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Re: Re: Solving a Integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg78496] Re: [mg78389] Re: Solving a Integral
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Tue, 3 Jul 2007 05:26:36 -0400 (EDT)
  • References: <f5dkds$no$1@smc.vnet.net> <f5qi0d$49f$1@smc.vnet.net> <13143474.1183201922172.JavaMail.root@m35> <op.turq64t3qu6oor@monster.ma.dl.cox.net>
  • Reply-to: drmajorbob at bigfoot.com

Oops... there are a couple of places I used "second" when it should be  
"power".

I'd changed the name but not completely.

Bobby

On Sat, 30 Jun 2007 23:38:54 -0500, DrMajorBob <drmajorbob at bigfoot.com> 
wrote:

> FWIW, here's a more drawn-out thought process for Maxim's derivation:
>
> The problem is
>
> constant Integrate[integrand, {x, 0, Infinity}]
>
> where
>
> constant = beta^-alpha/Gamma[alpha];
> integrand = \[ExponentialE]^-x/beta /(1 + \[ExponentialE]^(-b - a x))
>     x^(-1 + alpha);
>
> The integrand divides into a power of x and a part that looks sort of 
> geometric:
>
> almostGeometric = Take[integrand, 2]
> power = Last@integrand
>
> \[ExponentialE]^-x/beta/(1 + \[ExponentialE]^(-b - a x))
>
> x^(-1 + alpha)
>
> and the geometric part we need is
>
> geometric = almostGeometric/\[ExponentialE]^(b + a x - x/beta)
>
> \[ExponentialE]^(-b - a x)/(1 + \[ExponentialE]^(-b - a x))
>
> Just checking our memory:
>
> Normal@Series[z/(1 + z), {z, 0, 10}]
>
> z - z^2 + z^3 - z^4 + z^5 - z^6 + z^7 - z^8 + z^9 - z^10
>
> The substitution rule, the summand, and the sum are:
>
> rule = z -> \[ExponentialE]^(-b - a x);
> summand[n_] = (-1)^n z^(n + 1);
> sum=Sum[summand[n], {n, 0, Infinity}]
>
> z/(1 + z)
>
> So the integrand is
>
> integrand == second almostGeometric/geometric (sum /. rule)
>
> True
>
> Integrating the nth term (including the constant out front), we get:
>
> int[n_] =
>   Assuming[{alpha > 0, beta > 0, a > 0, b < 0, n >= 0},
>    Integrate[
>     constant  second almostGeometric/
>       geometric (summand[n] /. rule), {x, 0, \[Infinity]}]]
>
> (-1)^n \[ExponentialE]^(-b n) (1 + a beta n)^-alpha
>
> and the sum of these is
>
> Sum[int[n], {n, 0, Infinity}]
>
> LerchPhi[-E^(-b), alpha, 1/(a*beta)]/(a*beta)^alpha
>
> I couldn't get the INDEFINITE integral this way, BTW.
>
> Bobby
>
> On Sat, 30 Jun 2007 05:03:37 -0500, dimitris <dimmechan at yahoo.com> wrote:
>
>>
>> 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
>>
>> Amazing solution!
>> Mind (more...) + Mathematica (less...) triumph!
>>
>> Dimitris
>>
>>
>>
>
>
>



-- =

DrMajorBob at bigfoot.com


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