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

 Messages: [ Previous | Next ]
 Nasser Abbasi Posts: 6,677 Registered: 2/7/05
Re: integral for fun
Posted: Feb 28, 2014 6:51 PM

On 2/28/2014 3:39 PM, Axel Vogt wrote:

>
> Maple 17 actually gives it 'immediately' in terms of hypergeometric 0F7
>

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

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