Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2007
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2007

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

Search the Archive

Re: Indefinite Integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg77676] Re: Indefinite Integral
  • From: Valeri Astanoff <astanoff at gmail.com>
  • Date: Thu, 14 Jun 2007 06:27:15 -0400 (EDT)
  • References: <f4j0ij$j7n$1@smc.vnet.net>

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