Small $\delta$ behavior of rational integral I need to do the following integral, and extract the leading small $\delta$ behavior:
$$I(\delta, A)=\int_0^1 dx \, x^n \frac{1}{(x^2 A+\delta)^\nu},\qquad 0< n < \nu,  \enspace A\in\mathbb{R}$$
$n$ and $\nu$ are integers.  I can't set $\delta$ to zero straightaway because the resulting integral $\int_0^1 x^{n-2\nu}$ does not converge because of the singularity near $x=0$.  So my first try is to add and subtract the small $x=0$ behavior:
$$I(\delta, A) = \int_0^1 dx\,x^n \frac{1}{\delta^\nu} +\int_0^1 dx x^n \left[\frac{1}{(x^2 A+\delta)^\nu}-\frac{1}{\delta^\nu}\right]$$
I was pretty sure the first term which gives the leading $\delta\rightarrow 0$ behavior until sample examples on Mathematica revealed that if $n$ is even, the behavior seems to be like $1/\delta^{(2\nu-n-1)/2}$.  I can't understand how the fractional power showed up.
 A: The term that you have subtracted does not make the integral finite for $\delta \to 0$. By comparing the two terms in the bracket of the denomiator, you obtain $A x_* ^2 = \delta$ or $x_* = \sqrt{\delta/A}$ such that the main contribution of the integral is given by $x < x_*$. 
We use this result and introduce the new variables $x= x_* y$ in order to obtain
$$I(\delta, A) = A^{-(n+1)/2} \delta^{(n+1)/2-\nu} 
\int_0^{\sqrt{A/\delta} } \frac{dy \,y^n}{(1+y^2)^\nu}. $$
If $2 \nu > n+1> 0$, we can set the upper limit of integration to $\infty$ and obtain
$$ I(\delta, A) =   A^{-(n+1)/2} \delta^{(n+1)/2-\nu}  \Biggl[\underbrace{\int_0^{\infty } \frac{dy \,y^n}{(1+y^2)^\nu}}_{\text{const}} - \underbrace{\int_{\sqrt{A/\delta}}^{\infty } \frac{dy \,y^n}{(1+y^2)^\nu}}_{O(\delta^{\nu-(n+1)/2})}\Biggr] .$$
If $n+1 \leq 0$, the integral diverges for all $\delta$.
If $n+1 \geq 2 \nu $, the behavior is regular in the sense that the limit $\delta \to 0$ exists and thus
$$I(\delta ,A) = A^{-\nu} \int_0^1 dx\,x^{n-2\nu} + O(\delta) $$ 
