# Why do I get contradictory assumptions and how can I avoid them?

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!

• So is it $\sqrt{2a}$ or $2\sqrt{a}$ or just $2a$ (but then why Sqrt[2a] in the mathematica code)? Aug 2, 2016 at 13:40
• @LLlAMnYP with out the square root. Sorry for the misunderstood Aug 2, 2016 at 13:41
• 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. Aug 2, 2016 at 13:43
• @LLlAMnYP you're right Aug 2, 2016 at 13:46

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.

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]


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

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)  *)