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I am trying to evaluate the following integral, but the result I got is imaginary part. Do you know if there is a way to get a better evaluation of difficult integrals?

In S = Integrate[1/(Pi*Sqrt[(1/a) - (1/x)]*Sqrt[(1/x) - (1/b)]*(1 - B*x^2)), {x, a,b}] 

where, a, b >0, b>a.

If you have any suggestions on how to evaluate this, let me know. Thank you in advance.

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  • $\begingroup$ Should B be b? $\endgroup$
    – N.J.Evans
    Commented Sep 7, 2018 at 13:14
  • $\begingroup$ No, they are different. $\endgroup$ Commented Sep 7, 2018 at 13:16

1 Answer 1

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If you restrict B to be real

S = Assuming[b > a > 0 && Element[B, Reals], Integrate[
    1/(Pi*Sqrt[(1/a) - (1/x)]*Sqrt[(1/x) - (1/b)]*(1 - B*x^2)),
    {x, a, b}] // Simplify]

(* ConditionalExpression[(Sqrt[
    a b] (Sqrt[-(-1 + a Sqrt[B]) (-1 + b Sqrt[B])] Log[-1 - a Sqrt[B]] + 
      Sqrt[-(1 + a Sqrt[B]) (1 + b Sqrt[B])] Log[1 - a Sqrt[B]] - 
      Sqrt[-(1 + a Sqrt[B]) (1 + b Sqrt[B])] Log[-1 + b Sqrt[B]] - 
      Sqrt[-(-1 + a Sqrt[B]) (-1 + b Sqrt[B])] Log[1 + b Sqrt[B]] - 
      Sqrt[-(1 + a Sqrt[B]) (1 + b Sqrt[B])] Log[Sqrt[B] - a B] - 
      Sqrt[-(-1 + a Sqrt[B]) (-1 + b Sqrt[B])] Log[Sqrt[B] + a B] + 
      Sqrt[-(1 + a Sqrt[B]) (1 + b Sqrt[B])] Log[Sqrt[B] - b B] + 
      Sqrt[-(-1 + a Sqrt[B]) (-1 + b Sqrt[B])]
        Log[Sqrt[B] + b B]))/(2 Sqrt[-(-1 + a Sqrt[B]) (-1 + b Sqrt[B])]
     Sqrt[-(1 + a Sqrt[B]) (1 + b Sqrt[B])] Sqrt[
    B] π), ((B > 0 && 1/b^2 >= B) || B < 0 || 
    1/a^2 <= B) && (a >= Re[1/Sqrt[B]] || b <= Re[1/Sqrt[B]] || 
    Sqrt[B] ∉ Reals)] *)

Or for the more restrictive case of B > 0

S = Assuming[b > a > 0 && B > 0, Integrate[
    1/(Pi*Sqrt[(1/a) - (1/x)]*Sqrt[(1/x) - (1/b)]*(1 - B*x^2)),
    {x, a, b}] // Simplify]

(* ConditionalExpression[(
 Sqrt[a b] (-Sqrt[1 - (a + b) Sqrt[B] + a b B] + Sqrt[
    1 + (a + b) Sqrt[B] + a b B]))/(2 Sqrt[B (-1 + a^2 B) (-1 + b^2 B)]), 
 1/b^2 > B] *)

EDIT: Example

S /. {a -> 1, b -> 2, B -> 1/8} // FullSimplify

(* 4 Sqrt[2/7 (5 - Sqrt[7])] *)

% // N

(* 3.28059 *)
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