Login is not through the link at top, it is through
the link in the header of the contribution to which
I want to reply - just as in the first screen shot
and takes me to what you see in the 3rd shot.

The width of the reply box is just as wide as the
heading for the contributions. Find a shot at
www.mapleprimes.com/files/102_screen_width.gif
where this is from 17'' using IE 6.0 while all the
others where 15'' flat screen using Mozilla 1.7.12.
heading for the contributions is also the visible
width for contributions.

Thank you for your answer. I did not ask to include ps or pdf functionality, I said it would allow to print through such. For login see 3 screen shots: choose to be in a thread, klick on login to reply and then the 3rd shows you are at the main URL. You will find the URLs each in the screen shots. http://www.mapleprimes.com/files/102_reply_in_thread.zip The 1st shot also shows the problem with the side bars. Fine if you have no problems, so no reason to change. And for the reply box there is no way to change width. But height. It needs some time until the screen stabilizes again and decides to have a hight for the box - and is fun to wait again if you move the mouse cursor into it. Finally it shrinked to 1 line space :-) So far for my last feedback on the page, I do not want to see perfect solutions to be improved and probably waste your time. Or mine.

The problem with the singularity remains (same steps as already shown).
But I can imagine the following: the problem vanishes, if the terms does
not occure, i.e. the coefficients are such that they cancel it out.

A bit vague ... but let us try, your coeffs do have only 4 - 6 places
and as the aricle is 'old', it just may that the authors works with a
low numerical precision.

The singularity is a residuum (look at the series), so

  'residue(K(y), y=t0) = 0'; 
  solve(%,a):
  `a`=completesquare(%,c);
                                         2
                                (c + 2 b)
                            a = ----------
                                    3
Now evaluate with your parameters:

  eval(%, [a = 0.1e-5, b = 0.1066590211e-4, c = 0.1e-2]);

                           -5                  -6
                     0.1 10   = 0.3477062180 10

Which - within that low precision - says: there is no residue. And
working with low precision for quadrature may actually work. Can
you try to find out a more precise value for the params? May be
you find out what the author has assumed/said/etc. Otherwise it
is false I think. If the author or Journal is of known reputation
you should be able to exclude that by searching for later citations.

PS: Ken, the board mail did not accept answering
your mail, but anyway: here you have my answer. 

I agree with Alejandro. And it would allow to print as ps or pdf
which holds graphics ...

Additionally: if I say "login to reply" in a contribution it would be
good to directly jump into it - concurrently after login I am thrown
onto the main page and have to search the thread gain (which may be
nasty, if it is an elder thread).

Also I would prefer a simple way to switch off the stuff at the left
right on the screen (say by a button on/off) - this covers half of
the screen. At least it should vanish, if I am reading _in_ a thread,
then it is really unimportant and disturbing (ok, a bit can be done
through personal settings, but I read without login - and always do
enter the board through bookmarked www.mapleprimes.com/tracker to see
just the latest contributions).

And, well, the box to enter text for reply should be as large as the
screen - it is really to small.

It would be easier to give the integral with symbolics and parameters
separately (rounding errors and conditions you know).

I do not see, why the integral exists and think it is not convergent
(BTW: again I can not see it completely displayed on that board, it
still has all the noisy stuff at the boarders):

Int((y^2-1)^(1/2)/y/(1-1.5*y^2+(y^2-1)^(1/2)*(3*y^2-1)^(1/2))*(1-1.5*y^2)*
(.1e-5-(.1066590211e-4*y+.1e-2*(y^2-1)^(1/2))^2),y = 1 .. 1.4037);

J:=convert(%,rational); # try to make it 'exact' 
 
K:=unapply(integrand(J),y); 
plot(K(y),y=1..2); # to see possible problems

so try to find the singularity (caused by zeros in the denumerator)

K2:= unapply(denom(K(y)),y);
K1:= unapply(numer(K(y)),y);

K2(y)=0: [solve(%,y)];
                               1/2       1/2
                            2 3       2 3
                        [0, ------, - ------]
                              3         3

We only need one of these:

t0:=2/sqrt(3); evalf(%);

Then try to find out what happens there:

series( K(t),t=t0,3); 

That Laurent series starts with constant * 1/( t - t0) and that
t0 is in your integration path.

This shows: you integrate through the singularity like you would do for
Int( 1/(t-3/2), t=1..2) or Int( 1/t, t=-1..1), which does not exist.

May be your posted example was badly chosen or has a typo?

still unclear what you have in mind for your example
posted first - the last case is 'easy', as one can 
'see' the pattern, while for the origianl it is really
unclear what you expect

still unclear what you have in mind for your example
posted first - the last case is 'easy', as one can 
'see' the pattern, while for the origianl it is really
unclear what you expect

Unclear to me what you mean ... what is the result you want (by paper
and pencil)? 1/2*e+1/2*f = g; theE:=solve(%,e); eval(y, e=theE); shows
that the expressions just interchange g and e. So what do you expect
to _see_? What is your actual question behind the stuff?

Unclear to me what you mean ... what is the result you want (by paper
and pencil)? 1/2*e+1/2*f = g; theE:=solve(%,e); eval(y, e=theE); shows
that the expressions just interchange g and e. So what do you expect
to _see_? What is your actual question behind the stuff?

Strange, I would expect 'normal' to do that ... and I do
not like frontend (to complicated for me) ... anyway:

y:=a*(1/2*e+1/2*f)^3+b*(1/2*e+1/2*f)^2+c*(1/2*e+1/2*f)+d;
A:=numer(y);B:=denom(y);

'y=A/B'; is(%); # now check it


Not clear what you mean by substitute, there is no pattern
to which it applies (try subs, algsubs, applyrule ...).

If you want just eliminate your d then try this

a*e^3+b*e^2+c*e+d = 0; isolate(%,d); subs(%,A);

Otherwise - as you suggest - try to set up a desired form
and 'solve for what is needed to hold'.

Strange, I would expect 'normal' to do that ... and I do
not like frontend (to complicated for me) ... anyway:

y:=a*(1/2*e+1/2*f)^3+b*(1/2*e+1/2*f)^2+c*(1/2*e+1/2*f)+d;
A:=numer(y);B:=denom(y);

'y=A/B'; is(%); # now check it


Not clear what you mean by substitute, there is no pattern
to which it applies (try subs, algsubs, applyrule ...).

If you want just eliminate your d then try this

a*e^3+b*e^2+c*e+d = 0; isolate(%,d); subs(%,A);

Otherwise - as you suggest - try to set up a desired form
and 'solve for what is needed to hold'.

Alejandro, May be you are still interested in a solution, Maple can give an analytic one in terms of elliptic function:

Int(cosh(v)/sqrt(sech(v)-sech(vm)),v=0..vm);
J:=integrand(%);

convert(J,exp);
'int(%,v)';: 
theI:=simplify(%) assuming (0 < v):

'limit(theI,v=vm,left) - eval(theI,v=0)';
simplify(%) assuming 0 < vm:
theSol:=%:
# indets(theSol,function): convert(%,list);

Due to approximation errors one has to use the real part only.

Digits:=30: 'eval(theSol,vm=0.01)'; evalf(%): Re(%); Digits:=14:
gives 2.22156642743968396085047891865 and thus fits into the old
results.

Digits:=30: 'eval(theSol,vm=0.001)'; evalf(%): Re(%); Digits:=14:
gives 2.22144271864023726509396238200, your vm can become small.

Digits:=30: 'eval(theSol,vm=0.5)'; evalf(%): Re(%); Digits:=14:
gives 2.54837900700785977030637219126, 'confirm' it by integration:
'eval(Int(cosh(v)/(sech(v)-sech(vm))^(1/2), v = 0 .. vm, method =
 _d01ajc),vm=0.5)'; evalf[10](%); returns 2.548379007.

Of course there is no need to work with 30 Digits for evaluating.

However I have not checked whether the anti-derivative is always continous (the expression is lenghty), the examples make it only plausible ...

even if that is convenient i configure my browser to always clean up as far as possible (where the rest is done by a (freeware?) program like ClearProg.exe). May be this needs some additional logins ... but i prefer it for security reasons (and s.th. like a small fire wall), even if i have to view the site without my prefered setting. This sounds like a little paranoia, but can avoid troubles.

ääähh ... this is surprising. At least. Did I miss it in the docu? What if I have f_n and n is assigned a value later? And if I type f3 I can see it as indexed - the same table behaviour?

why this almost reminds me of the posting from India below ... http://www.mapleprimes.com/forum/i-am-looking-for-freelance-maple-10-developers-/programmers