1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
    Jun 21 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. $\endgroup$ Jun 21 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? $\endgroup$ Jun 21 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. $\endgroup$ Jun 21 at 7:31

1 Answer 1

2
$\begingroup$

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])] + 
  ArcTan[(((p-x1)*Cos[2*y])/(p+x1))*((Sqrt[(p-x1)^2+
  a*p*x1]*Sin[2*y])/(Sqrt[p]*Sqrt[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]))*
 ArcTan[(Sqrt[p^2-(2-a)*p*x1+x1^2]*Sin[2*y])/(Sqrt[p]*Sqrt[x1]*
  Sqrt[4-a*Sin[2*y]^2])]- 
 (1/(2*(p-x1)))*(EllipticF[2*y,a/4]-((p+x1)/(2*x1))*
 EllipticPi[-((p-x1)^2/(4*p*x1)),2*y,a/4])
$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Jun 22 at 7:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.