# Primitive function/Definite integral/Examples and remarks/Section

## Notation

Let ${\displaystyle {}I}$ denote a real interval, and

${\displaystyle F\colon I\longrightarrow \mathbb {R} }$

a primitive function for a function ${\displaystyle {}f\colon I\rightarrow \mathbb {R} }$. Suppose that ${\displaystyle {}a,b\in I}$. Then one sets

${\displaystyle {}F|_{a}^{b}:=F(b)-F(a)=\int _{a}^{b}f(t)\,dt\,.}$

This notation is basically used for computations, in particular, when we want to determine definite integrals.

Using known results about the derivatives of differentiable functions, we obtain a list of primitive functions for some important functions. In general however, it is difficult to find a primitive function.

The primitive function of ${\displaystyle {}x^{a}}$, where ${\displaystyle {}x\in \mathbb {R} _{+}}$ and ${\displaystyle {}a\in \mathbb {R} }$, ${\displaystyle {}a\neq -1}$, is ${\displaystyle {}{\frac {1}{a+1}}x^{a+1}}$.

## Example

Suppose that the distance between two masses (thought of as mass points) ${\displaystyle {}M}$ and ${\displaystyle {}m}$ is ${\displaystyle {}R_{0}}$. Because of gravitation, this system contains a certain potential energy. How is this potential energy changing, when we move these masses to a distance ${\displaystyle {}R_{1}\geq R_{0}}$?

The needed energy is force times path, where the force itself depends on the distance between the masses. Due to the gravitation law, the force, given the distance ${\displaystyle {}r}$ between the masses, equals

${\displaystyle {}F(r)=\gamma {\frac {Mm}{r^{2}}}\,,}$

where ${\displaystyle {}\gamma }$ denotes the constant of gravitation. Therefore, the energy needed to increase the distance from ${\displaystyle {}R_{0}}$ to ${\displaystyle {}R_{1}}$, equals

${\displaystyle {}E=\int _{R_{0}}^{R_{1}}\gamma {\frac {Mm}{r^{2}}}\,dr=\gamma Mm\int _{R_{0}}^{R_{1}}{\frac {1}{r^{2}}}\,dr=\gamma Mm{\left(-{\frac {1}{r}}|_{R_{0}}^{R_{1}}\right)}=\gamma Mm{\left({\frac {1}{R_{0}}}-{\frac {1}{R_{1}}}\right)}\,.}$

Hence it is possible to assign a value to the difference between the potential energies for the two distances ${\displaystyle {}R_{0}}$ and ${\displaystyle {}R_{1}}$, though it is not possible to assign an absolute value to the potential energy for a given distance.

The primitive function of the function ${\displaystyle {}{\frac {1}{x}}}$ is the natural logarithm.

The primitive function of the exponential function is the exponential function itself.

The primitive function of ${\displaystyle {}\sin x}$ is ${\displaystyle {}-\cos x}$, the primitive function of ${\displaystyle {}\cos x}$ is ${\displaystyle {}\sin x}$.

The primitive function of ${\displaystyle {}{\frac {1}{1+x^{2}}}}$ is ${\displaystyle {}\arctan x}$, due to fact.

The primitive function of ${\displaystyle {}{\frac {1}{1-x^{2}}}}$ (for ${\displaystyle {}x\in {]{-1},1[}}$) is ${\displaystyle {}{\frac {1}{2}}\ln {\frac {1+x}{1-x}}}$, because we have

{\displaystyle {}{\begin{aligned}{\left({\frac {1}{2}}\cdot \ln {\frac {1+x}{1-x}}\right)}^{\prime }&={\frac {1}{2}}\cdot {\frac {1-x}{1+x}}\cdot {\frac {(1-x)+(1+x)}{(1-x)^{2}}}\\&={\frac {1}{2}}\cdot {\frac {2}{(1+x)(1-x)}}\\&={\frac {1}{(1-x^{2})}}.\end{aligned}}}

Caution! Integration rules are only applicable for functions, which are defined on the whole interval. In particular, the following is not true

${\displaystyle {}\int _{-a}^{a}{\frac {dt}{t^{2}}}\,dt=-{\frac {1}{x}}|_{-a}^{a}=-{\frac {1}{a}}-{\frac {1}{a}}=-{\frac {2}{a}}\,,}$

since we integrate over a point where the function is not defined.

## Example

We consider the function

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

given by

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

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

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

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

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

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

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

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

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