I am in trouble with the following integral:

$$rtartaruga(r)=\int \frac{\Sigma(r') }{\Delta(r') \sqrt{\Delta th}} d r'$$ Where: $$\Delta(r) =\left(a^2+r^2\right) \left(\frac{r^2}{l^2}+1\right)-2 m r$$ $$\Sigma(r) =\sqrt{\text{$\Delta $th} \left(a^2+r^2\right)^2-a^2 \Delta \sin ^2(\theta )}$$ And $\Delta th$ is a constant in $r$: $\text{$\Delta $th}=1-\frac{a^2 \cos ^2(\theta )}{l^2}$.

In the following region of parameters: $|a|<l$ and $m>m_c$ (where $m_c$ is a certain positive constant $\Delta (r)$ has two positive roots and the integrand is divergent in these points.

I would really like to have an analytical solution for the above integral so i try to evaluate it symbolically:

\[CapitalSigma] = 
  Sqrt[(r^2 + a^2)^2 \[CapitalDelta]th - 
    a^2 \[CapitalDelta] Sin[\[Theta]]^2];

\[CapitalDelta] = (r^2 + a^2) (1 + r^2/l^2) - 2 m r;

\[CapitalDelta]th = 1 - a^2/l^2 Cos[\[Theta]]^2;

 mc = l/(3 Sqrt[6]) (Sqrt[(1 + a^2/l^2) + 12/l^2 a^2] + 2 a^2/l^2 + 2)*
   Sqrt[(Sqrt[(1 + a^2/l^2) + 12/l^2 a^2] - a^2/l^2 - 1)];


 rtartaruga[r_] = 
  r, Assumptions -> {{r, m, \[Theta], a, l} \[Element] Reals, 
    Abs[a] < l, m > mc, r > 0, l > 0, \[Theta] >= 0, \[Theta] <= Pi}, 
  PrincipalValue -> True]

The output is a very long one and includes special functions. When i try to evaluate it at some point:

a = 1/2;
l = 2;
m = 1;
\[Theta] = Pi/4;
Clear[a, l, m, \[Theta]]  

I get:

-0.274199 - 0.47465 I

But the result has to be real as the integrand!

I know that there are other similar questions on this issue, and i understand that the problem is about branch cuts on the complex plane, but practically i can't find a way to get out my problem (if it's possible).

May someone help me please?

Thanks in advance!

  • $\begingroup$ If I am reading this correctly, it is an antiderivative as opposed to a definite integral. So the result need not be real valued. Also the two options being given will be ignored (possibly the Assumptions will have a minor effect, I'm not absolutely certain about that one). $\endgroup$ – Daniel Lichtblau Jan 17 at 19:02
  • $\begingroup$ Hi! First of all thank you for your reply.The indefinite integral of a real function could has a constant imaginary part, right? The problem is that this is not my case:ln[81]:= a = 1/2; l = 2; m = 1; [Theta] = Pi/4; N[rtartaruga[[0.4], 10] N[rtartaruga[[0.7], 10] N[rtartaruga[[5], 10] N[rtartaruga[[10], 10] N[rtartaruga[20], 10] Clear[a, l, m, [Theta]] Out[85]= -0.274199 - 0.47465 I Out[86]= -0.166399 - 0.492226 I Out[87]= 4.822826746 + 3.423249825 I Out[88]= -2.238523542 + 0.*10^-10 I Out[89]= -2.045605814 + 0.*10^-10 I $\endgroup$ – SuperBaba Jan 18 at 10:25
  • $\begingroup$ Due to branch cuts it could be piecewise constant actually. $\endgroup$ – Daniel Lichtblau Jan 18 at 16:50
  • $\begingroup$ Thank you very much, now it's all clear to me! $\endgroup$ – SuperBaba Jan 26 at 13:57

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