How is it possible that the function $F(x)$ defined by :
$$
F(x)=\left\{\begin{array}{ll} x\sqrt{x}\sin\frac{1}{x}, & x> 0 \\
0, & x=0\end{array}\right.
$$
$$
F'(x)=\left\{\begin{array}{ll} \frac{3}{2}\sqrt{x}\sin\frac{1}{x}-\frac{1}{\sqrt{x}}\cos\frac{1}{x}, & x> 0 \\
0, & x=0\end{array}\right.
$$
has a derivative, which is not Riemann- integrable on any interval $[0,|b|]$, the function is continuous and every continuous function should have a R-Integral?
Best Answer
Lebesgue's criterion for Riemann integrability states that a function $f: [a, b] \longrightarrow \Bbb R$ is Riemann integrable if and only if it is bounded and continuous almost everywhere on $[a,b]$.
In the present case, $|F(x)| \leq |b| \sqrt{|b|}$ on $[0, |b|]$ and $F$ is also continuous on $[0, |b|]$, so by Lebesgue's criterion $F$ is Riemann integrable.
On the other hand, $F'$ is not bounded in any neighborhood of $x = 0$, so by Lebesgue's criterion it is not Riemann integrable.
The point of this example is that you can't always integrate a function just because that function is a derivative of some other function.