# Ngô Quốc Anh

## April 10, 2015

### Existence of antiderivative of discontinuous functions

Filed under: Uncategorized — Ngô Quốc Anh @ 0:17

It is well-known that every continuous functions admits antiderivative. In this note, we show how to prove existence of antiderivative of some discontinuous functions.

A typical example if the following function

$f(x)=\begin{cases}\sin \frac 1x, & \text{ if } x \ne 0,\\ 0, & \text{ if }x=0.\end{cases}$

By taking to different sequences $x_k = 1/(2k\pi)$ and $y_k = 1/(\pi/2 + 2k\pi)$ we immediately see that $f$ is discontinuous at $x=0$. However, we will show that $f$ admits $F$ as its antiderivative.

To this end, we first consider the following function

$G(x)=\begin{cases}x^2\cos \frac 1x, & \text{ if } x \ne 0,\\ 0, & \text{ if }x=0.\end{cases}$

First we show that $G$ is differentiable. Clearly whenever $x \ne 0$, we obtain

$\displaystyle G'(x)=\sin \frac 1x + 2x \cos \frac 1x.$

At $x=0$, we easily check that

$\displaystyle G'(0)=\lim_{x \to 0} \frac{G(x)-G(0)}{x} = \lim_{x \to 0} x\cos \frac 1x = 0$

thanks to $|\cos \frac 1x| \leqslant 1$. Thus, we have just shown that $G$ is differentiable with derivative

$G'(x)=\begin{cases} \sin \frac 1x + 2x \cos \frac 1x, & \text{ if } x \ne 0,\\ 0, & \text{ if }x=0.\end{cases}$

Then we decompose $G'$ into two parts as follows $G' = f + h$ with

$h(x)=\begin{cases} 2x \cos \frac 1x, & \text{ if } x \ne 0,\\ 0, & \text{ if }x=0.\end{cases}$

Observe that $h$ is continuous, hence there exists its antiderivative $H$, i.e. $H'(x)=h(x)$ for any $x \in \mathbb R$. Hence, in view of the decomposition above, there holds

$\displaystyle f = (G-H)'.$

In other words, we have just proved that $G-H$ is the antiderivative of the function $f$.

Remark: In this problem, the antiderivative of $f$ cannot be written in terms of elementary functions, so as the function $H$. The only point in this problem is that we convert the discontinuous function $f$ (at $late x=0$) to a continuous function $h$ so that its antiderivative does naturally exist.