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Re: Integral Sum

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  • Subject: [mg131328] Re: Integral Sum
  • From: Bob Hanlon <hanlonr357 at gmail.com>
  • Date: Sat, 29 Jun 2013 04:53:50 -0400 (EDT)
  • Approved: Steven M. Christensen <steve@smc.vnet.net>, Moderator
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  • References: <20130628081253.9924069D8@smc.vnet.net>

rule = Integrate[expr1_, {z_, a_, b_}] +
    Integrate[expr2_, {z_, a_, b_}] :>

   Integrate[Simplify[expr1 + expr2], {z, a, b}];


Integrate[2*Exp[-z[1]*x]*x*V[x]/(z[1]*(z[1] + z[j])^2), {x, 0, Infinity}] +
  Integrate[
   Exp[-z[1]*x]*x*V[x]/(z[1]^2*(z[1] + z[j])^2), {x, 0, Infinity}] /. rule


Integrate[(x*V[x]*(1 + 2*z[1]))/
     (E^(x*z[1])*(z[1]^2*(z[1] + z[j])^2)),
   {x, 0, Infinity}]



Bob Hanlon




On Fri, Jun 28, 2013 at 4:12 AM, <losze1cj at gmail.com> wrote:

> Does any know why mathematica won't combine these integrals
> Integrate[
>   2*Exp[-z[1]*x]*x*V[x]/(z[1]*(z[1] + z[j])^2), {x, 0, Infinity}] +
>  Integrate[
>   Exp[-z[1]*x]*x*V[x]/(z[1]^2*(z[1] + z[j])^2), {x, 0, Infinity}]
> and how I can get it to?
>
> I am assuming that V[x] is simply a polynomial in x, so that the integrals
> are convergent.
>
> I would like to see this expression simplify to be
> Integrate[
>  2*Exp[-z[1]*x]*x*V[x]/(z[1]*(z[1] + z[j])^2) +
>   Exp[-z[1]*x]*x*V[x]/(z[1]^2*(z[1] + z[j])^2), {x, 0, Infinity}], so that
> I can ultimately get to the simplified expression
> FullSimplify[
>  Integrate[
>   2*Exp[-z[1]*x]*x*V[x]/(z[1]*(z[1] + z[j])^2) +
>    Exp[-z[1]*x]*x*V[x]/(z[1]^2*(z[1] + z[j])^2), {x, 0, Infinity}]].
>
> Any tips?
> Thanks.
>
>


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