If you are interested in such problems I'd recommend the book

Bressoud D., Wagon S. - A course in computational number theory, Wiley, 2000

It also contains Mathematica code but you could try to convert it into Maple.

@Adam Ledger 

@aamirkhan 

Yes, it seems that there exists an elementary integral wrt s. Your f_3 is a sum of >6000 terms, and each can be integrated simbolycally (but I did not verify all of them). Maple seems to refuse to integrate globally. You could use map(int,...)  but I don't think that the resulting HUGE expression will be useful.
Note also that using floats in symbolic computations is not usually a good idea.

This post reminds me of the following interesting formula for the nth prime due to Willan:

nthprime := n ->
1 + sum(floor(n^(1/n)*sum(floor(cos(Pi*((x-1)!+1)/x)^2),x=1..m)^(-1/n)),m=1..2^n);

It is correct but cannot be practically used for n>9 say.

@Markiyan Hirnyk 

Let's come back to our horses :-)

A:=plot3d(1, x=0..4, y=0..3, view = [0..4, 0..3, 0..2],
      image =cat(kernelopts(mapledir),"/data/images/fjords.jpg")):
with(plots):with(plottools):
f := transform( (x,y,z) -> [x^2-y^2, 2*y*x]):
display(f(A), axes = none);

@Mac Dude 

To avoid the interpolation one may take the inverse transform (provided it is bijective).
So, let A be a subset of the rectangle R(A), and f:A-->B  the transform (bijective).
B subset of a rectangle R(B).
Simply take the color(b) := color(g(b)), b in B, where g is the inverse mapping.

@Markiyan. Unortunately plottools:-transform can be applied only to the plot structure, not to the background.
One may try plot3d(...,image=filename).

@Mac Dude 

Conformal mappings are easy to construct:

(x,y) |--> (u,v) = (Re f(x+iy), Im f(x+iy)),

f being holomorphic such as f(z)=z^2, f(z)=1/z, f(z)=sin(z) etc.

[ The example is essentially (x,y) |--> (x, y^3) which is of course not conformal;
a conformal example would be e.g.  (x, y) |-->  (x^2-y^2, 2*x*y) ]
The origin does not matter, a translation does not affect the conformality.

Of course the image will have to be clipped (into a rectangle); if f(z_0) = oo, it wil be of course clipped.
The example can be easily adapted.

Thanks, I have inserted.

@Preben Alsholm 

@rlopez 

I think that the (global) bivariate limits for rational functions were introduced in Maple 2015 only.
I remember this because I was really impressed by the new algorithm.

limit( f(x,y), {x=0,y=0}); # (1)

is the limit (multi) at (0,0).
If it exists, it equals

limit( f(r*cos(t), r*sin(t)), r=0); # (2)

But it is possible that (2) exists and (1) does not (because (2) is not necessarily uniform wrt t).
For example, consider f(x,y) = x^2/y  or  f(x,y) = x^2*y/(x^4+y^2).

@Preben Alsholm 

I would vote for it, but I am not affiliated to Maplesoft :-)

Actually Maple uses residues in computing integrals, see e.g.

int(ln(x)/(x^6-2*x+4), x=0..infinity);

@Marko Riedel 

In this case it seems that using residues is the only solution:

where L = log, R = rational function without zeros in [0,infinity)
(this is what you did, J(n) being so expressed via J(n-1), J(n-2),... ).

@Marko Riedel 

tedious to code?

pw:=(n,a) -> add(NumberTheory:-Moebius(n/d)*a^d, d in NumberTheory:-Divisors(n)):

Test:
StringTools:-Generate(4, "01"):
select( StringTools:-IsPrimitive, %);
pw(4,2)=nops(%);

@MortenZdk 

The fact that your command did not work is more a Maple's fault than yours. The Explore command is rather new and must still be polished. In Maple 2015 not even the f:=a*x ...  version did not work, it had to be entered as

Explore(plot(a*x, ...

So, keep studying and try the examples in the Help (you can turn them into worksheets so that you can experiment). Also, User's Manual is a good start.

Of course it is possible. But:

1. This is equivalent to det(J)=0
2. In solving such systems, some solutions will be lost.
    See e.g.  solve(x*y+z*t);

Since you did not provide an example, I am not able to help more.

@cuongtd