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MathGroup Archive 2007

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Re: Re: Indefinite Integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg77613] Re: [mg77558] Re: Indefinite Integral
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Thu, 14 Jun 2007 05:13:33 -0400 (EDT)
  • References: <f4j0ij$j7n$1@smc.vnet.net> <200706131121.HAA06594@smc.vnet.net>

dimitris wrote:
> Hello Joel.
> 
> I use Mathematica 5.2 in the following.
> 
> Here is the integrand.
> 
> In[1]:=
> int[a_, b_, p_, x_] := Exp[(-p)*x]*(Sin[x]/(a^2*Sin[x]^2 +
> b^2*Cos[x]^2)^(3/2))
> 
> As you may notice (otherwise you would not have posted the query!)
> Mathematica returns the indefinite integral unevaluated.
> 
> Here are two subcases of the parameters that Mathematica
> returns an antiderivative.
> 
> In[17]:=
> Integrate[int[a, a, p, x], x]
> Out[17]=
> -((Cos[x] + p*Sin[x])/(E^(p*x)*((a^2)^(3/2)*(1 + p^2))))
> 
> In[18]:=
> Integrate[int[a, b, 0, x], x]
> Out[18]=
> -((Sqrt[2]*Cos[x])/(a^2*Sqrt[a^2 + b^2 + (-a^2 + b^2)*Cos[2*x]]))
> 
> But I am sure you have also checked this.
> 
> When Mathematica does not succeed, we should not give up
> but trying to help it!
> Here (I assume that you have consult the well known collection
> of tables in literature; I didn't have time to search them for your
> integral) I
> tried various approaches as substitutions, integration by parts etc
> but with
> unsatisfactory results.
> 
> For the end I left one approach which you should always have in mind
> for parametric integrals like yours!
> 
> This approach consists of the following steps:
> 
> Take the indefinite integral with respect to a and b.
> 
> In[27]:=
> Int=Integrate[int[a, b, p, x], a, b]
> Out[27]=
> -((Sqrt[a^2 + b^2 + (-a^2 + b^2)*Cos[2*x]]*Csc[x]*Sec[x]^2)/
> (E^(p*x)*(Sqrt[2]*a*b)))
> 
> (*Check*)
> 
> In[34]:=
> Simplify[D[%, a, b] == int[a, b, p, x]]
> Out[34]=
> True
> 
> Now take the antiderivative of Int wrt x.
> 
> In[95]:=
> Integrate[Int, x]
> 
> Out[95]=
> -((1/(Sqrt[2]*a*b))*((Sqrt[a^2 + b^2 + (-a^2 + b^2)*Cos[2*x]]*Sec[x])/
> E^(p*x) +
>     (3*a^2 + b^2)*p*Integrate`TableMatchTrig1[((-(1/2))*I*Csc[2*x])/
> E^((-I + p)*x), x, 0] +
>     4*b^2*Integrate`TableMatchTrig1[((1/2)*Csc[2*x])/E^((-I + p)*x),
> x, 0] +
>     3*a^2*p*Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((I +
> p)*x), x, 0] +
>     b^2*p*Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((I + p)*x),
> x, 0] +
>     4*b^2*Integrate`TableMatchTrig1[((1/2)*Csc[2*x])/E^((I + p)*x), x,
> 0] -
>     a^2*p*Integrate`TableMatchTrig1[((-(1/2))*I*Csc[2*x])/E^((-3*I +
> p)*x), x, 0] +
>     b^2*p*Integrate`TableMatchTrig1[((-(1/2))*I*Csc[2*x])/E^((-3*I +
> p)*x), x, 0] -
>     a^2*p*Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((3*I +
> p)*x), x, 0] +
>     b^2*p*Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((3*I +
> p)*x), x, 0]))
> 
> Not exactly a readable result; but still a result!

Umm, yeah. But pretty clearly a buggy result.


> I am wondering what version 6 returns here!

Mine returns the integral unevaluated.


> Take now the derivatives wrt to a,b of the last expression and you
> get
> 
> In[135]:=
> res=FullSimplify[D[%, a, b]] (*wait a couple of minutes for the
> simplification!*)

What you are doing at this point could be categorized as "garbage out, 
garbage in". Not recommended procedure...


> Out[135]=
> (1/Sqrt[2])*((4*Cos[x]*Sin[x]^2)/(E^(p*x)*(a^2 + b^2 + (-a^2 +
> b^2)*Cos[2*x])^(3/2)) +
>    (1/a^2 + 3/b^2)*p*Integrate`TableMatchTrig1[((-(1/2))*I*Csc[2*x])/
> E^((-I + p)*x), x, 0] +
>    (4*Integrate`TableMatchTrig1[((1/2)*Csc[2*x])/E^((-I + p)*x), x,
> 0])/a^2 +
>    (1/(a^2*b^2))*((3*a^2 +
> b^2)*p*Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((I + p)*x), x,
> 0] +
>      4*b^2*Integrate`TableMatchTrig1[((1/2)*Csc[2*x])/E^((I + p)*x),
> x, 0] -
>      (a - b)*(a + b)*p*(Integrate`TableMatchTrig1[((-
> (1/2))*I*Csc[2*x])/E^((-3*I + p)*x), x, 0] +
>        Integrate`TableMatchTrig1[((1/2)*I*Csc[2*x])/E^((3*I + p)*x),
> x, 0])))
> 
> And this is the desired result! A RESULT? This mesh?

Well, garbage back out. No surprise at this step.


> Now if I could see your face I am sure that you would look quite
> curious
> about how someone could use this result!
 >
> You asked for a closed form solution in Mathematica!
> That is, what Mathematica (in collaboration with mind...) returns!

I'd say this one was not a better day for either partner.


> I sent my reply to Daniel Lichtblau of WRI for more details.
> 
> Best Regards
> Dimitris
 >
> 
>  /  Joel Storch       :
> 
>>I would like a closed-form expression for
>>
>>Integrate[Exp[-p x] Sin[x]/(a^2 Sin[x]^2+b^2 Cos[x]^2)^(3/2),x]
>>
>>where a,b & p are positive constants


Not much more to say so I'll paraphrase (read: shamelessly plagiarize) 
Gertrude Stein: A bug is a bug is a bug.


Daniel Lichtblau
Wolfram Research





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