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Topic: integral for fun
Replies: 25   Last Post: Mar 6, 2014 4:10 PM

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Nasser Abbasi

Posts: 6,677
Registered: 2/7/05
Re: integral for fun
Posted: Feb 28, 2014 6:51 PM
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On 2/28/2014 3:39 PM, Axel Vogt wrote:

> Maple 17 actually gives it 'immediately' in terms of hypergeometric 0F7
> (and your numerical value)

Mathematica actually finds the indefinite integral with no
problem. The problem comes when evaluating this result at
the limits of integration in order to find the final value.

The limit has to be used, becuase directly evaluation of x=Infinity
or x=-Infinity producdes indeterminate form(s).

When trying the limit, Mathematica is taking for ever to finish,
so I stopped it after many hrs. (same what happens it seems
with the OP)

int = Integrate[Cos[x]^4/(1 + x^8), x];
Infinity::indet: Indeterminate expression 1+-\[Infinity]+\[Infinity] encountered. >>


Limit[int,x->Infinity] (* takes long time....*)

And I could not figure a way around it in the small amount of time I played
with it. Here is "int" fyi: It contains many expressions involving
the CosIntegral and SinIntegral (It seems Maple used hypergeometric 0F7

(3/32)*ArcTan[Sec[Pi/8]*(x - Sin[Pi/8])]*
Cos[Pi/8] + (3/32)*
ArcTan[Sec[Pi/8]*(x + Sin[Pi/8])]*
Cos[Pi/8] - (3/64)*Cos[Pi/8]*
Log[1 + x^2 - 2*x*Cos[Pi/8]] +
(3/64)*Cos[Pi/8]*Log[1 + x^2 +
2*x*Cos[Pi/8]] +
(3/32)*ArcTan[(x - Cos[Pi/8])*Csc[Pi/8]]*
Sin[Pi/8] + (3/32)*ArcTan[(x + Cos[Pi/8])*
Csc[Pi/8]]*Sin[Pi/8] -
(3/64)*Log[1 + x^2 - 2*x*Sin[Pi/8]]*
Sin[Pi/8] + (3/64)*
Log[1 + x^2 + 2*x*Sin[Pi/8]]*Sin[Pi/8] +
CosIntegral[-2*(-1)^(1/8) + 2*x] +
.... too large to post it all

Maple 17 also used Si and Ci, but it uses summation in
its output, hence it looks much smaller than the output above

z:=int(cos(x)^4/(1 + x^8), x);

cos(_R1)))/_R1^7, _R1 = RootOf(_Z^8+65536)))+(1/4)*
cos(_R1)))/_R1^7, _R1 = RootOf(_Z^8+256)))+(3/8)*
(sum(_R*ln(x+8*_R), _R = RootOf(16777216*_Z^8+1)))


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