A change of variables is mathematicaly a diffeomorphism, so it it acts between spaces of same dimension.
E.g. you cannot have a diffeomorphism from R to R^2.

What result would you expect? f is a function of one variable and it seems that you want to treat it as having two.

@Markiyan Hirnyk 

Yes, the result is correct.

#  Sum( binomial(i,k)*(Sum(1/j!,j=i+1 .. infinity)),i=k..infinity);
#  Maple cannot find the final sum. Change the summation order:
Sum((Sum(  binomial(i,k)*   1/j!,  i=k .. j-1)),j=k+1..infinity);            
value(%);

@Markiyan Hirnyk 

But I have nothing new. I have said that the code executed in 2016.2 gives correct plots; the code was the same in your last worksheet. So, what is the problem? Normal people do not ask the same question twice.

@Markiyan Hirnyk 

Your worksheet is 2017.0 (probably beta).

After deleting the plot and reexecuting in 2016 it's OK.

RectangularWindow, as the help page says multiply the signal by 1, i.e. does nothing.
The reasons to do so seem to be:
- the user can see this as a comment, and knows that no windowing is applied
- the command does some checks and conversions (if needed).

As for the TriangleWindow, listplot was forgotten for Re(u), i.e.
listplot(Re( u ))
must be used.

Define:
n:=4;
dist:=add( (- m*x[i] - q + y[i])^2, i=1..n);
Then solve for m,q:

diff(dist,m)=0, diff(dist,q)=0;

@acer 

Of course, if sigma is very small, increasing Digits will be mandatory.

@sand15athome 

You have two solutions: switch to software floats or simplify the function using Heaviside (in the latter case evalhf has no problems).
You can inspect `f*g`(x). You will see that with my sugestion it does not contain Dirac and is much simpler (and probably also faster).

So, you have a continuous function Phi(t). It seems to be (almost) periodic.
What do you mean by a probability density associated to it?

@sideshow 

The solution is not unique. E.g.

Sol := proc (T) options operator, arrow; fsolve({FM(q, Jo, T) = 1, FQ(q, Jo, T) = q}, {Jo=0.5 .. 1.5, q=0.5 .. 1}) end proc:Sol(0.3);
              {Jo = 1.124786970, q = 0.7332828274}

@leafgreen 

Your function f(x) = 1/ln(x + 1/x)  does not have an elementary antiderivative (= indefinite integral). This can be proved using https://en.wikipedia.org/wiki/Risch_algorithm.
Maple applies it for this kind of functions. Unfortunately such algorithm does not exist for definite integrals.

Most such integrals cannot be expressed with elementary (or even special) functions.
Why do you think that this one must be an exception?

@sideshow 

fsolve gives  {Jo = -1.082524999, T = -.4330099990}.

What does not work well?

@sideshow 

I have considered the system

FM(q, Jo, T) = 1,   FQ(q, Jo, T) = 2.

That's because the original
FM(q, Jo, T) = 0,   FQ(q, Jo, T) = 0,
obviously has no solution. And the Mathematica code contains something similar.
You must set the system you want to solve, I cannot guess it.