I am facing integrals of the form
$$I=\int d^{2\omega} x \frac{1}{(x-x_1)^{2a_1}}\frac{1}{(x-x_2)^{2a_2}} ... \frac{1}{(x-x_n)^{2a_n}}\tag{1}$$
where the $x$’s are $d$-dimensional vectors in the context of dimensional regularization, with $d=2\omega\approx 4$. I can solve such integrals only when $n=2$, thus I tried to derive a formula to do partial fraction decomposition of the integrand. Here is my ansatz as well as the solution for the coefficients:
$$\begin{align}\frac{1}{(x-x_1)^{2a}(x-x_2)^{2b}}&=\frac{A}{(x-x_1)^{2a}}+\frac{B}{(x-x_2)^{2b}}\\&\overset{!}{=}\frac{A(x-x_2)^{2b}+B(x-x_1)^{2a}}{(x-x_1)^{2a}(x-x_2)^{2b}} \tag{2}\end{align}$$
I can obtain $A$ and $B$ by setting $x=x_1$ or $x_2$ since this must hold for all $x$:
$$\begin{align} A&=\frac{1}{(x_1-x_2)^{2b}} \\ B&=\frac{1}{(x_1-x_2)^{2a}}\end{align}\tag{3}$$
So all in all I get the following decomposition:
$$\frac{1}{(x-x_1)^{2a}(x-x_2)^{2b}}=\frac{1}{(x_1-x_2)^{2b}(x-x_1)^{2a}}+\frac{1}{(x_1-x_2)^{2a}(x-x_2)^{2b}} \tag{4}$$
Armed with this, I can solve my integrals! But two things bother me:
- In the case $x_1=x_2$, I have on the left-hand side of $(4)$:
$$\frac{1}{(x-x_1)^{2(a+b)}}\tag{5}$$
while on the right-hand side I have a divergence! However in my case, $x_1=x_2$ never happens so if this case is not covered by $(4)$, it’s fine. But it makes me doubt that my expression is right.
- More problematic (in my case), taking the limit $x \to x_1$ on both sides does not seems to give the same result. This might be important, since I integrate over all $x$ and thus the point $x=x_1$ is relevant.
So what are your suggestions? Can I use $(4)$? If not, what would be a suitable expression for decomposing the fraction? Don’t forget that I am dealing with $4$-vectors!