MathGroup Archive 2003

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

Search the Archive

Re: Re: Integration


>Should we as users of Mathematica demand from Wolfram either retract
>his false statement or make Mathematica actually do the job?
>
>Alex

As you do not specify, I am not sure what is the statement under
consideration. Here are two possibilities I found in The Mathematica
Book.


"Integrate can evaluate essentially all indefinite integrals and
most definite integrals listed in standard books of tables."



"In the most convenient cases, integrals can be done purely in terms of
elementary functions such as exponentials, logarithms, and trigonometric
functions. In fact, if you give an integrand that involves only such
elementary functions, then [...] if the corresponding integral can be
expressed in terms of elementary functions, then Integrate will
essentially
always succeeed in finding it."

Note that this is a section on indefinite integration. Different claims
are made for definite integrals in the next section, 3.5.8.


As for your example, from a perusal of MathGroup archives I gather it
is as below.

f = ((2*Pi*b)/aa)*(3*ee^2*(xx/lam^2 + I*yy/(lam^2 - ee^2))*
      (1/(lam*(lam^2 - ee^2)*Sqrt[aa^2*lam^2 - l1^2]*
	  Sqrt[aa^2*lam^2 - l2^2])) - 
     (xx/lam^2 + (I*yy)/(lam^2 - ee^2))^3*((aa^2*lam^3*(lam^2 - ee^2))/
       (Sqrt[aa^2*lam^2 - l1^2]*Sqrt[aa^2*lam^2 - l2^2])^3)); 

Here is what I get for the antiderivative.

In[4]:= InputForm[Timing[indefint=Integrate[f,lam]]]

Out[4]//InputForm=
{80.02*Second, (2*b*Pi*(Sqrt[-l1^2 + aa^2*lam^2]*Sqrt[-l2^2 +
aa^2*lam^2]*
     ((xx*(3*l1^2*l2^2 - aa^2*ee^2*xx^2))/(2*l1^4*l2^4*lam^2) + 
      (aa^2*(-(aa^2*ee^2*xx) + l1^2*xx + I*l1^2*yy)^3)/
       (l1^4*(-(aa^2*ee^2) + l1^2)^2*(l1^2 - l2^2)^2*(-l1^2 +
aa^2*lam^2)) - 
      (aa^2*(aa^2*ee^2*xx - l2^2*xx - I*l2^2*yy)^3)/(l2^4*(aa^2*ee^2 -
l2^2)^2*
        (l1^2 - l2^2)^2*(-l2^2 + aa^2*lam^2)) + 
      ((I/2)*yy*(-3*aa^4*ee^4 + 3*aa^2*ee^2*l1^2 + 3*aa^2*ee^2*l2^2 - 
         3*l1^2*l2^2 - aa^2*ee^2*yy^2))/((aa^2*ee^2 - l1^2)^2*
        (aa^2*ee^2 - l2^2)^2*(-ee^2 + lam^2))) + 
    (Sqrt[(l1 - aa*lam)*(l2 - aa*lam)*(l1 + aa*lam)*(l2 + aa*lam)]*
      Sqrt[l1^2 - aa^2*lam^2]*Sqrt[l2^2 - aa^2*lam^2]*
      (-(l1^5*l2^5*(6*l1^4*l2^4*(xx - I*yy) + 6*aa^8*ee^8*(xx -
(2*I)*yy) + 
          aa^2*ee^2*l1^2*l2^2*(-12*l2^2*xx - 3*l1^2*(4*xx - (5*I)*yy) + 
            (15*I)*l2^2*yy + 6*xx*yy^2 + (2*I)*yy^3) + 
          aa^6*ee^6*(-12*l2^2*xx - 3*l1^2*(4*xx - (7*I)*yy) +
(21*I)*l2^2*yy + 
            6*xx*yy^2 - (4*I)*yy^3) + aa^4*ee^4*(l1^4*(6*xx - (9*I)*yy)
+ 
            l1^2*(12*l2^2*(2*xx - (3*I)*yy) + (-6*xx + I*yy)*yy^2) + 
            l2^2*(l2^2*(6*xx - (9*I)*yy) + (-6*xx + I*yy)*yy^2)))*
         Log[(2*Sqrt[l1^2 - aa^2*lam^2]*Sqrt[l2^2 - aa^2*lam^2])/
            (-ee^2 + lam^2) + (-2*l1^2*l2^2 - 2*aa^4*ee^2*lam^2 + 
             aa^2*(l1^2 + l2^2)*(ee^2 + lam^2))/(Sqrt[aa^2*ee^2 - l1^2]*
             Sqrt[aa^2*ee^2 - l2^2]*(ee^2 - lam^2))]) - (aa^2*ee^2 -
l1^2)^(5/2)*
        (aa^2*ee^2 - l2^2)^(5/2)*(3*aa^4*ee^4*(l1^2 + l2^2)*xx^3 - 
         aa^2*ee^2*l1^2*l2^2*xx*(3*l1^2 + 3*l2^2 + 2*xx*(xx + (3*I)*yy))
- 
         6*l1^4*l2^4*(xx - I*yy))*Log[-((aa^2*(l1^2 + l2^2))/(l1*l2)) + 
          (2*(l1*l2 + Sqrt[l1^2 - aa^2*lam^2]*Sqrt[l2^2 -
aa^2*lam^2]))/lam^2]))/
     (4*ee^2*l1^5*Sqrt[aa^2*ee^2 - l1^2]*l2^5*Sqrt[aa^2*ee^2 - l2^2]*
      (aa^4*ee^4 - aa^2*ee^2*l1^2 - aa^2*ee^2*l2^2 + l1^2*l2^2)^2*
      Sqrt[-((l1 - aa*lam)*(l1 + aa*lam))]*Sqrt[-((l2 - aa*lam)*(l2 +
aa*lam))]*
      Sqrt[(l1^2 - aa^2*lam^2)*(l2^2 - aa^2*lam^2)])))/aa}

To make sense of the definite integral you will need to provide
reasonable
Assumptions to restrict parameter ranges. Otherwise the integration code
will almost certainly bog down in nontrivial computations involving
parameter conditions that get spawned in the process.


Daniel Lichtblau
Wolfram Research


  • Prev by Date: Re: Re: Integration
  • Next by Date: Re: Trying to use Mathematica as "word processor" for my math homework
  • Previous by thread: Re: Re: Re: Re: Integration
  • Next by thread: Re: Re: Re: Integration