Re: Solving a Integral
- To: mathgroup at smc.vnet.net
- Subject: [mg78210] Re: Solving a Integral
- From: m.r at inbox.ru
- Date: Tue, 26 Jun 2007 04:08:45 -0400 (EDT)
- References: <f5dkds$no$1@smc.vnet.net>
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
>
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> 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