Leibniz integral rule (singular) Definte $I(\epsilon):=\int_{
\epsilon}^1\frac{\,\mathrm{d}x}{\sqrt{x-\epsilon}}$ for $\epsilon<0$
Want to show that $\frac{\mathrm{d}I}{\mathrm{d}\epsilon}=\lim_{\Delta\epsilon\rightarrow0}\frac{I(\epsilon+\Delta\epsilon)-I(\epsilon)}{\Delta\epsilon}$ exists for $\epsilon <0$ without explicitly computing the definite integral.
Is there a singular version of Leibniz integral rule? Thanks.

By the Leibniz integral rule (really a version of the chain rule), we have
$$\frac{dI}{d\epsilon} = \int_\epsilon ^1 \frac{\partial}{\partial \epsilon}\left(\frac{1}{\sqrt{x-\epsilon}}\right)dx - \frac{1}{1-\epsilon}\frac{d\epsilon}{d\epsilon} = \int_\epsilon ^1 \frac{\partial}{\partial \epsilon}\left(\frac{1}{\sqrt{x-\epsilon}}\right)dx - \frac{1}{1-\epsilon}. $$
From here, just calculate the integral and take the limit as $\epsilon\to0^-$.
