I'm evaluating an complex integral and this is a part of it: $$\lim_{\substack{\epsilon\to 0 \\ \delta\to 0}}\,\int_{-\delta}^{\delta}\frac{\left(\epsilon e^{\mathrm{i}\varphi}\right)^{(s-1)}e^{\epsilon e^{\mathrm{i}\varphi}}}{1-e^{\epsilon e^{\mathrm{i}\varphi}}}\mathrm{i} \epsilon e^{\mathrm{i}\varphi}\,\mathrm{d}\varphi$$ I think it is very easy to solve this. I would just plug in $\epsilon$ and this integral, will converge fast to zero. The problem is that we also have $\delta$. Let's ignore the integral and just look at the function, which it contains. There is no way that a $\delta$-term is going to infinity (so we could plug in $\epsilon$ without missing something).
But does this also hold for the integral of function, so we can be sure a $\delta$ term does not go to infinity?
Generally my question would be: Does the integral of a function, which has no singularities (the function), also have no singularities?