This is http://www.mapleprimes.com/questions/201767-Fsolve-Cannot-Solve-This-Equations-And ?

Actually you do not need that package (cf. my answer in that thread: the root of the the 'simple' equation is unique).

The reason for your doubts may be caused by the input: you only have some 3 or 4 digits in some cases (like Pi=3.14 for you ep0?) - which may prevent the system to react on changes over other variables (even if you increase precision, as suggested by Carl Love). You may wish to feed symbolics and/or exact values as far as possible.

I have no solution. But do not think that 'brute force' may work.

May be one should look at it as L(constant + convexFct/(x-1) ), where
L = log[x] (more like an operator) and constant to let it be defined?

It 'smells' like a converse to "when convex(concave(x)) is convex ?"

Would you mind to talk about the origin of the question?

eliminate is the key, so thanks would be proper to Carl

dito:

  u(r,theta,z,t)*V[0](diff(u(r,theta,z,t),r)*V[0]/r[0])
# ..................??

That makes it a function, you forgot a multiplication sign?

Each lhs rhs has 'v' as factor for multiplicity 1, each rhs lhs is some alpha.

Setting v=0 gives: all alpha have to be zero, any parameter else.

Else dividing by v this gives a new system with alpha/v (or more easy
to read: setting v=1 etc).

Now 'eliminate' works (sys = the above system of equations):

  map('q -> q/v', sys);
  eliminate(%, {a, b, g, p, u}):
  simplify(%, size);

That ignores needed conditions on the alphas

It would be fine, if an Admin would convert that part about hdb vs. help and Maple 18 to be a Post (and delete my suggestion after that).

I think it is really worth to an own topic.

Already for selections it is a bit risky - you have to aware of what Maple 'knows' up to that calls. And what it does after them.

It would be better if you provide your code (as an upload or in readable form).

One wants the order of a differentiation. However Maple displays it in
a special way, which can be seen printing it:

  diff(h(x), x$2); lprint(%);

      diff(diff(h(x),x),x)

So it is an iterated use. For each diff multiply by a dummy variable.
Then the degree for the dummy counts the number of differentiations.
It does not depend on the function(s) used for the operator D.

The rest of the code is a matter of taste. 

In simple form for your case just use:

  eval(eq, diff = '(f,x) ->diff(f,x)*Delta' );    # polynom in Delta
  coeff(%, Delta, 4);                  # extract desired coefficient


                     / 2         \ / 2         \
                     |d          | |d          |
                     |--- a[1](x)| |--- a[2](x)|
                     |  2        | |  2        |
                     \dx         / \dx         /



For better selections one may use the analysis on the variables used,
as it is shown in Alejandro's first 3 lines.

I will need some time to write it down (perhaps with some gaps)

You may 'verify' it with your last result, dividing it by ln(2)*Pi/32,then using 'identify', however one needs to reduce to Digits=10.

Edited for a sketch / justification:

First care for Int(ln(abs),x), using Change(%, 1/(2^x) = xi, xi) and
Change(%, 2^(-I*(y+I))*xi = x, x) transforms it to

  1/ln(2)*Int(ln(abs(x-1))/x,x = 0 .. 2^(1/2)*2^(-I*y));

                          1/2  (-I y)
                         2    2
                        /
                  1    |              ln(| x - 1 |)
                -----  |              ------------- dx
                ln(2)  |                    x
                      /
                        0

For that there is a formula in Lewin "Polylogarithms and associated
functions" on page 28. NB: his Dilog(z) is polylog(2,z) = dilog(1-z).
I take that (have not checked his references from 1932).

The task now writes as

Int((12*y^2-1)/(4*y^2+1)^3/ln(2) *
    Re(dilog(1-2^(1/2+I*y))), y = 0 .. infinity);

Now observe that *this* dilog =: h0 is periodic, period = 2*Pi/ln(2), 
and switch to sum the integrals over the periods.

That sum has a formal solution in terms of Int( h0*Psi(2,...) ).

Partial integration w.r.t. h0 makes it Int(Psi(1,...) * D(h0)) + some
constant. That constant vanishes due to periodics of h0 and D(h) = log.

Now again do partial integration w.r.t. to that log part.

The result is B2 + A2, where
 
B2 := 1/32*ln(2)*Pi*coth(1/4*ln(2))
A2 := 1/32*I*ln(2)*
  Int((2^(1/2)*cos(2*tau*Pi)-2)/(-3+2*2^(1/2)*cos(2*tau*Pi)) *
      (Psi(1/4*I*ln(2)/Pi+tau)-Psi(-1/4*I*ln(2)/Pi+tau)), 
      tau = 0 .. 1);

B2 turns out to be 1/32*(2*2^(1/2)+3)*Pi*ln(2).

For the remaining integral note that Psi and conjugation permutes, so
it writes as 

J:= Int(-2*Im(Psi(1/4*I*ln(2)/Pi+tau))* 
    (2^(1/2)*cos(2*tau*Pi)-2)/(-3+2*2^(1/2)*cos(2*tau*Pi)),tau = 0 .. 1)

Numerically it can be determined to high accuracy, identified to -2*Pi.

Thus B2 + A2 = 1/32*(2*2^(1/2)+3)*ln(2)*Pi+1/32*ln(2)*J = asserted.

Ok, I lost a minus sign on my way ...


For a formal proof J = -2*Pi I think one can use formulas to be found
in Prudnikov & Marichev (they also have something like Lewin, above),
but I am not sure in which cases they assume to be over the Reals.

I will see.

Can you get more corrrect decimal places?

BTW: what do you want to express by "lofty goal" ?

I do not think that direct way gives many digits or would one allow to guess the very value.

Yes, it is much more easy and does not involve calling a very expensive function

It was time consuming, yes (going false routes).

But it is only some Numerics and using Maple in good manner for the task,
if I would be asked to judge my answer.

New task: Use w(s) = 1-2^(1-s) instead of Zeta(s), find a symbolic answer
(I did not succeed)

That would give the task for Dedekind's Eta instead of Zeta. That seems to
be more 'friendly' for Numerics. But would certainly not answer the very claim.


PS: I will not pay 40 € to Springer for 3 pages

http://axelvogt.de/temp/MP_integral_2dim_zeta.pdf
http://axelvogt.de/temp/MP_integral_2dim_zeta.mws

PS: do you have some reference for your task?

For x = fixed the integrand decays fast (looking at asympt w.r.t. y). But does that 'fit' on 2-dim segments? As I indicated: I give up for that.