Re: (Not trivial) Definite Integration of a rational function

• To: mathgroup at smc.vnet.net
• Subject: [mg74772] Re: (Not trivial) Definite Integration of a rational function
• From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
• Date: Wed, 4 Apr 2007 04:11:18 -0400 (EDT)
• References: <euslob\$9nf\$1@smc.vnet.net>

```"dimitris" <dimmechan at yahoo.com> wrote:
> I think it is time to open a new thread about integration, isn't it?
>
> Anyway...Here we go..(Mathematica 5.2 is used).
>
> f = HoldForm[Integrate[(2*x)/((x + 1)*(x^3 + 3*x^2 + 2*x + 1)), {x, 0,
> Infinity}]]
[snip]
> Attempts to get an analytic result failed.
[snip]
> I wonder why Mathematica since it gets the indefinite integral, it
> fails to evaluate the definite integral. Any ideas?

Well, it succeeds in getting the following definite integral:

Integrate[(2*x)/((x + 1)*(x^3 + 3*x^2 + 2*x + 1)), {x, 0, x1},
Assumptions -> x1 > 0]

Then take the limit of that result as x1 -> Infinity and Simplify. You
should get

(2/23)*(Log[-Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 3]]*Root[1 + 2*#1 + 3*#1^2 +
#1^3 & , 3]*(7 + 2*Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 3]) -
Log[-Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 1]]*(5 + 4*Root[1 + 2*#1 + 3*#1^2
+ #1^3 & , 3] + 2*Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 3]^2 +
2*Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 2]*(2 + 6*Root[1 + 2*#1 + 3*#1^2 +
#1^3 & , 3] + 3*Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 3]^2) + Root[1 +
2*#1 + 3*#1^2 + #1^3 & , 2]^2*(2 + 6*Root[1 + 2*#1 + 3*#1^2 + #1^3 & ,
3] + 3*Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 3]^2)) + Log[-Root[1 + 2*#1 +
3*#1^2 + #1^3 & , 2]]*(4*Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 2] + 2*Root[1
+ 2*#1 + 3*#1^2 + #1^3 & , 2]^2 -
3*(1 + 2*Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 1]*Root[1 + 2*#1 + 3*#1^2 +
#1^3 & , 3]*(2 + Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 3]) +
Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 1]^2*Root[1 + 2*#1 + 3*#1^2 + #1^3
& , 3]*(2 + Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 3]))))

which is correct.

David

```

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