Re: Re: Solving a Integral
- To: mathgroup at smc.vnet.net
- Subject: [mg78450] Re: [mg78389] Re: Solving a Integral
- From: DrMajorBob <drmajorbob at bigfoot.com>
- Date: Mon, 2 Jul 2007 06:47:14 -0400 (EDT)
- References: <f5dkds$no$1@smc.vnet.net> <f5qi0d$49f$1@smc.vnet.net> <13143474.1183201922172.JavaMail.root@m35>
- Reply-to: drmajorbob at bigfoot.com
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