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MathGroup Archive 2006

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Re: error function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg65996] Re: [mg65985] error function
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Wed, 26 Apr 2006 04:37:55 -0400 (EDT)
  • References: <200604250919.FAA07113@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

ann wrote:
> Hello everybody,
>  I would be greatful if someone can help me with the following
> problem..
> 
> 
> i have a task of integrating a complicated funtion  which upon
> execution gives the result with Erf[] how to get it solved.
> 
> To say the problem in detail , i have to do the following operation
> 
> Integrate[ (F^2 + C*G^2 ) * Real part of (F*complex conjugate of F *
> exp(I*theta) ] with the limits [-infinity to infinity] where F and G
> are of the form
> 
> F=x1 * exp{ (-(t-x2)^2/x3^2 + i/2 * x4(t-xc) + i x4* (t-xc) + i x5 and
> g is of the same form with x replaced with y.
> 
> now this results in the erf[of another complicated function]
> 
> how to get it solved..or to how should i help mathematica to get it
> completely solved??
 >
> expecting suggestions
> 
> thankyou in advance
> 
> ann.

I'm at a bit of a loss to understand what it is you tried to do, and 
what you would like to see happen. What do you wish to solve? What was 
your specific input? Output?

Let's start with something that might represent the integral you have in 
mind, making certain assumptions in translating to viable Mathematica 
format.

ff = x1 * Exp[-(t-x2)^2/x3^2 + I/2*x4*(t-xc) + I*x4*(t-xc) + I*x5]
gg = y1 * Exp[-(t-y2)^2/y3^2 + I/2*y4*(t-yc) + I*y4*(t-yc) + I*y5]

I assume i was meant as I(=Sqrt[-1]), and that t is the variable of 
integration. Moreover where you use Exp[I*theta] I assume you had in 
mind t (so effectively you will be taking a cosine transform). I also 
assume that all constants are real valued and will impose that they are 
positive to make the task easier for Integrate's proviso mechanism (I 
did not check to see if this made a difference, either in speed or in 
the result).

Then the integral can be found as below.

ii = Integrate[(ff^2+gg^2)*
   ComplexExpand[Re[ff*Conjugate[ff]+gg*Conjugate[gg]]]*Cos[t],
   {t,-Infinity,Infinity},
   Assumptions->{x1>0,y1>0,x3>0,x2>0,y2>0,y3>0,x4>0,
     y4>0,xc>0,yc>0,x5>0,y5>0}]

 From this I obtain the result below.

((E^((3*x3^2*x4*y3^2)/(2*(x3^2 + y3^2))) +
     E^(((2*I)*(x3^2*y2 + x2*y3^2))/(x3^2 + y3^2)))*Sqrt[Pi/2]*x1^2*y1^2)/
   (2*E^((16*x2^2 - 8*x2*(4*y2 + I*(-1 + 3*x4)*y3^2) +
       8*(2*y2^2 - I*(2*x5 - 3*x4*xc)*y3^2) +
       x3^2*((-16*I)*x5 + (8*I)*y2 + y3^2 + 9*x4^2*y3^2 +
         6*x4*((4*I)*xc - (4*I)*y2 + y3^2)))/(8*(x3^2 + y3^2)))*
    Sqrt[x3^(-2) + y3^(-2)]) +
  ((E^(((2*I)*(x3^2*y2 + x2*y3^2))/(x3^2 + y3^2)) +
     E^((3*x3^2*y3^2*y4)/(2*(x3^2 + y3^2))))*Sqrt[Pi/2]*x1^2*y1^2)/
   (2*E^((16*x2^2 - 8*x2*(4*y2 + I*y3^2*(-1 + 3*y4)) +
       8*(2*y2^2 - I*y3^2*(2*y5 - 3*y4*yc)) +
       x3^2*((-8*I)*y2*(-1 + 3*y4) + (y3 + 3*y3*y4)^2 +
         (8*I)*(-2*y5 + 3*y4*yc)))/(8*(x3^2 + y3^2)))*
    Sqrt[x3^(-2) + y3^(-2)]) +
  (E^(I*y2*(-1 + 3*y4) - (y3 + 3*y3*y4)^2/16 + (2*I)*y5 - (3*I)*y4*yc)*
    (E^((2*I)*y2) + E^((3*y3^2*y4)/4))*Sqrt[Pi]*y1^4*y3)/4 +
  (E^(I*x2*(-1 + 3*x4) - (x3 + 3*x3*x4)^2/16 + (2*I)*x5 - (3*I)*x4*xc)*
    Sqrt[Pi]*x1^4*x3*(2*(E^((2*I)*x2) + E^((3*x3^2*x4)/4))*
      Sqrt[(8*x2 + I*x3^2*(1 + 3*x4))^2] + E^((2*I)*x2)*
      (8*x2 + I*x3^2*(1 + 3*x4))*Erf[Sqrt[(8*x2 + I*x3^2*(1 + 3*x4))^2]/
        (4*x3)] - E^((2*I)*x2)*(8*x2 + I*x3^2*(1 + 3*x4))*
      Erf[Sqrt[((2*x2)/x3 + (I/4)*(x3 + 3*x3*x4))^2]]))/
   (8*Sqrt[(8*x2 + I*x3^2*(1 + 3*x4))^2])

We'll check at a specific set of values for the parameters.

subs = {x1->1.1,y1->1.7,x3->0.4,x2->1.9,y2->0.2,y3->0.5,x4->0.8,
   y4->1.6,xc->1.1,yc->0.8,x5->1.2,y5->0.9};

In[8]:= ii /. subs

Out[8]= 1.16009 - 2.10437 I

In[9]:= NIntegrate[(ff^2+gg^2)*
   ComplexExpand[Re[ff*Conjugate[ff]+gg*Conjugate[gg]]]*Cos[t] /. subs,
   {t,-Infinity,Infinity}]
Out[9]= 1.16009 - 2.10437 I

So it passes a basic sanity check.

In the development version of Mathematica I get a different, Erf-free 
result:

(E^(I*x2*(-1 + 3*x4) - (x3 + 3*x3*x4)^2/16 + (2*I)*x5 - (3*I)*x4*xc)*
    (E^((2*I)*x2) + E^((3*x3^2*x4)/4))*Sqrt[Pi]*x1^4*x3)/4 +
  (E^(I*y2*(-1 + 3*y4) - (y3 + 3*y3*y4)^2/16 + (2*I)*y5 - (3*I)*y4*yc)*
    (E^((2*I)*y2) + E^((3*y3^2*y4)/4))*Sqrt[Pi]*y1^4*y3)/4 +
  ((E^((3*x3^2*x4*y3^2)/(2*(x3^2 + y3^2))) +
     E^(((2*I)*(x3^2*y2 + x2*y3^2))/(x3^2 + y3^2)))*Sqrt[Pi/2]*x1^2*x3*y1^2*
    y3)/(2*E^((16*x2^2 - 8*x2*(4*y2 + I*(-1 + 3*x4)*y3^2) +
       8*(2*y2^2 - I*(2*x5 - 3*x4*xc)*y3^2) +
       x3^2*((-16*I)*x5 + (8*I)*y2 + y3^2 + 9*x4^2*y3^2 +
         6*x4*((4*I)*xc - (4*I)*y2 + y3^2)))/(8*(x3^2 + y3^2)))*
    Sqrt[x3^2 + y3^2]) + ((E^(((2*I)*(x3^2*y2 + x2*y3^2))/(x3^2 + y3^2)) +
     E^((3*x3^2*y3^2*y4)/(2*(x3^2 + y3^2))))*Sqrt[Pi/2]*x1^2*x3*y1^2*y3)/
   (2*E^((16*x2^2 - 8*x2*(4*y2 + I*y3^2*(-1 + 3*y4)) +
       8*(2*y2^2 - I*y3^2*(2*y5 - 3*y4*yc)) +
       x3^2*((-8*I)*y2*(-1 + 3*y4) + (y3 + 3*y3*y4)^2 +
         (8*I)*(-2*y5 + 3*y4*yc)))/(8*(x3^2 + y3^2)))*Sqrt[x3^2 + y3^2])

It gives the same numeric value on substitution as above.

It is not clear to me if either of these results are what you seek but I 
would guess they at least provide a start for either further computation 
or refining the question.


Daniel Lichtblau
Wolfram Research



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