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

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clicliclic@freenet.de

Posts: 997
Registered: 4/26/08
Re: integral for fun
Posted: Mar 2, 2014 5:29 AM
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Jonas Matuzas schrieb:
> 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)))
> >
> > -----------------------
> >

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


Typically, Mathematica and Maple would here split the integration
interval and use convolution of Meijer-G functions for INT(COS(c*x)/(1 +
x^8), x, 0, inf), I think. As Axel has noted, COS(x)^8 = 1/8*cos(4*x) +
1/2*cos(2*x) + 3/8 can be decomposed accordingly. The Meijer-G technique
allows to deal a large range of elementary and non-elementary factors in
the integrand and can be implemented as a sequence of transformation and
simplification rules. The 0F7 expressions were presumably obtained in
this way, although their ultimate reducibility to elementary
constituents may be difficult to detect, and the reduction difficult to
effect. See <http://en.wikipedia.org/wiki/Meijer_G-function> for some
background.

Martin.



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