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Topic: Integral Sum
Replies: 1   Last Post: Jun 29, 2013 4:46 AM

 Bob Hanlon Posts: 906 Registered: 10/29/11
Re: Integral Sum
Posted: Jun 29, 2013 4:46 AM

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@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.
>
>