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I'm trying to solve this integral

$$P\int_{-\sqrt 2 a}^{\sqrt2 a} dx \frac{\sqrt{1-(1-(x/a)^2)^2}}{x-y}$$

where P stands for Cauchy's Principal Value. So I code this

    Integrate[Sqrt[1-(1-(x/a)^2)^2]/(x-y),{x,-Sqrt[2]a,Sqrt[2]a},PrincipalValue->True]

But I get a message saying

Warning: contradictory assumption(s). False && 0 < x < 1/4096 encountered

So, how can I avoid this and most important how can I solve this integral

Thanks!

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    $\begingroup$ So is it $\sqrt{2a}$ or $2\sqrt{a}$ or just $2a$ (but then why Sqrt[2a] in the mathematica code)? $\endgroup$ – LLlAMnYP Aug 2 '16 at 13:40
  • $\begingroup$ @LLlAMnYP with out the square root. Sorry for the misunderstood $\endgroup$ – Daniel Aug 2 '16 at 13:41
  • $\begingroup$ I believe you wanted to integrate over the range where the function is real-valued, correct? This would be -Sqrt[2] a, +Sqrt[2] a. $\endgroup$ – LLlAMnYP Aug 2 '16 at 13:43
  • $\begingroup$ @LLlAMnYP you're right $\endgroup$ – Daniel Aug 2 '16 at 13:46
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Integrate[Sqrt[1 - (1 - (x)^2)^2]/(x - y), {x, -Sqrt[2], Sqrt[2]}, PrincipalValue -> True]

Returns

ConditionalExpression[-y (2 Sqrt[2] + Sqrt[2 - y^2] Log[2] + 
2 Sqrt[2 - y^2] Log[y/(2 + Sqrt[4 - 2 y^2])]), 0 < Re[y] <= Sqrt[2] && Im[y] == 0]

or in TeXForm

$$ -y \left(2 \sqrt{2-y^2} \log \left(\frac{y}{\sqrt{4-2 y^2}+2}\right)+\sqrt{2-y^2} \log (2)+2 \sqrt{2}\right) $$

This shouldn't be hard to generalize to arbitrary a.

| improve this answer | |
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You can try supplying some assumptions of your own.

Integrate[
  Sqrt[1 - (1 - (x/a)^2)^2]/(x - y), {x, -Sqrt[2] a, Sqrt[2] a},
  Assumptions -> a > 0 && 0 < y < a Sqrt[2],
  PrincipalValue -> True]

result

If these assumption hold in your problem space, then the above is an answer.

| improve this answer | |
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Clear all variables or start with a fresh kernel

Integrate[
  Sqrt[1 - (1 - (x/a)^2)^2]/(x - y), {x, -Sqrt[2] a, Sqrt[2] a}, 
  PrincipalValue -> True] // FullSimplify

(*  ConditionalExpression[
   (1/(a^2*Sqrt[-2*a^2 + y^2]))*
     (Re[y]*((-Sqrt[2])*a*
             Sqrt[-2*a^2 + y^2] + 
           a*Sqrt[2 - (4*a^2)/y^2]*
             Re[y] - 
           (ArcCos[(Sqrt[2]*a)/y] - 
                I*(Log[Sqrt[2]*a - y] - 
                     Log[-y] - Log[2*a - 
                         Sqrt[2]*y] + 
                     Log[2*a - I*Sqrt[2]*
                           Sqrt[-2*a^2 + y^2]]))*
             (2*a^2 - Re[y]^2) + 
           Pi*(-2*a^2 + Re[y]^2))), 
   a > 1/Sqrt[2] && -2*a^2 < Re[y] < 
       (-Sqrt[2])*a && y == Re[y]]  *)

Assuming[{a > 1/Sqrt[2], -2 a^2 < y < -Sqrt[2] a}, 
 Integrate[
   Sqrt[1 - (1 - (x/a)^2)^2]/(x - y), {x, -Sqrt[2] a, Sqrt[2] a}, 
   PrincipalValue -> True] // FullSimplify]

(*  -((2*y*(Sqrt[2]*a + 
            Sqrt[-2*a^2 + y^2]*
              ArcSin[(Sqrt[2]*a)/y]))/a^2)  *)
| improve this answer | |
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