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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