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.

See also: Wiki

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