Re: Integral Sum
- To: undisclosed-recipients:;
- 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
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-outx@smc.vnet.net
- Delivered-to: mathgroup-newsendx@smc.vnet.net
- Newsgroups: comp.soft-sys.math.mathematica
- 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.
>
>
- References:
- Integral Sum
- From: losze1cj@gmail.com
- Integral Sum