I wish to compute Cauchy principal value integral $$P\int_0^1\frac{f(x)}{(x-y)^2}dx$$ numerically, but using the PV prescription $$P\frac{1}{x^2}=\frac{1}{2}(\frac{1}{(x+i \epsilon)^2}+\frac{1}{(x-i\epsilon)^2})$$ could introduce extra imaginary part into the integral while I want it to remain real.
I also tried $$P\int_0^1\frac{f(x)}{(x-y)^2}dx=\int_0^{y-\epsilon}\frac{f(x)}{(x-y)^2}dx+\int_{y+\epsilon}^1\frac{f(x)}{(x-y)^2}dx-\frac{2}{\epsilon}f(y)$$ which gives relatively good result but numerically it's large minus large so I have to keep the intergration really accurate, and that's hard for me.
Another way I read in some papers is $$P\int_0^1\frac{f(x)}{(x-y)^2}dx=\int_0^1\frac{f(x)-f(y)-(x-y)f'(y)-\dots}{(x-y)^2}dx$$ but I can't produce the same result with the former one using this method, I don't know where I did wrong.
Does anyone know any good PV prescription in this case?
The prescription to use (and works in all cases) is the definition $$P\int_{0}^1 \frac{f(x)}{(x-y)^2}dx=\lim_{\epsilon \to 0} \left[\int_{0}^{y-\epsilon}\frac{f(x)}{(x-y)^2}dx +\int_{ y+\epsilon}^{1} \frac{f(x)}{(x-y)^2}dx\right],$$ i.e., the singularity at $x=y$ is approached symmetrically from both sides.
In you case, the principal value is divergent unless $f(y) =0$.