Primitive function/Definite integral/Examples and remarks/Section

Notation

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

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

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

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

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

Example

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

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 ${}r$ between the masses, equals

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

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

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

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

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

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

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

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

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

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

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

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

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

Example

We consider the function

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

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

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