Piecewise Continuous Function does not necessarily have Improper Integrals
Theorem
Let $f$ be a real function defined on a closed interval $\left[{a \,.\,.\, b}\right]$, $a < b$.
Let $f$ be a piecewise continuous function:
Let $f$ be a real function defined on a closed interval $\left[{a \,.\,.\, b}\right]$.
$f$ is piecewise continuous if and only if:
- there exists a finite subdivision $\left\{ {x_0, x_1, \ldots, x_n}\right\}$ of $\left[{a \,.\,.\, b}\right]$, where $x_0 = a$ and $x_n = b$, such that:
- for all $i \in \left\{ {1, 2, \ldots, n}\right\}$, $f$ is continuous on $\left({x_{i − 1} \,.\,.\, x_i}\right)$.
Then it is not necessarily the case that $f$ is a piecewise continuous function with improper integrals:
$f$ is piecewise continuous with improper integrals if and only if:
- there exists a finite subdivision $\left\{{x_0, x_1, \ldots, x_n}\right\}$ of $\left[{a \,.\,.\, b}\right]$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \left\{ {1, 2, \ldots, n}\right\}$:
- $(1): \quad f$ is continuous on $\left({x_{i − 1} \,.\,.\, x_i}\right)$
- $(2): \quad$ the improper integrals $\displaystyle \int_{ {x_{i - 1} }^+}^{ {x_i}^-} f \left({x}\right) \rd x$ all exist.
Proof
Consider the function:
- $f \left({x}\right) = \begin{cases} 0 & : x = a \\ \dfrac 1 {x - a} & : x \in \left({a \,.\,.\, b}\right] \end{cases}$
Since $\dfrac 1 {x - a}$ is continuous on $\left({a \,.\,.\, b}\right)$, $f$ is continuous on $\left({a \,.\,.\, b}\right)$.
Therefore, $f$ is a piecewise continuous function for the (finite) subdivision $\left\{{a, b}\right\}$ of $\left[{a \,.\,.\, b}\right]$.
We now consider whether the improper integral $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ exists.
Let $c$ be a point in $\left({a \,.\,.\, b}\right)$.
From the definition of improper integral, the existence of $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ requires that $\displaystyle \lim_{\gamma \mathop \to a^+} \int_\gamma^c \dfrac 1 {x - a} \rd x$ exists.
We have
\(\displaystyle \int_\gamma^c \frac 1 {x - a} \rd x\) | \(=\) | \(\displaystyle \ln \left({x - a}\right) \vert _\gamma^c\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \ln \left({c - a}\right) - \ln \left({\gamma - a}\right)\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \ln \left({c - a}\right) + \ln \left({\frac 1 {\gamma - a} }\right)\) |
The last right hand side approaches $\infty$ as $\gamma$ approaches $a$ from above.
So $\displaystyle \int_{a^+}^c \dfrac 1 {x - a} \rd x$ does not exist.
Therefore, $\displaystyle \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ does not exist either.
Accordingly, $f$ is not a piecewise continuous function with improper integrals.
$\blacksquare$