Piecewise Continuous Function does not necessarily have Improper Integrals

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Piecewise_Continuous_Function_does_not_necessarily_have_Improper_Integrals&oldid=348004"