= \int^r_{r_0}\frac{x^n {\rm d}x}{\sqrt{f(x)}}, \quad f(x):=(x-x_1)(x-x_2)(x-x_3), $$ Here, $r_0>0, n=1,0,-1$, and $x_{1,2,3}$ are roots of cubic equation $f(x)=0$.

I am interested in a case when $f(x)=0$ has one real root $x_1>0$ and two complex roots $x_2$ and $x_3$ (since it's cubic equation, they are also complex conjugate, $x_3=\bar{x_2}$).

I want to know if $J_{-1}$ can be written without imaginary number $i$, because this integral appear in a problem of physics, $J_n$ should be real function. 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 $x_2=a+ ib$ and thus $x_3=\bar{x_2}=a-ib$. Then we have $f(x)=(x-x_1)((x-a)^2+b^2)$.

Now, transform the variable from $x$ to $y$ with $$ x=x_1+p \tan^2y, \quad p:=\sqrt{(x_1-a)^2+b^2}. $$ After some manipulation, $J_n$ reduces to $$ \frac{\sqrt{p}}{2}J_n(r)=\int\frac{(x_1+p\tan^2y)^n {\rm d}y}{\sqrt{1-A\sin^2y\cos^2y}}, $$ where $A:=2(p-(x_1-a))/p$ and $0<A<2$ by definition.

For $n=0$ and $+1$, it is no problem. For $n=0$, Mathematica returns $$ \frac{\sqrt{p}}{2}J_0(r)=\frac{1}{2}F(2y\mid \frac{A}{4}), $$ where $F(\phi\mid m)$ is the elliptic integral of the first kind, EllipticF. For $n=1$, Mathematica returns a combination of elliptic integrals and elementary functions (I do not express them here).

Note that there is no imaginary number $i$ so far.

For $n=-1$, however, integration of $J_{-1}$, i.e.,

Integrate[(x_1 + p Tan[y]^2)^-1 (1 - A Sin[y]^2 Cos[y]^2)^(-1/2), y]

returns a messy form with $i$: complex expression

where $\Pi(n,\phi\mid m)$ is the incomplete elliptic integral of the third kind. I found that Plot of $J_{-1}$ are indeed real under certain range of $a, p, A$. But, I guess there should be a simpler form in $J_{-1}$ without $i$.

Maybe the above transformation is not good for $J_{-1}$? Is there some identities that eliminate $i$?

I am not comfortable that in the messy form, $\sqrt{A-4}$ appears, although $0<A<2$. I am beginner of elliptic integral, any comments are welcome. Thank you.

  • $\begingroup$ You can always try the usual suspects, ie (Full)Simplify, PowerExpand, ComplexExpand etc. but honestly this question seems more suited for a math forum. $\endgroup$
    – MarcoB
    Commented Jun 21, 2022 at 3:40
  • $\begingroup$ @MarcoB The above messy one is already simplified form. Yes, as you suggest, it seems purely mathematical question. I will also post it on math stack exchange. Thank you. $\endgroup$ Commented Jun 21, 2022 at 4:58
  • $\begingroup$ "I want to know if ... can be written without imaginary number $i$, because this integral appear in a problem of physics, ... should be real..." - not that there isn't an affirmative answer for your particular problem, but are you aware of casus irreducibilis? $\endgroup$ Commented Jun 21, 2022 at 7:19
  • $\begingroup$ @J.M. Yes, I know casus irreducibilis in mathematics. So, I know my sentence about "result should be real" is actually not appropriate. But, I am still expecting a possibility that $J_{-1}$ can be written in real functions. $\endgroup$ Commented Jun 21, 2022 at 7:31

1 Answer 1


I replaced all cos^2 with 1 - sin^2 to arrive at

Integrate[1/((x1 + (p - x1)*Sin[y]^2)*Sqrt[1 - a*Sin[y]^2 + 
 a*Sin[y]^4])- Sin[y]^2/((x1 + (p - x1)*Sin[y]^2)*
 Sqrt[1 - a*Sin[y]^2 + a*Sin[y]^4]), y]

The first integral calculates without further manipulation, the second with the sin^2 in the nominator is given by

Integrate[(x1 + (p - x1)*Sin[y]^2)/((x1 + (p - x1)*Sin[y]^2)*
 Sqrt[1 - a*Sin[y]^2 + a*Sin[y]^4]), y] - 
Integrate[x1/((x1 + (p - x1)*Sin[y]^2)*Sqrt[1 - 
 a*Sin[y]^2 +a*Sin[y]^4]), y] = 
Integrate[((p - x1)*Sin[y]^2)/((x1 + (p - x1)*Sin[y]^2)*
 Sqrt[1 - a*Sin[y]^2 + a*Sin[y]^4]), y]

where the leftmost integral may be simplified to

Integrate[1/Sqrt[1 - a*Sin[y]^2 + a*Sin[y]^4], y]

Now the rightmost integral gives in! Then apply transformations for negative squared modulus of EllipticPi and EllipticF and negative parameter of the EllipticPi . The completely real result can be seen in the Plot command below, still somewhat messy. Maybe someone is able to simplify the ArcTans further, I wasn' t.

a = 0.7; p = 0.3; x1 = 2.3; 
Plot[{NIntegrate[1/(Sqrt[1 - a*Cos[yy]^2*Sin[yy]^2]*
 (x1 + p*Tan[yy]^2)), {yy, 0.3, y}], 
 ((a*p - 2*(p - x1))/(4*((p - x1)^2 + a*p*x1)))*
 EllipticF[2*y, a/4]+(p/(2*Sqrt[p*x1]*Sqrt[(p-x1)^2+a*p*x1]))*
 (ArcTan[(Sqrt[(p - x1)^2 + a*p*x1]*Sin[2*y])/(Sqrt[p*x1]*
  Sqrt[4 - a*Sin[2*y]^2])] + 
  (((4 - a)*p*(p - x1))/(4*(p + x1)*((p - x1)^2 + a*p*x1)))*
 EllipticPi[((p - x1)^2 + a*p*x1)/(p + x1)^2, 2*y, a/4] - 
  (((a*p - 2*(p - x1))/(4*((p - x1)^2 + a*p*x1)))*
 EllipticF[2*yy,a/4] + 
  (p/(2*Sqrt[p*x1]*Sqrt[(p - x1)^2 + a*p*x1]))*
  (ArcTan[(Sqrt[(p - x1)^2 + a*p*x1]*
   Sin[2*yy])/(Sqrt[p*x1]*Sqrt[4 - a*Sin[2*yy]^2])] + 
   ArcTan[(((p - x1)*Cos[2*yy])/(p+x1))*
    ((Sqrt[(p - x1)^2 + a*p*x1]*Sin[2*yy])/
    (Sqrt[p]*Sqrt[x1]*Sqrt[4 - a*Sin[2*yy]^2]))]) + (((4 - a)*
     p*(p - x1))/(4*(p + x1)*((p - x1)^2 + a*p*x1)))*
  EllipticPi[((p - x1)^2 + a*p*x1)/(p + x1)^2, 2*yy, a/4] /. 
 yy -> 0.3)}, {y, 0.3, Pi/2}, PlotStyle -> {Blue, Dashed}]

enter image description here

Edit: Using Rubi and letting the parameter of EllipticPi staying negative the result of the integral is simpler:

(Sqrt[p]/(2*Sqrt[x1]*Sqrt[(p - x1)^2 + a*p*x1]))*
  • $\begingroup$ Wow! This is really a nice expression. I didn't expect the integral can be reduced to such a compact form. Is it also possible to reduce $J_{-1}$ to your last expression by Mathematica? Sorry, I am not familiar with Rubi. Anyway, your reply is helping me a lot. $\endgroup$ Commented Jun 22, 2022 at 7:56

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