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Mathematica isn't evaluating this integral and I'm wondering if there's a bug in how I'm asking:

the function I want to integrate is

$$f(\theta) = \cos^2(\theta)\sqrt{A \cos^2{\theta} + B \sin^2{\theta} + m^2} $$

And I want to do

$$\int\limits_0^{2\pi} d\theta f(\theta)$$

I'm asking Mathematica to

f = Cos[th]^2 Sqrt[A Cos[th]^2 + B Sin[th]^2 + m^2]

Integrate[f,{th,0,2Pi}, Assumptions->{Element[A,Reals], Element[B,Reals], Element[m,Reals]}] 

and not getting an output, though the indefinite integral spits out something cumbersome. Any ideas?

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  • 2
    $\begingroup$ Please give a complete example as Mathematica code next time (including how you defined f) $\endgroup$ – Szabolcs Mar 12 at 18:24
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Decreasing powers and making an assumption to deal with the reals , one obtains

Integrate[(Cos[2 \[Theta]] + 1)/2*Sqrt[a*(Cos[2 \[Theta]] + 1)/2+b*(-Cos[2 \[Theta]]+1)/2 + 
m^2], {\[Theta], 0, 2*Pi}, Assumptions -> m^2 > RealAbs[a] + RealAbs[b]]

(1/(3 (a - b)))2 (Sqrt[(a + m^2)/(b + m^2)] Sqrt[ b + m^2] ((2 a - b + m^2) EllipticE[(a - b)/( a + m^2)] - (b + m^2) EllipticK[(a - b)/(a + m^2)]) + Sqrt[a + m^2] Sqrt[(b + m^2)/( a + m^2)] ((2 a - b + m^2) EllipticE[(-a + b)/( b + m^2)] - (a + m^2) EllipticK[(-a + b)/(b + m^2)]))

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  • $\begingroup$ The result is numerically confirmed for $\left\{m=\sqrt{5},a=-1,b=-2\right\} $. The answer by Bill Watts produces the same value. $\endgroup$ – user64494 Mar 12 at 21:08
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You can help Mathematica out a little by using a Jacobi elliptic function substitution:

(Cos[t]^2 Sqrt[A (1 - Sin[t]^2) + B Sin[t]^2 + m^2 // Simplify] /. 
 t -> JacobiAmplitude[u, (A - B)/(A + m^2)]) D[JacobiAmplitude[u, (A - B)/(A + m^2)], u]

This yields

   JacobiCN[u, (A - B)/(A + m^2)]^2 JacobiDN[u, (A - B)/(A + m^2)]
   Sqrt[A + m^2 + (-A + B) JacobiSN[u, (A - B)/(A + m^2)]^2]

which means the integral of interest is

expr = Sqrt[A + m^2] Integrate[JacobiCN[u, (A - B)/(A + m^2)]^2
                               JacobiDN[u, (A - B)/(A + m^2)]^2, u];

(expr /. u -> 4 EllipticK[(A - B)/(A + m^2)]) - (expr /. u -> 0) /. 
EllipticE[JacobiAmplitude[4 EllipticK[m_], m_], m_] :> 4 EllipticE[m]

which yields the result

(Sqrt[A + m^2] (8 Sqrt[2] (A + m^2) (2 A - B + m^2) Sqrt[(2 A + 2 m^2)/(A + m^2)] 
 EllipticE[(A - B)/(A + m^2)] - 8 Sqrt[2] (A + m^2) (B + m^2) Sqrt[(2 A + 2 m^2)/(A + m^2)]
 EllipticK[(A - B)/(A + m^2)]))/(6 (A - B) (2 A + 2 m^2))

(Note that I have tacitly assumed A > B > 0.)

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Plotting the function with some values for A, B, and m shows the function to be continuous from 0 to 2 π at least when the values are real. Integrate can handle the indefinite integral.

f = Cos[θ]^2 Sqrt[A Cos[θ]^2 + B Sin[θ]^2 + m^2]

F[θ_] = Integrate[f, θ]//Simplify

int = F[2 π] - F[0] // Simplify
(*(4 Sqrt[A +  m^2] ((2 A - B + m^2) EllipticE[(A - B)/( m^2 + A)] - (B + m^2) EllipticK[(A - B)/(m^2 + A)]))/(3 (A - B))*)

Maybe Integrate cannot tell if there are circumstances where the function might be discontinuous in the interval and would not commit to a definite integral result. The are undoubtedly values you could pick that make the function complex or discontinuous.

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  • $\begingroup$ It would be useful to test the answer numerically, say for $\{m^2=\frac 1 {100}, A=-3,B=1\}$. If I am not mistaken, I obtain $-0.107351+4.2018 i$ from your formula and $0.107356\, +4.2018 i$ numerically integrating. The real parts differ in sign. $\endgroup$ – user64494 Mar 12 at 21:20

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