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

 Messages: [ Previous | Next ]
 Jonas Matuzas Posts: 8 Registered: 2/28/14
Re: integral for fun
Posted: Mar 3, 2014 2:30 AM

On Sunday, March 2, 2014 12:29:38 PM UTC+2, clicl...@freenet.de wrote:
> 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.

yes.. I know these functions.
Still calculating!!! it is already 4'th day has started (start was Friday morning). I am using 6'th fastest cpu in the world http://www.cpubenchmark.net/singleThread.html
Is is worth to wait for the end ?
|My question is- what is he (mathematica) doing ?
Is it fall in eternal loop ?

Date Subject Author
2/28/14 Jonas Matuzas
2/28/14 Jean-Michel Collard
3/1/14 Jonas Matuzas
2/28/14 Axel Vogt
2/28/14 Axel Vogt
2/28/14 Richard Fateman
2/28/14 Axel Vogt
2/28/14 Nasser Abbasi
2/28/14 Richard Fateman
3/1/14 Axel Vogt
3/1/14 Axel Vogt
3/1/14 Jonas Matuzas
3/1/14 Richard Fateman
3/1/14 Axel Vogt
3/1/14 Nasser Abbasi
3/3/14 Waldek Hebisch
3/4/14 Richard Fateman
3/4/14 Waldek Hebisch
3/4/14 Richard Fateman
3/6/14 Waldek Hebisch
3/1/14 Jonas Matuzas
3/2/14 clicliclic@freenet.de
3/3/14 Jonas Matuzas
3/1/14 Jonas Matuzas
3/1/14 Jonas Matuzas
3/1/14 Jonas Matuzas