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