## 1201 Reputation

17 years, 241 days

## that last one is very...

that last one is very efficient  indeed.

## ...uneval does the trick....

...uneval does the trick.

thanks

## I see. I never encountered...

I see. I never encountered this kind of problem before.

Thanks, acer

## Try this...

But why do you want to do that ?

f:=x->x^2;

y:=f(x);

## Try this...

But why do you want to do that ?

f:=x->x^2;

y:=f(x);

## note, this is a slow...

note, this is a slow forum, you usually don't get an answer 1 min. after your 1st post.

## this one is for those loving...

this one is for those loving brackets:

`with(ListTools):`

L:=[a,b,c,d,c,d,e,f,a]:

[{FindRepetitions(L)[]}[]];

[a, c, d]

## one could use numboccur as...

one could use numboccur as part of the task.

(If allowed)

## see below...

setDiff:=proc(s1::set,s2::set) s1 minus s2; end proc;

## Well i'm referring now...

Well i'm referring now only to the arctan Term, the rational terms are not difficult to transform to the time domain. I don't want to say the remaining Integrals are easy to solve by hand, but at least they have a form that Maple can handle  properly.

Edit: sorry, i left out constants in the Exponential terms (last line), but after correction Maple can still handle it.

## @Sandor: thanks but...

but in this case, Maple does a bad job with this transformation of the arctan term.  I solved this by hand with a convolution integral in the time domain and got a nice solution. The problem with Maple is, its solutions often contain complex terms that -when combined correctly - simplify to real terms. A result with  Ei  Integrals containing complex arguments is just useless for further processing (plotting for example).

## finally it works...

Altough i don't get it why it has to be Y(s) and not Y, but obviously, you're right.

## try this:...

with(plots):with(plottools):
c:=pieslice([0,0], 3, 0..3/4*Pi, color=blue):
d:=pieslice([0,0], 1, 0..3/4*Pi, color=white):
display([d,c], axes=none, scaling=constrained);

## yes jakubi, that kind of...

yes jakubi, that kind of workaround is  a good solution. I figured out that even rectangles are sufficient for my task. I want to sketch the graph of a rational function with poles,zeros etc. and one step to do this is to exclude (rectangular) areas where no function values  can be found. These areas can be described  by inequalities und i thought it would be easy to plot them. On the other hand: this is one rare opportunity to use the drawing portion of maple :-)