I would like to expand integrals of the form $$ - \int_1^{1-\varepsilon} \frac{f(x)dx}{\sqrt{1-x^2}} $$ where I know that $f(x)$ is well-behaved around $x=1$ but otherwise is a free function here (if you need more info about what $f(x)$ could be, it will be ratios of polynomials and fractional powers of polynomials that all do not have a zero at $x=1$). I don't mind expressing a result using derivatives of $f(x)$.
For example, if $f(x)=1$, the integral is $\frac{\pi}{2} - \arcsin(1-\varepsilon)$ and I can find, from applying $\arcsin$ to $\sin(x-\pi/2) \approx 1 - x^2/2$, an expansion for the $\arcsin$ and show that an approximation to my integal is $\sqrt{2 \varepsilon}$. In the general case, the best thing I can come up with immediately is then pulling out $f(1)$, the value of $f$ at the pole, to get $f(1) \sqrt{2 \varepsilon}$. What if I want better accuracy?
All I could think of is to Taylor expand $f(x)$, integrate each term by hand (Mathematica does the general moments of this inverse square root with the Hypergeometric function, but that doesn't seem too appealing) and then perhaps try to Taylor expand the inverse functions for each term? (To be honest, right now I'm even confused how to expand arcsin at 1 without going to the inverse function...)
In practice, $\varepsilon < 0.1$ and I'm probably okay with a leading- or next-to-leading order expression. I'm curious now how this can be done elegantly and in general! Complex analysis is of course fine.
Thanks!
You can clean this up considerably still. The cleanest way I can think of to look at what you're interested in is
$$I(\varepsilon;f)=\int_0^\varepsilon x^{-1/2} f(x) dx$$
by absorbing the $\sqrt{1+x}$ into the $f(x)$ and moving the singularity (which incidentally is not a pole) to $x=0$. This $f(x)$ is now continuous and does not vanish at $x=0$.
Probably the most straightforward way to do small $\varepsilon$ asymptotics for this is indeed to do a partial Taylor expansion of $f(x)$ at $x=0$, resulting in the asymptotic expansion
$$I \sim \sum_{k=0}^n \int_0^\varepsilon \frac{f^{(k)}(0)}{k!} x^{k-1/2} dx=\sum_{k=0}^n \frac{f^{(k)}(0)}{k!} \frac{\varepsilon^{k+1/2}}{k+1/2}.$$
Now the only messy thing is dealing with derivatives of $f$ when it has this $\sqrt{2+x}$ baked into it, but that's just mechanical.