Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2007
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Mind+Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg78068] Re: Mind+Mathematica
  • From: Michael <mcauxeu at gmail.com>
  • Date: Fri, 22 Jun 2007 06:39:18 -0400 (EDT)
  • References: <f5die5$r69$1@smc.vnet.net>

On Jun 21, 3:03 am, dimitris <dimmec... at yahoo.com> wrote:
> The integral
>
> Integrate[Sin[z]*Sin[z^3 + z], {z, 0, Infinity}]
>
> (as I was informed)
>
> gives a incorrectly divergent message.
> The integral however is convergent.
>
> The following is part of my response to another forum.
> Demonstrate how vital is to help Mathematica sometimes.
>
> In[2]:=
> $Version
>
> Out[2]=
> "5.2 for Microsoft Windows (June 20, 2005)"
>
> In[3]:=
> int=Integrate[Sin[z]*Sin[z^3 + z], {z, 0, Infinity}](*the integral
> stays unevaluated*)
>
> Out[3]=
> Integrate[Sin[z]*Sin[z + z^3], {z, 0, Infinity}]
>
> In[3]:=
> int2 = (int /. Integrate[f_, x_] :> Integrate[#1, {z, 0, Infinity}]
> & ) /@ Expand[Sin[z]*TrigExpand[Sin[z^3 + z]]]
>
> Out[3]=
> (1/72)*(2*Sqrt[6]*Pi*(BesselI[1/3, (4*Sqrt[2/3])/3] - BesselJ[1/3,
> (4*Sqrt[2/3])/3]) +
>     3*Gamma[1/3]*(2*Sqrt[3] - Sqrt[2]*BesselI[-(1/3), (4*Sqrt[2/3])/
> 3]*Gamma[2/3] - Sqrt[2]*BesselJ[-(1/3), (4*Sqrt[2/3])/3]*Gamma[2/3]))
> + Integrate[Cos[z]*Sin[z]*Sin[z^3], {z, 0, Infinity}]
>
> In[4]:=
> int3 = (1/2)*Integrate[Sin[2*z]*Sin[z^3], {z, 0, Infinity}]
>
> Out[4]=
> (Pi*(AiryAi[-(2/3^(1/3))] - AiryAi[2/3^(1/3)]))/(4*3^(1/3))
>
> In[5]:=
> FullSimplify[int2 /. Integrate[x___] :> int3]
>
> Out[5]=
> (-2*3^(1/6)*Pi*AiryAi[2/3^(1/3)] + Gamma[1/3])/(4*Sqrt[3])
>
> In[6]:=
> N[%, 40]
>
> Out[6]=
> 0.295741225849781931593673891336119670357883693300484102195`40.
>
> Brought to you by M^2
> (Man+Mathematica!)
>
> Dimitris
>
> PS
> I spent almost two hours to figure out a workaround.
> How ancient Greeks said:
> "It is not easy to get Goods"
>
> PS2
> Enjoy Mathematics and Mathematica!

Hi,

This is an interesting example from Dimitris, especially on how one
needs to (surprisingly) 'coax' an answer using "int3" in version 5.2.

For posterity, in version 6, I get:

(\[Pi] (AiryAi[-2/3^(1/3)] - AiryAi[2/3^(1/3)]))/(
 4 3^(1/3)) + (-3^(
    1/6) \[Pi] (AiryAi[-2/3^(1/3)] + AiryAi[2/3^(1/3)]) +
  Gamma[1/3])/(4 Sqrt[3])

upon using the definition given above for "int2".  Strangely, the
message

"Integrate::idiv: Integral of Sin[z] Sin[z+z^3] does not converge on \
{0,\[Infinity]}. >>"

does appear (twice); however, I would not have expected a returned
answer if this integral truely is divergent.

Regards,

Michael



  • Prev by Date: Re: Solving a Integral
  • Next by Date: Re: v5.2 preferred for stability over v6.0
  • Previous by thread: Re: Mind+Mathematica
  • Next by thread: Re: Mind+Mathematica