Is this a bug in MuPAD?

Question

I have u0 as step function (piecewise definition):

piecewise([0 < t, 1], [t < 0, 0])

Now I compute:

int(u0(t)|t=k, k=-infinity..t)

which gives

piecewise([0 < t, t], [t <= 0, 0])

In a textbook I found the answer:

piecewise([0 =< t, t], [t < 0, 0])

Is this a bug in MuPAD?

A possible reason you may have been down voted is that your code is not runnable. Define the `u0(t)` function as I suggest in [this answer](http://stackoverflow.com/a/27750287/2278029). Your title is also very poor and wouldn't help anyone find your question to answer their own future query. This is a community. Don't just think of yourself when asking. This is your second question in a row that you've suggested there might be a bug rather than the more likely case of you making a mistake or not understanding how to use MuPAD (see [Occam's razor](http://en.wikipedia.org/wiki/Occam's_razor)). — horchler, Jan 05 '15 at 23:52
No it is not. That is not runnable code because the *the code itself* does not define `u0`. If one tries to run your first two lines of code as written they will get an error. — horchler, Jan 06 '15 at 16:20
if u cant define before like I explained u0 then u have no business trying to understand my question , I am sorry , but you are trolling — TraceKira, Jan 06 '15 at 20:28
*I* understand what you mean, but maybe your down voters didn't (you yourself didn't know how to define `u0` as a function/prcedure in [own previous question](http://stackoverflow.com/q/27742672/2278029)). I'm sorry that you took offense, but it's not very nice to accuse someone who's trying to help you improve your question and help other reproduce your issue of being a troll. We try to be nice to each other and be helpful on this site. Take care. — horchler, Jan 06 '15 at 21:19

score 1 · Answer 1 · answered Jan 05 '15 at 21:43

1

I does not matter if you use f(t)=t or f(t)=0 for t=0. In both cases the result is 0

answered Jan 05 '15 at 21:43

Daniel

1 Answers1