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MathGroup Archive 2007

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Re: Solving a Integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg78390] Re: Solving a Integral
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Sat, 30 Jun 2007 06:04:08 -0400 (EDT)
  • References: <f5dkds$no$1@smc.vnet.net><f5qi0d$49f$1@smc.vnet.net>

m... at inbox.ru       :
> On Jun 21, 5:37 am, ehrnsperge... at pg.com wrote:
> > I need help in solving the following integral:
> >
> > Integral = 1/(beta^alpha* Gamma[alpha]) *
> > Integrate[x^(alpha-1)*Exp[-x/beta]/(1+Exp[-a*x-b]),{x,0, infinity},
> > Assumptions: (alpha> 0)||(beta > 0)||(a > 0)||(b <0)]
> >
> > The Integral is approximately 1/(beta^alpha* Gamma[alpha])
> > *1/(1+Exp[-a*alpha*beta-b]) + Order[alpha*beta^2]
> >
> > However, I would like to have an exact analytical solution, and I am
> > failing to convince Mathematica to give me the solution. Is there a way to
> > ask Mathematica to give the solution as a series expansion of my
> > approximate solution?
> >
> > Thanks so much for your help,
> >
> > Bruno
> >
> > Dr. Bruno Ehrnsperger
> > Principal Scientist
> >
> > Procter & Gamble Service GmbH
> > Sulzbacherstr.40
> > 65824 Schwalbach
> > Germany
> >
> > fon +49-6196-89-4412
> > fax +49-6196-89-22965
> > e-mail: ehrnsperge... at pg.com
> > internet:www.pg.com
> >
> > Gesch=E4ftsf=FChrer: Otmar W. Debald, Gerhard Ritter, Dr. Klaus Schumann,
> > Willi Schwerdtle
> > Sitz: Sulzbacher Str. 40, 65824 Schwalbach am Taunus, Amtsgericht:
> > K=F6nigstein im Taunus HRB 4990
>
> Expand the integrand into a series of exponents:
>
> In[1]:= 1/(beta^alpha Gamma[alpha]) x^(alpha - 1) E^(-x/beta)/
>     (1 + E^(-a x - b)) ==
>   1/(beta^alpha Gamma[alpha]) Sum[
>     (-1)^k x^(alpha - 1) E^((-a k - 1/beta) x - b k),
>     {k, 0, Infinity}] // Simplify
>
> Out[1]= True
>
> In[2]:= Assuming[{alpha > 0, beta > 0, a > 0, k >= 0},
>   Integrate[(-1)^k x^(alpha - 1) E^((-a k - 1/beta) x - b k),
>     {x, 0, Infinity}]]
>
> Out[2]= (-1)^k E^(-b k) (beta/(1 + a beta k))^alpha Gamma[alpha]
>
> In[3]:= 1/(beta^alpha Gamma[alpha]) Sum[
>   % /. (x_/y_)^p_ :> (1/Expand[y/x])^p, {k, 0, Infinity}]
>
> Out[3]= a^-alpha beta^-alpha LerchPhi[-E^-b, alpha, 1/(a beta)]
>
> Maxim Rytin
> m.r at inbox.ru

No problem.
I understand how you found the sum.

Dimitris



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