@Preben Alsholm Yes^^ It seems to be iterating Risch algorithm forever, that's why the result is so big!

@Preben Alsholm So does FTOC use only lookup tables for the indefinite integral or does it also have access to series expansions (as in meijerg I suppose?)

Or how is it evaluating the indefinite integral?

@Preben Alsholm

## ftoc and ftocms uses the Fundamental Theorem of Calculus.

Out of curiosity: How do these methods proceed?

I presume they first convert the integrand into a corresponding series expression then integrate termwise (indefinitely) and convert the series back to an elementary expression? Finally the fundamental theorem is used?

@Preben Alsholm " ## NOTE added: Today I cannot reproduce the wrong ftoc and ftocms results for f itself. They now  agree with the correct  meijerg result. "

When was this added? So now it works for you? What did you do apart from this trick CarlLove mentioned?

@Preben Alsholm Hey again. I just wanted to say that I tried with Maple17 as well, and it seems the bug has been around for a long time.

@Preben Alsholm 

So is it possible to say, what other similar expressions relying on these methods will return an error/wrong result?

Can it be fixed?

@Axel Vogt I think the problem is that it once evaluates the crucial expression to

sqrt(I*(omega0^2-t))

and the other time to

sqrt(I)*sqrt(omega0^2-t)

These are however not the same.

@Mariusz Iwaniuk Hey and Thanks. The point is, is it possible to reexpress Ei1 by Ei or somewhat different, in order to get rid of the I*Pi term? I'd like to have this function be represented by manifest real functions.

I have a different expression which is not yet simplified:

(-(exp(j*z))^2*Ei(1, j*z)+ln(-z)-ln(z)+Ei(1, -j*z))/(2*j*exp(j*z))

The problem is that, Ei(1,z) jumps by 2Pi when crossing the negative real axis.

In fact it seems that Ei(1,-j*z)=-Ei(j*z)-I*Pi so why is this not simplified?

@Kitonum Hey thanks for this. Is it possible to use this in general when simplifying expressions or do I still in the end explicitly need to say

assuming R-sqrt(R^2+z^2)<0

?

@Mariusz Iwaniuk 

" Maple have some limitation about regarding the resolution of the case and falsehood."

Why is that? Why is it so difficult for it to realize that e.g. coulditbe(R-sqrt(R^2+z^2) < 0);#FAIL

is actually TRUE when assumptions are set?

Using with(MathematicalFunctions); and Assume() does not do any better or?

@Mariusz Iwaniuk So in my case this is relevant:

convert(signum(0, x,1), piecewise);

#piecewise(x < 0, -1, 0 <= x, 1)

However as R-sqrt(R^2+z^2)  is obviously <0 then why is it not simplified as -1 ?

@Carl Love Yes x>0.

@gkokovidis ok thx.

Assuming I have the following expression

f:=exp(I*3*x-x/2)/(1+I)

Now when using the procedure from Preben gives me a rather weird result.

I therefore went on and removed the x terms

remove(has,f,x)

then I use

PolarForm(%)

and finally

combine(select(has,f,x)*'%')

However I still can not combine the result i.e. even when I double quote ''%'' it expands out again.

@nm This seems awkward.. I actually have other exponential functions and combining this number also in the exponential seems to simplify everything significantly.

@Carl Love No I'm fine with it and I of course understand the reason behind it. Though when considering such sums which are only convergent within some domain the result I expected from maple evaluating it for me is always the analytic continuation. I just didn't expect it to be changed, because I thought everyone who is using maple knows the strictly speaking inequivalence here automatically.