Let the total variation be defined as:

$ \displaystyle{ TV=\int_{-\infty}^{\infty} \left |\frac{\partial f}{\partial x} \right | dx }. $

For the function,

$ f(x) = \cases{ 1 & $x<0$ \\ \sin \left( \pi \,x \right) & $0\leq x\leq 1$ \\ 2 & $x>1$\\ }, $

Maple software produces a result of $TV=\mathbf{2}$. But I think it must be $\mathbf{5}$. Maple is missing the left and right jump in the function and only integrating the $\sin(\pi x)$ !.

Am I missing something here? I think Maple will not do such a mistake, so I do not understand what is wrong in my interpretation. If someone understands the mistake then please help me out.

Here is an image of the function $f(x)$: function distribution

  • $\begingroup$ Are you familiar with the concept of distributions? $\endgroup$ – Jacob Manaker Feb 28 '17 at 21:15
  • $\begingroup$ @DougM Yes, but the jump at $x=0$ must be added in the integration right? Similarly the jump at $x=1$ must get added, isn't it? $\endgroup$ – Sourabh Bhat Feb 28 '17 at 21:35
  • 1
    $\begingroup$ That's the wrong definition of total variation for discontinuous functions. One way to rectify it is to add on the absolute values of the size of all discontinuities. $\endgroup$ – Bowditch Feb 28 '17 at 21:50
  • $\begingroup$ @Bowditch Thank you. Will the differentiation not result in Dirac delta functions at the discontinuities? The integration of which will result in expected solution? Maple does not even produce a warning, that was a bit surprising. $\endgroup$ – Sourabh Bhat Feb 28 '17 at 22:17

Sometimes Maple coerced to do this kind of thing sucessfully, by using its Heaviside and Dirac forms.

However care must be taken. (It may be, say, that it has to be able to pull the Dirac calls out of the abs before it can integrate them as desired.)


f := piecewise( x < 0, 1, x <= 1, sin(Pi*x), 2 );

                          {     1              x < 0
                     f := { sin(Pi x)         x <= 1
                          {     2            otherwise

F := convert( f, Heaviside ):



Fig := abs( diff( F, x ) ):                      



simplify( Int( Fig, x=-infinity..infinity ) ):


   +Dirac(x)+2*Dirac(-1+x),x = -infinity .. infinity)



And another (simple) example:


g := piecewise( x < 0, 3+Pi, x <= 1, sin(Pi*x), 7+Pi );

                          {  3 + Pi            x < 0
                     g := { sin(Pi x)         x <= 1
                          {  7 + Pi          otherwise

G := convert( g, Heaviside ):



Gig := abs( diff( G, x ) ):



simplify( Int( Gig, x=-infinity..infinity ) ):


   +(7+Pi)*Dirac(-1+x)+(3+Pi)*Dirac(x),x = -infinity .. infinity)


                            12 + 2 Pi
  • $\begingroup$ I know you from MaplePrime thats why I am asking you this. Why there is not a stackexchange website for maple? $\endgroup$ – zhk Mar 2 '17 at 16:05
  • $\begingroup$ I don't know why the "Maple use community" has not created such a site. The site www.mapleprimes.com has existed for quite a long time (almost 12 years) and perhaps the Maple user community has become used to trying it first. In contrast IIRC the Mathematica stackexchange site already had high traffic when the later community.wolfram.com was created, so perhaps its user community was not very motivated to switch en masse. I note that mapleprimes does not censor mention of competing products, or product criticism -- its code-of-conduct relates mostly to manners, etc. $\endgroup$ – acer Mar 2 '17 at 18:06
  • $\begingroup$ I tried once myself but it didn't work. It's my wish that maple should have such website. If you open one, I will support you? $\endgroup$ – zhk Mar 2 '17 at 18:08

You are correct that Maple does not understand your commands as you intend them, but I suspect not for the reasons you think. I don't think there is a way to fix this, unless Maple supports Stieltjes integration, or has a built-in to calculate total variation.

Your definition of total variation is correct in a distributional sense. What does that mean?

For each integrable $ f \in \mathbb{R}\to\mathbb{R} $, we can define $ f^* \in C_0(\mathbb{R}\to\mathbb{R})\to\mathbb{R} $, a continuous linear functional defined on the set of all continuous functions vanishing at infinity, as follows: $$ f^*(g) = \int_{\mathbb{R}}{f(x)g(x)\,dx} $$ If $ f $ is differentiable, we have the useful relation that, for functions $ g $ vanishing sufficiently rapidly at infinity $$ \int_{\mathbb{R}}{-f(x)g'(x)\,dx} = (-f(x)g(x))|_{x\to-\infty}^{\infty} - \int_{\mathbb{R}}{-f'(x)g(x)\,dx} = 0 + \int_{\mathbb{R}}{f'(x)g(x)\,dx} $$ In other words, the functional $ T : L^1(\mathbb{R}\to\mathbb{R})\to C_0(\mathbb{R}\to\mathbb{R})\to\mathbb{R}$; $$ T(f)(g) = \int_{\mathbb{R}}{-f(x)g'(x)\,dx} $$ satisfies $ T(f) = f'^* $ whenever the right hand side is well-defined. It seems reasonable to say even when the RHS is not well-defined, $ T(f) $ represents some sort of generalized derivative of $ f $. Using this $ T $ to extend the derivative is traditionally called working "in a distributional sense."

It seems (from a cursory internet search) that Maple has only limited support for distributions (namely, the only feature is Dirac's delta as a built-in). So Maple is likely defining the derivative of your $f$ as follows: $$ f'(x) = \begin{cases} 0 & x < 0\text{ or }1<x \\ \pi \cos{\pi x} & 0<x<1 \end{cases} $$ Note that this is not defined at 0 and 1, and integrates to 2.

We might do better and say $$ f'(x) = \begin{cases} 0 & x < 0\text{ or }1<x \\ \pi \cos{\pi x} & 0<x<1 \\ -\infty & x = 0 \\ \infty & x = 1 \end{cases} $$ But even then, we're stuck, because integrals don't care about single points. Your function is complicated, so let's take a simple example, and apply the Riemann Integral (what I'm saying also applies if you know the Lebesgue Integral—one-point sets are Lebesgue-Null). Let $ g_a : L^1(\mathbb{R}\to\mathbb{R}) $; $$ g_a(x) = \begin{cases} 0 & x \neq 0 \\ a & x = 0 \end{cases} $$ Then $$ \int_{-1}^1{g_a(x)\,dx} = \lim_{P \to [0,1]}{\sum_{z\in[x,y]\in P}{g_a(z)(y-x)}} $$ where P is in the directed set of partitions under inclusion. Since g vanishes away from the origin, we can drop the terms corresponding to intervals that do not include 0; then $$ \left|\int_{-1}^1{g_a(x)\,dx}\right| = \lim_{\max{(-x,y)}\to0}{|g_a(z)(y-x)|} \leq \lim_{\max{(-x,y)}\to0}{|a|(y-x)} = 0 $$ as $ |g_a|\leq|a| $ everywhere. So $\int_{-1}^1{g_a(x)\,dx}=0$. Worse, even when we take an infinite discontinuity, we still have a null integral. $ g_a \to g_{\infty} $ uniformly as $ a \to \infty $, so $$ \int_{-1}^1{g_{\infty}(x)\,dx} = \int_{-1}^1{\lim_{a\to\infty}{g_a(x)}\,dx} = \lim_{a\to\infty}{\int_{-1}^1{g_a(x)\,dx}} = \lim_{a\to\infty}{0} = 0 $$

Any one-point discontinuity can be represented as the sum of $ g_k $ and a continuous function. The first integrates to 0, so the discontinuity may be dropped. Maple is ignoring the infinities that pop up in your derivative and just integrating over them to 0.

  • $\begingroup$ I was expecting that two Dirac delta functions will get created during differentiation at the discontinuities and the integration of which will result in required result. Seems like Maple simply ignores these discontinuities, and does not even give a warning. Thanks for your elaborate answer, I really appreciate it. $\endgroup$ – Sourabh Bhat Feb 28 '17 at 22:08

For piecewise functions, let Maple handle the discontinuities like we would do by hand when evaluating Lebesgue-Stieltjes integrals on functions with bounded variation.

f := unapply(piecewise(x<0,1,0 <= x and x <= 1,sin(Pi*x),2),x);
integrand := abs(diff(f(x),x));

points := discont(f(x),x); ## List of discontinuities of f

thesum := seq(abs(limit(f(x),x = points[i],left) - 
                  limit(f(x),x = points[i],right))
            ,i = 1..numelems(points));

theint := int(integrand,x = -infinity..infinity);

add([thesum,theint]); ## returns 5

A point to note is that Maple will catch all discontinuities of $\sin(\pi x)$ if not for the otherwise unnecessary and statement in the function definition.


Maple is just doing a simple Riemann integral here. That area under the curve calculation just ignores the simple undefined values at 0 and 1.


Mathematica gives the same answer as Maple, presumably because it performs Riemann (not Stieltjes) integration:

f[x_] := Piecewise[
        {{1, x < 0}, 
         {Sin[π x], 0 <= x <= 1}, 
         {2, x > 1}}];
Integrate[Abs[D[f[x], x]], {x, -\[Infinity], \[Infinity]}]

(* 2 *)


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