I am dealing with an integral in a limit of the following shape:


$$\lim\_{\epsilon \to 0} \int\_0^{\frac{\pi}{2}} dx \frac{2 \epsilon}{1-(1-\epsilon^2)\sin^2(x)}$$


Formally, assuming that $x=\arcsin(\frac{1}{\sqrt{1-\epsilon^2}})$ is a possible value on the integration path, we obtain a non zero residue (independent of $\epsilon$) and therefore a residual contribution to the integral.


However, we only have a residual contribution if the singularity is actually on the integration path. Since $\sin(x)$ on the interval $[ 0,\frac{\pi}{2} ]$ strictly increases from $0$ to $1$, the only possibility for the denominator to vanish in this case is when $x=\frac{\pi}{2}$ (so that $\sin(x)=1$) and $\epsilon=0$. That would correspond to a singularity on the upper boundary of the integration interval.


Now in the given limit I can imagine two ways of argumentation:


1. In the limit $\epsilon \rightarrow 0$ the singularity approaches the upper border of the integration interval. So we have to avoid the singularity using a quarter of a circle into the complex plane, which would give us a residual contribution corresponding to $\frac{1}{4}$ of the residue. The Principal Part of the integral then simply vanishes, since $\epsilon=0$ sets the integrand to zero. (But the residual part still contributes and therefore the integral is not zero.)
2. For any arbitrarily small real $\epsilon$ there is no singularity on the integration interval. The singularity only seems to appear on the border of the interval if the limit has already been carried out and $\epsilon=0$. But in this case the integrand as such is zero, so the integration can yield nothing else but zero. (No additional contribution from the residual part).


Which of the argumentations above is correct? Or is there another argumentation to resolve this? Does the residue contribute or does it not?



Residue at an integration border in case of a limit?
Of course, you cannot put $\epsilon=0$ under the integral.
You must evaluate it for $\epsilon\neq 0$ and then pass to the limit.


The correct way to find this integral is to spread it from $-\pi/2$ to $\pi/2$,
using that your function is even, and then make the change of the variable
$z=\exp(ix)$. This will reduce to an integral of a rational function over a closed
contour, $|z|=1$ (because $\sin^2$ has period $\pi$) which is then easy to evaluate.


