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

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Jonas Matuzas

Posts: 8
Registered: 2/28/14
Re: integral for fun
Posted: Mar 1, 2014 8:20 AM
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On Saturday, March 1, 2014 1:51:38 AM UTC+2, Nasser M. Abbasi wrote:
> 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];
>
> int/.x->Infinity
>
> Infinity::indet: Indeterminate expression 1+-\[Infinity]+\[Infinity] encountered. >>
>
>
>
> While
>
>
>
> 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
>
>
>
> https://reference.wolfram.com/mathematica/ref/SinIntegral.html
>
> http://reference.wolfram.com/mathematica/ref/CosIntegral.html
>
>
>
> ------------------------------------
>
> (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] +
>
> (1/2)*(-(((-1)^(7/8)*(Cos[2*(-1)^(1/8)]*
>
> CosIntegral[-2*(-1)^(1/8) + 2*x] +
>
> Sin[2*(-1)^(1/8)]*SinIntegral[
>
> .... 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);
>
>
>
> (1/32)*(sum((8192*(Si(-4*x+_R1)*sin(_R1)+Ci(4*x-_R1)*
>
> cos(_R1)))/_R1^7, _R1 = RootOf(_Z^8+65536)))+(1/4)*
>
> (sum((32*(Si(-2*x+_R1)*sin(_R1)+Ci(2*x-_R1)*
>
> cos(_R1)))/_R1^7, _R1 = RootOf(_Z^8+256)))+(3/8)*
>
> (sum(_R*ln(x+8*_R), _R = RootOf(16777216*_Z^8+1)))
>
> -----------------------
>
>
>
> --Nasser


thank you for your analysis . You think Matematics is calculating limits? I think it should use Residues theory...



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