Mathematica 9 is now available
Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2008

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

Search the Archive

Re: integration

  • To: mathgroup at smc.vnet.net
  • Subject: [mg92599] Re: integration
  • From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>
  • Date: Tue, 7 Oct 2008 07:06:51 -0400 (EDT)

Dear Experts,
I have been trying to simplify(integrate) the following function, but
M6 seems to give a complex answer which i cann't understand.. please
help.

x[s_]=\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(s\)]\(Cos[
\*FractionBox[\(r\ t\ \((\(-\[Kappa]0\) + \[Kappa]1 + r\ \[Kappa]1)\)
+ \((1 + r)\)\ S\ \((\[Kappa]0 - \[Kappa]1)\)\ \((\(-Log[S]\) + Log[S
+ r\ t])\)\),
SuperscriptBox[\(r\), \(2\)]]] \[DifferentialD]t\)\)

Regards,
RG

Hi, RG,
your problem seems to be in Mathematics, rather than in Mathematica. The integral you need (as it is now) is too cumbersome. 
Make the integrand simpler by hiding and simplifying your notations, and the result will become more understandable. 
For example, below is your integral along with the result in which I denoted some terms by a, b and c to make it shorter:
In[2]:= \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(s\)]\(Cos[\ 
   t\ a + b + c*Log[S + r\ t]] \[DifferentialD]t\)\)

Out[2]= If[(Re[S/(r s)] >= 0 && S/(r s) != 0) || Re[S/(r s)] <= -1 || 
  Im[S/(r s)] != 0, -(1/(2 a))
   S^(-\[ImaginaryI] c) (-((\[ImaginaryI] a S)/
      r))^(-\[ImaginaryI] c) (S^(2 \[ImaginaryI] c)
        Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a S)/
         r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
        a^2 S^2)/r^2)^(\[ImaginaryI] c)
        Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a S)/
        r] (\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r])) + (1/(
  2 a))(r s + S)^(-\[ImaginaryI] c) (-((\[ImaginaryI] a (r s + S))/
     r))^(-\[ImaginaryI] c) ((r s + S)^(2 \[ImaginaryI] c)
       Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a (r s + S))/
        r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
       a^2 (r s + S)^2)/r^2)^(\[ImaginaryI] c)
       Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a (r s + S))/
       r] (\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r])), 
 Integrate[Cos[b + a t + c Log[t r + S]], {t, 0, s}, 
  Assumptions -> ! ((Re[S/(r s)] >= 0 && S/(r s) != 0) || 
      Re[S/(r s)] <= -1 || Im[S/(r s)] != 0)]]


This is your result after simplification:

In[3]:= -(1/(2 a))
   S^(-\[ImaginaryI] c) (-((\[ImaginaryI] a S)/
      r))^(-\[ImaginaryI] c) (S^(2 \[ImaginaryI] c)
        Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a S)/
         r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
        a^2 S^2)/r^2)^(\[ImaginaryI] c)
        Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a S)/
        r] (\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r])) + 
  1/(2 a)(r s + S)^(-\[ImaginaryI] c) (-((\[ImaginaryI] a (r s + S))/
     r))^(-\[ImaginaryI] c) ((r s + S)^(2 \[ImaginaryI] c)
       Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a (r s + S))/
        r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
       a^2 (r s + S)^2)/r^2)^(\[ImaginaryI] c)
       Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a (r s + S))/
       r] (\[ImaginaryI] Cos[b - (a S)/r] + 
        Sin[b - (a S)/r])) // FullSimplify

Out[3]= (1/(2 r))\[ExponentialE]^(-((\[ImaginaryI] (b r + a S))/
  r)) (\[ExponentialE]^(2 \[ImaginaryI] b) S^(1 + \[ImaginaryI] c)
     ExpIntegralE[-\[ImaginaryI] c, -((\[ImaginaryI] a S)/
      r)] - \[ExponentialE]^(2 \[ImaginaryI] b) (r s + S)^(
    1 + \[ImaginaryI] c)
     ExpIntegralE[-\[ImaginaryI] c, -((\[ImaginaryI] a (r s + S))/
      r)] + \[ExponentialE]^((2 \[ImaginaryI] a S)/
    r) (S^(1 - \[ImaginaryI] c)
        ExpIntegralE[\[ImaginaryI] c, (\[ImaginaryI] a S)/
        r] - (r s + S)^(1 - \[ImaginaryI] c)
        ExpIntegralE[\[ImaginaryI] c, (\[ImaginaryI] a (r s + S))/r]))

It is already comprehensible. Probably you may go still further by checking properties of the integral exponents. 
Have a look into the book: Abramowitz, M. & Stegun, I. A. Handbook of Mathematical Functions with formulas, 
graphs and mathematical tables. (National Bureau of Standards, 1964). Another way I would go is to try to transform it first, 
and to integrate afterwards. 

Success, Alexei


-- 
Alexei Boulbitch, Dr., Habil.
Senior Scientist

IEE S.A.
ZAE Weiergewan
11, rue Edmond Reuter
L-5326 Contern
Luxembourg

Phone: +352 2454 2566
Fax:   +352 2454 3566

Website: www.iee.lu

This e-mail may contain trade secrets or privileged, undisclosed or otherwise confidential information. If you are not the intended recipient and have received this e-mail in error, you are hereby notified that any review, copying or distribution of it is strictly prohibited. Please inform us immediately and destroy the original transmittal from your system. Thank you for your co-operation.




  • Prev by Date: Re: integration
  • Next by Date: RE: RE: Comparison between Mathematica and other
  • Previous by thread: Re: integration
  • Next by thread: Re: integration