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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?

HFPIs aren't actually on the syllabus, although it's briefly mentioned for a different example that Hadamard regularization gives the same answer as doing it with generalized functions. Well, I've tried every method I can think of - every substitution, every excuse for which complex value of a logarithm to take (depending on how you tackle it, you get artanh with an argument >1 in the result), some methods based on generalized functions (which my assignment otherwise doesn't work with), some which rely on Wikipedia's account of Hadamard regularization and some which do neither - but whatever I do, the answer isn't a positive real number.

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 $I=\int_0^{1/2}{(x^2-x)^{-2}dx}=+\infty$.

So what's going on here?

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