I'm trying to come up with the Taylor expansion of an integral expression. For simplicity, consider the toy integral
$$ I(\epsilon)=P\int_{-1}^{\epsilon^2}\frac{\epsilon}{x\sqrt{(\epsilon^2-x)(1-x)}}dx, $$
where $P$ denotes the Cauchy principal value, and $\epsilon^2<1$ . The key features of this integral are the following:
- The integrand has a simple pole within the region of integration.
- The upper limit of integration is a zero of the radicand in the denominator.
- In the limit $\epsilon\rightarrow0$ , the upper limit of integration approaches the simple pole in the integrand.
Other features of the integral are inessential. Now, what I would like is a Taylor expansion for $I(\epsilon)$ around $\epsilon=0$ . More precisely, I would like a systematic method for generating successive terms of the Taylor expansion of $I(\epsilon)$ , constructed directly from the integral expression. Of course, for this toy problem I can just do the integral explicitly and then expand:
$$ I(\epsilon)=\ln\left(\frac{1-\epsilon^2}{1+3\epsilon^2-2\epsilon\sqrt{2(1+\epsilon^2)}} \right) = 2\sqrt{2} \, \epsilon + \frac{\sqrt{2}}{3} \epsilon^3 + \mathcal{O}(\epsilon)^5, $$
but I'd like a method applicable to more general cases where the integral can't be written in a closed form. Right now, I only have a method for generating the first term in the Taylor series: noting that the dominant contribution to the integral comes from $x$ near $\epsilon^2$ , change to a new variable $y=\dfrac{x}{\epsilon^2}$ so that
$$ I(\epsilon)=P\int_{-1/\epsilon^2}^1\frac{\mathrm{sgn}(\epsilon)}{y\sqrt{(1-y)(1-\epsilon^2y)}}dy, $$
and thus
$$ I(0^+)=-I(0^-)=P\int_{-\infty}^1\frac{1}{y\sqrt{1-y}}dy=0=I(0). $$
However, this approach doesn't give me a way to systematically extract higher order terms. I'd appreciate any suggestions!
(By the way, this is related to, but different from, the questions asked here: Asymptotic expansion of an integral and A taylor series for an integral with a singularity)