Suppose we have a function $f:[b,d]\to\mathbb{R}$ that has a jump discontinuity at some point $b<a<d$ and continuous otherwise. Define $F(x) = \int_b^x$ Then, from other answers on the site, I know that

  • the jump discontinuity does not affect the value of $F(x)$
  • $F(x)$ is continuous (since the jump discontinuity did not affect the value of $F(x)$, and $F(x)$ would be continuous/differentiable otherwise)
  • $F(x)$ is not differentiable (at $a$)

When i think of a function not being differentiable I normally imagine a kink or a jump.

But I don't see how/why $F(x)$ would have a kink/jump (if that is the problem for differentiability). It can't jump because the discontinuity doesn't affect the value of the integral. For a kink I have no intuition

I tried computing an example where $$g:[-2,3]\to \mathbb{R} = \begin{cases} 1 & x\not = 0\\ 5 & x=0\end{cases}$$

Then we have, for$G(x)\equiv \int_{-2}^xg$ , that $$ G(x) = \begin{cases} \int_{-2}^{x} 1 = (x) +2 =2+x & x< 0\\ \\lim_{\epsilon\to 0} \bigg(\int_{-2}^{-\epsilon}1ds + \int_{\epsilon}^x 1 ds = -\epsilon +2 + x-\epsilon = 2+x-2\epsilon\bigg ) = 2+x & x\geq 0 \end{cases} $$

So the integral is $G(x) = 2+x$, which is continuous and differentiable. So I am messing up the calculation somewhere, but I don't see where? (perhaps I cannot write the integral as the limit as $\epsilon\to 0$, precisely because the discontinuity?)

  • $\begingroup$ No, $F$ is differentiable at $a$ (from the right). $\endgroup$
    – zhw.
    May 19, 2018 at 17:06
  • $\begingroup$ The problem is that $F(x)$ is only defined for $x \geq a$. Differentiability may not be defined at the endpoint of a half-open interval or the endpoints of a closed interval. Typically, differentiable at a point means differentiability from the right and from the left. $\endgroup$
    – D.B.
    May 19, 2018 at 17:08
  • $\begingroup$ You could say that $F(x)$ is differentiable on $(a,x)$. $\endgroup$
    – D.B.
    May 19, 2018 at 17:09
  • 1
    $\begingroup$ To gain intuition for why there is a kink, just compute $F(x)$ with (for example) $f(x)=0$ for $x \le a$ and $f(x)=1$ for $x>a$. $\endgroup$ May 19, 2018 at 17:22
  • 1
    $\begingroup$ The specific value assigned at $a$ does not affect the value of $F$, but the amount $f$ jumps by affects $F.$ $\endgroup$
    – zhw.
    May 19, 2018 at 17:37

2 Answers 2


In my mind "the" example of a kink is

$F(t) = 0$ for $t \leq 0$

$F(t) = t$ for $t \geq 0$.

This is zero for a while and then suddenly starts to increase. It is also clearly the integral of

$f(t) = 0$ for $t \leq 0$

$f(t) = 1$ for $t > 0$.

This is a source of kinks in higher derivatives too, because you can easily take e.g.

$G(t) = 0$ for $t \leq 0$

$G(t) = t^3$ for $t \geq 0$.

Clearly continuous. And 1st derivative is continuous. Second derivative is continuous but with a kink.


This could be clarified by using a physical interpretation for integration. In particular, if you integrate velocity, you get displacement. A discontinuity of the integrand would then by a point at which the velocity suddenly changes - which would lead to a kink in the object's path.

For instance, suppose you have a function $$f(x)=\begin{cases} -1 &\text{if }x\leq 0 \\ 1 & \text{if }x> 0 \end{cases}$$ If you integrate this, you get a function $F(x)=|x|+C$. You can check this by direct computation. This has a kink at $x=0$, which is exactly where the integrand switches from negative to positive. That is, if you image this as the position of an object, before $x=0$, the object is moving downwards at a constant rate, and after $x=0$, it is moving upwards, which causes the position to not be differentiable at $x=0$.

This is all fairly evident from a graph of $|x|$. Note that its derivative is $-1$ for all negative values and $1$ for all positive values and a change in the direction occurs where the discontinuity in the integrand was.

enter image description here

In response to the edit to your question: This argument does not apply to removable discontinuities like the $$f(x)=\begin{cases}1 & \text{if }x\neq 0 \\ 5 &\text{if }x=0\end{cases}$$ function you give. It is only a property of jump discontinuities. Your integration of the function is correct. For physical interpretation, it is completely inconsequential if an object tries to move with a velocity of $5$ for absolutely zero time - there needs to be some interval (or, more precisely, a set of positive measure) on which the integral has change in order to reflect any change in the result. From the mathematical side, if you've defined integration by Riemann sums, note that the contribution to the integral of the one place where $f$ takes the value $5$ becomes decreasingly large as you use finer and finer partitions, thus does not matter at all to the integral.

  • $\begingroup$ Among other issues, I had my discontinuities mixed up. Thanks for the thorough answer. $\endgroup$
    – user106860
    May 19, 2018 at 18:09

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