Recently we learned the Leibniz-Newton formula, that linked the primitive with the integral. I know that Leibniz-Newton can be applied when the function is both integrable and primitivable, but not necessarily continuous. Our teacher drew the circles below. Although I can easily find examples of functions that are only integrable and not primitivable, I cannot find a function that is only primitivable and not integrable.
-Could you find such a function (primitivable, but not integrable)?
By a primitivable function $f$, I mean it admits a primitive $F$ so that $F$ is derivable and $F'=f$. The integration is in Riemann's sense.
I think the crucial point is that there exist functions which are derivable, but whose derivative is not continuous (and possibly unbounded). Therefore, there exist discontinuous (and possibly unbounded) functions which are primitivable. Using the Riemann integral, a discontinuous integrand is not a problem if the number of points of discontinuity is finite and if the function is bounded, but it is a problem if the integrand function is unbounded.
Consider the following case $$ f(x) = \begin{cases} 2 x \, \sin \bigg( \dfrac{1}{x^2} \bigg) - \dfrac{2}{x} \, \cos \bigg( \dfrac{1}{x^2} \bigg)& \text{if $x \neq 0$}\\ 0& \text{if $x = 0$} \end{cases} $$ This function is discontinuous in $x = 0$ and unbounded, since it oscillates wildly around $x = 0$, and it is not Riemann-integrable (let's say on $[-1,1]$ for definiteness). However, it is primitivable, since it is the derivative of the function $$ F(x) = \begin{cases} x^{2} \, \sin \bigg( \dfrac{1}{x^2} \bigg)& \text{if $x \neq 0$}\\ 0& \text{if $x = 0$} \end{cases} $$
So this is an example of a primitivable function which is not Riemann-integrable. Note the crucial role of the unboundedness. I hope this answer may help.