This is a strange post to find in the "Announcements" section.

The integrand

 f:=arccos(x)/(1+x^4); 

and its numerator, in particular, are smooth and bounded functions within the
interval [0,1]. So, for the evaluation of the integral J

j:=u->Int(u,x=0..1):
J:=j(f);

I will expand arccos(x) in a Taylor series and integrate term-by-term:

(fn,fd):=(numer,denom)(f):
fns:=convert(fn,FormalPowerSeries);
tk:=op([2,1],fns):

We get two pieces:

g1:=op(1,fns)/fd;
g2:=subsop(1=tk/fd,op(2,fns));

          Pi
g1 := ----------
              4
      2 (1 + x )

So, the integral of g1 is:

J1:=j(g1);
J1v:=value(J1);

While the integral for the series becomes formally:

subsop(1=j(op(1,g2)),g2);
value(%):
J2s:=map(simplify,%);
sk:=op(1,J2s):

So, formally, J=J1v+J2s. Here, the terms with k odd are problematic. So, for a
numerical check of this result, I will sum even and odd terms separately.
First, the sum of the even terms:

eval(sk,k=2*m);
sm:=simplify(%) assuming m::posint;
Se:=Sum(sm,m=0..infinity);

I have not got a closed form for this sum, but there is no problem to evaluate
it numerically.

Now, for the odd terms, evaluation of these terms produce error messages: "Error, (in
LerchPhi) numeric exception: division by zero". And 'limit', with factorials
replaced by 'GAMMA' calls:

skc:=convert(sk,GAMMA,factorial);

remains unevaluated (both "ordinary" or MultiSeries). Clearly there are
singularities, but it seems to me that they are "mild" in the sense that the limit
exist. Eg, a plot like

plot(skc,k=0..5);

shows a quite smooth curve. So, in the assumption that these limits exist, I
evaluate a numerical approximation of these limits as a mean of the value of
the terms with k shifted a bit below and a bit above, and add a number of
them:

Digits:=15:
So:=1/2*(add(evalf(eval(sk,k=2*i+1+10^(-6))),i=0..50)+
add(evalf(eval(sk,k=2*i+1-10^(-6))),i=0..50));

                        So := -.348302048743784e-1

Then, adding the numerical evaluation of these three terms:

evalf(J1v+Se)+So;
                        0.922570909713502

I get a result that compares quite well with:

evalf(J);
                          0.922548175702252

Clearly, there is a "delicate" balance between the value of 'Digits', the
magnitude of the "shift" and the number of terms in 'So'.

HI

is my first time in high mathematics quarters of maples...

Has been a bit frustrating trying solving the Integral the old way, by paper and triks, substitutions and so... even more frustrating when I gave a try with matematica: not a second and here the solution (do know if is the right one, but it seams old fashon enough to be so): 

(1) input: [Integral]ArcTan (x)/(1+x^4) dx
 
(2)output: 1/2 ArcTan (ArcTan[x²])
 
(3)output: diff (integral output)->((ArcTan x)/(1+x⁴)) just to verify I were awake...
 
(4) imput: evaluate Int, x=1..4
 
(5) Output: 1/4 ArcTan ((-Sqr(2)*p+8 ArcTan[16]+Sqr(2)*(ArcTan[(4 *Sqr(2)/15]+ArcTanh[(4*Sqr(2)/17]))
 
What do you thinck, it is just allucination or ...
in any case how to know if a math prog gibes the right answers, in such unhandly cases?!
 
Thanks for the great insight which comes out from your discussion, I'm notv a mathematician though i have great passion for it ... as it happens quite often about the prittiest girl in the school...
 
Behave, if not
 
be good!
 
Federiclet(ITA)

In spite of the megabugaerror,

I am still alive

^_^

be good

f

Even if it was my intention to give a somwwhat detailed error report I put together
what was done above do give the result in Maple.

  Int( arccos(x)/(x - a), x = 0 .. 1 ) =
= Int(-t*sin(t)/(-cos(t)+a),t = 0 .. 1/2*Pi) =
= Int(ln(-cos(t)+a),t = 0 .. 1/2*Pi) - 1/2*Pi*ln(a) =
= Int(ln(-x+a)/(1-x^2)^(1/2),x = 0 .. 1) - 1/2*Pi*ln(a) = # let Maple evaluate it
= ln( (1+sqrt(1-1/(a^2)))/2 )*Pi/2 - 1/a*hypergeom([1/2, 1, 1],[3/2, 3/2],1/(a^2))

Let me call E the function given by the last expression.

Then it makes sense to use (complex) partial fraction for the original problem
using convert(1 / ( 1+x^4),fullparfrac) and this then writes as

  'Sum(-1/4*alpha*E(alpha),alpha = RootOf(1+Z^4))';
  allvalues(%): simplify(%,size); 

  evalf(%);

                     -----
                      \
                       )      (-alpha * E(alpha) / 4)
                      /
                     -----
              alpha = RootOf(1+Z^4)

  1/2 hypergeom([1/2, 1, 1], [3/2, 3/2], I) + 1/2 hypergeom([1/2, 1, 1], [3/2, 3/2], -I)

                         0.9225481756 + 0. I

which is the numerical value for  Int( arccos(x) / ( 1+x^4) , x=0 .. 1).


One can write the result as Re( hypergeom([1/2, 1, 1], [3/2, 3/2], I) ), but still
in terms of hypergeometrics 3F2, which I do not like too much - but have nothing
better.

May be one can do Int(ln(-x+a)/(1-x^2)^(1/2),x = 0 .. 1) through contour integrals,
but I am too long out of practicing that.

In fact, the calculation of the integral over the interval (-1,1):

J:=Int(arccos(x)/(1+x^4),x=-1..1);

is quite simple using the symmetries of the integrand, ie its even denominator
invariant under the change x=-u and the transformation of arccos(x) under this
change:

with(IntegrationTools):
Change(J,x=-u);
Expand(%):
subsop(1=value(op(1,%)),%);
combine(%,ln);

As the second term is -J, it turns out that the first term is 2*J. Ie the
value of J is:

op(1,%)/2;
evalf(%);

2.723675968

Geometrically, the curve of arccos(x) divides the rectangle of height Pi and
base on the interval (-1,1) in two halves. So, because of the even
denominator, J the integral below the curve is equal to the integral above the
curve up to Pi.

This does not solve the original integral but it is nice and value(J) also
produces a wrong and horrible sum of dilogs:

evalf(value(J));

-.7657561321-.4274644056*I