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!
Sqrt[2a]
in the mathematica code)? $\endgroup$-Sqrt[2] a, +Sqrt[2] a
. $\endgroup$