MathGroup Archive 2007

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

Search the Archive

Re: Indefinite Integral


On 11 juin, 10:19, Joel Storch <jsto... at earthlink.net> wrote:
> 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

Good day,

Just in case you didn't try it, an idea that could be useful
is to expand the periodic factor in Fourier trig series
to get an approximate expression of the integral (very messy)
and its asymptotic value (less messy):


In[1]:=<<Calculus`FourierTransform`

In[2]:=fs=FourierTrigSeries[Sin[x]/(a^2*Sin[x]^2+
b^2*Cos[x]^2)^(3/2),x,5,FourierParameters->{0,1/(2Pi)}] ;

(order 5 seems to be sufficient for non pathological cases)

In[3]:=F[x_] = Integrate[Exp[(-p)*x]*fs, x];

In[4]:=FF = Simplify[Limit[F[x] - F[0], x -> Infinity]]

Out[4]=(2*((-b^2)*(9*b^4*(65 + 2*p^2 + p^4) + 2*a^2*b^2*(495 +
1246*p^2 + 175*p^4) + a^4*(4185 + 3890*p^2 + 281*p^4))*
EllipticE[1 - a^2/b^2] + a*((-b)*(9*b^4*(65 + 2*p^2 + p^4) +
2*a^2*b^2*(495 + 1246*p^2 + 175*p^4) + a^4*(4185 + 3890*p^2 +
281*p^4))*EllipticE[1 - b^2/a^2] + (9*b^4*(145 + 98*p^2 +
17*p^4) + 3*a^4*(675 + 518*p^2 + 35*p^4) + 2*a^2*b^2*(1215 +
1982*p^2 + 191*p^4))*(a*EllipticK[1 - a^2/b^2] +
b*EllipticK[1 - b^2/a^2]))))/(3*a^2*b*(a^2 - b^2)^3*
(225 + 259*p^2 + 35*p^4 + p^6)*Pi)


Check asymptotic value when b -> a :

In[5]:=Limit[FF,b->a]//Simplify

Out[5]=1/(a^3*(1 + p^2))

In[6]:=Integrate[Exp[(-p)*x]*(Sin[x]/(a^2*Sin[x]^2 +
a^2*Cos[x]^2)^(3/2)), {x, 0, Infinity}]

Out[6]=1/(a^3*(1 + p^2))

hth

V.Astanoff



  • Prev by Date: Re: Overly complicated reductions?
  • Next by Date: Re: Tabbed notebooks?
  • Previous by thread: Re: Re: Indefinite Integral
  • Next by thread: indefinite integral