Suppose we have the following piecewise defined function for $v,s >0$: $$ \begin{align} f(v,s) = \begin{cases} v/s \ &\text{for} \ v < s \\ 1 \ &\text{for} \ v \geq s \end{cases} \end{align} $$ And the following function integrating over the previously mentioned function: $$ F(v) = \int_{\underline{s}}^{\overline{s}} f(v,s) \ ds $$ Where $\underline{s},\ \overline{s}$ represent some finite bounds to the integral with $\underline{s} < \overline{s}$. While I would say that $f(v,s)$ is not differentiable at the point $v=s$, I somehow can not wrap my head around $F(v)$ and would really appreciate any help.
My questions are basically:
- Is $F(v)$ differentiable in $v$?
- If yes, where?
- If not, why not?
Thank you guys!
Have you heard of the Fundamental Theorem of Calculus? The integrand is continuous.