Primitive function and Riemann-Integral/1 over x sin 1 over x^2/Example

We consider the function

$f\colon \mathbb {R} \longrightarrow \mathbb {R} ,t\longmapsto f(t),$

given by

${}f(t):={\begin{cases}0{\text{ for }}t=0,\\{\frac {1}{t}}\sin {\frac {1}{t^{2}}}{\text{ for }}t\neq 0\,.\end{cases}}\,$

This function is not Riemann-integrable, because it it neither bounded from above nor from below. Hence, there exist no upper step functions for ${}f$. However, ${}f$ still has a primitive function. To see this, we consider the function

${}H(t):={\begin{cases}0{\text{ for }}t=0,\\{\frac {t^{2}}{2}}\cos {\frac {1}{t^{2}}}{\text{ for }}t\neq 0\,.\end{cases}}\,$

This function is differentiable. For ${}t\neq 0$, the derivative is

${}H'(t)=t\cos {\frac {1}{t^{2}}}+{\frac {1}{t}}\sin {\frac {1}{t^{2}}}\,.$

For ${}t=0$, the difference quotient is

${}{\frac {{\frac {h^{2}}{2}}\cos {\frac {1}{h^{2}}}}{h}}={\frac {h}{2}}\cos {\frac {1}{h^{2}}}\,.$

For ${}h\mapsto 0$, the limit exists and equals ${}0$, so that ${}H$ is differentiable everywhere (but not continuously differentiable). The first summand in ${}H'$ is continuous, and therefore, due to fact, it has a primitive function ${}G$. Hence ${}H-G$ is a primitive function for ${}f$. This follows for ${}t\neq 0$ from the explicit derivative and for ${}t=0$ from

${}H'(0)-G'(0)=0-0=0\,.$