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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
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