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

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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

Failed to parse (unknown function "\begin{cases}"): {\displaystyle {{}} H(t) := \begin{cases} 0 \text{ for } t <table class="metadata plainlinks ambox ambox-notice" style=""> <tr> <td class="mbox-image"><div style="width: 52px;"> [[File:Wikiversity logo 2017.svg|50px|link=]]</div></td> <td class="mbox-text" style=""> '''[[m:Soft redirect|Soft redirect]]'''<br />This page can be found at <span id="SoftRedirect">[[mw:Help:Magic words#Other]]</span>. </td> </tr> </table>[[Category:Wikiversity soft redirects|Primitive function and Riemann-Integral/1 over x sin 1 over x^2/Example]] __NOINDEX__ 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

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

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