# Piecewise Continuous Function does not necessarily have Improper Integrals

## Theorem

Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$.

Let $f$ be a piecewise continuous function:

Let $f$ be a real function defined on a closed interval $\closedint a b$.

$f$ is piecewise continuous if and only if:

there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that:
for all $i \in \set {1, 2, \ldots, n}$, $f$ is continuous on $\openint {x_{i − 1} } {x_i}$.

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 $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \set {1, 2, \ldots, n}$:
$(1): \quad f$ is continuous on $\openint {x_{i − 1} } {x_i}$
$(2): \quad$ the improper integrals $\ds \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ all exist.

## Proof

Consider the function:

$\map f x = \begin{cases} 0 & : x = a \\ \dfrac 1 {x - a} & : x \in \hointl a b \end{cases}$

Since $\dfrac 1 {x - a}$ is continuous on $\openint a b$, $f$ is continuous on $\openint a b$.

Therefore, $f$ is a piecewise continuous function for the (finite) subdivision $\set {a, b}$ of $\closedint a b$.

We now consider whether the improper integral $\ds \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ exists.

Let $c$ be a point in $\openint a b$.

From the definition of improper integral, the existence of $\ds \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ requires that $\ds \lim_{\gamma \mathop \to a^+} \int_\gamma^c \dfrac 1 {x - a} \rd x$ exists.

We have

 $\ds \int_\gamma^c \frac 1 {x - a} \rd x$ $=$ $\ds \bigintlimits {\map \ln {x - a} } \gamma c$ $\ds$ $=$ $\ds \map \ln {c - a} - \map \ln {\gamma - a}$ $\ds$ $=$ $\ds \map \ln {c - a} + \map \ln {\frac 1 {\gamma - a} }$

The last right hand side approaches $\infty$ as $\gamma$ approaches $a$ from above.

So $\ds \int_{a^+}^c \dfrac 1 {x - a} \rd x$ does not exist.

Therefore, $\ds \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ does not exist either.

Accordingly, $f$ is not a piecewise continuous function with improper integrals.

$\blacksquare$