As part of an exercise in a grad course on "mathematical methods" (always such a helpful name), I've been asked to evaluate $I=\int_0^{1/2}{(x^2-x+c)^{-2}dx}$ as a Hadamard finite part integral for $0 \leq c<1/4$. The integrand is positive, so $I$ should be positive too, right?
Whatever methods I use, the closest I've gotten is this. When $c > 1/4$, $I = 4(4c-1)^{-3/2}\operatorname{arctan}{(4c-1)^{-1/2}}+1/{c(4c-1)}$, which is real. When $c < 0$, the identity $\operatorname{arctan} iz = i \operatorname{artanh} z$ again gives a real value, namely $I = 4(1-4c)^{-3/2}\operatorname{arctan}{(1-4c)^{-1/2}}+1/{c(4c-1)}$. (Maybe the first term needs a - sign due to me misusing powers of i, but I don't think so.) As I understand it, for $x > 1$ $\operatorname{artanh}x$ has multiple complex values, all with the same real part and in conjugate pairs. So can you "average out" the imaginary part? Well, even if you do, the result is negative for small positive $c$, and tends to $-\infty$ as $c \to 0$, even though we should have $I=\int_0^{1/2}{(x^2-x)^{-2}dx}=+\infty$.