Re: Re: Integration
- To: mathgroup at smc.vnet.net
- Subject: [mg44451] Re: [mg44437] Re: Integration
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Mon, 10 Nov 2003 04:52:21 -0500 (EST)
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- Sender: owner-wri-mathgroup at wolfram.com
>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
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