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(Not trivial) Definite Integration of a rational function
*To*: mathgroup at smc.vnet.net
*Subject*: [mg74749] (Not trivial) Definite Integration of a rational function
*From*: "dimitris" <dimmechan at yahoo.com>
*Date*: Tue, 3 Apr 2007 00:32:01 -0400 (EDT)
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}]]
We have
ReleaseHold[f /. Integrate[h_, {x_, a_, b_}] :> Plot[h, {x, a, 10}]]
(*plot*)
ReleaseHold[f /. Integrate -> NIntegrate] (*numerical estimation with
machine precision*)
0.3712169752602472
Attempts to get an analytic result failed.
Timing[ReleaseHold[f]]
{145.969*Second, Integrate[(2*x)/((1 + x)*(1 + 2*x + 3*x^2 + x^3)),
{x, 0, Infinity}]}
Timing[ReleaseHold[f /. Integrate[h___] :> Integrate[h,
GenerateConditions -> False]]]
{144.765*Second, Integrate[(2*x)/((1 + x)*(1 + 2*x + 3*x^2 + x^3)),
{x, 0, Infinity}, GenerateConditions -> False]}
(
BTW, Mathematica version 4 returns Infinity for this definite
integral. I think this happens because a partial fraction
decomposition is performed and the integrand is written
Apart[(2*x)/((x + 1)*(x^3 + 3*x^2 + 2*x + 1))]
-(2/(1 + x)) + (2*(1 + 2*x + x^2))/(1 + 2*x + 3*x^2 + x^3)
and after it integrates definitely each term of last sum
(Integrate[#1, {x, 0, Infinity}] & ) /@ %
Integrate::idiv: Integral of 1/(1 + x) does not converge on
{0,Infinity}.
Integrate::idiv: Integral of (1 + x)^2/(1 + 2*x + 3*x^2 + x^3) does
not \
converge on {0,Infinity}.
Integrate[-(2/(1 + x)), {x, 0, Infinity}] + Integrate[(2*(1 + 2*x +
x^2))/(1 + 2*x + 3*x^2 + x^3), {x, 0, Infinity}]
One additional curious thing is it doesn't show the relevant warning
message for convergence...
)
Let see the antiderivative
ReleaseHold[f /. Integrate[h_, {x_, a_, b_}] :> Integrate[h, x]]
Together[D[%, x]]
Plot[%%, {x, 0, 10}]; (*plot*)
2*(-Log[1 + x] + RootSum[1 + 2*#1 + 3*#1^2 + #1^3 & , (Log[x - #1] +
2*Log[x - #1]*#1 + Log[x - #1]*#1^2)/
(2 + 6*#1 + 3*#1^2) & ])
(2*x)/((1 + x)*(1 + 2*x + 3*x^2 + x^3))
I wonder why Mathematica since it gets the indefinite integral, it
fails to evaluate
the definite integral. Any ideas?
Since both the integrand and the antiderivative posses no
singularities/discontinuities
in the integration range I proceed to get the definite integral by
application of the Newton-Leibniz
formula
This time we get finally the definite integral
Timing[Limit[F, x -> Infinity] - Limit[F, x -> 0]]
Simplify[%[[2]]]
N[%]//Chop
{176.90699999999998*Second, -2*RootSum[1 + 2*#1 + 3*#1^2 + #1^3 & ,
(Log[-#1] + 2*Log[-#1]*#1 + Log[-#1]*#1^2)/(2 + 6*#1 + 3*#1^2)
& ]}
-2*RootSum[1 + 2*#1 + 3*#1^2 + #1^3 & , (Log[-#1] + 2*Log[-#1]*#1 +
Log[-#1]*#1^2)/(2 + 6*#1 + 3*#1^2) & ]
0.37121697526024766
But timings like this look quite unreasonable for integrals like this
(at least for me!).
Do you have other ideas of getting this definite integral (with better
time performances
if possible)?
It is also looks quite strange that in two other CAS (the one can be
freely downloaded from Internet)
I got directly the definite integral in a couple of seconds.
Any explanation about previous time performance of Limit will be
greatly appreciate.
Trying to write down the result
-2*RootSum[1 + 2*#1 + 3*#1^2 + #1^3 & , (Log[-#1] + 2*Log[-#1]*#1 +
Log[-#1]*#1^2)/(2 + 6*#1 + 3*#1^2) & ]
in a expanded format (i e no RootSum object; of course I am aware that
it will be lost the compacteness
that RootSum offers) I tried
Tr[Simplify[(-2*((Log[-#1] + 2*Log[-#1]*#1 + Log[-#1]*#1^2)/(2 + 6*#1
+ 3*#1^2)) & ) /@
(x /. Solve[(1 + 2*#1 + 3*#1^2 + #1^3 & )[x] == 0, x])]]
Chop[N[%]]
-((2*(12 + 2^(1/3)*(27 - 3*Sqrt[69])^(2/3) + 6*3^(1/3)*(2/(9 -
Sqrt[69]))^(2/3))*
Log[1 + (2/(27 - 3*Sqrt[69]))^(1/3) + ((1/2)*(9 -
Sqrt[69]))^(1/3)/3^(2/3)])/
(3*(6 + 2^(1/3)*(27 - 3*Sqrt[69])^(2/3) + 6*3^(1/3)*(2/(9 -
Sqrt[69]))^(2/3)))) -
(2*(8*I*(9 - Sqrt[69])^(2/3) + 2^(1/3)*3^(1/6)*(2*2^(1/3)*3^(1/6)*(-
I + Sqrt[3]) +
(9 - Sqrt[69])^(1/3)*(-9 - 3*I*Sqrt[3] + I*Sqrt[23] +
Sqrt[69])))*
Log[1 + (-1 + I*Sqrt[3])/(2^(2/3)*(27 - 3*Sqrt[69])^(1/3)) - ((1 +
I*Sqrt[3])*((1/2)*(9 - Sqrt[69]))^(1/3))/(2*3^(2/3))])/
(3*(4*I*(9 - Sqrt[69])^(2/3) + 2^(1/3)*3^(1/6)*(2*2^(1/3)*3^(1/6)*(-
I + Sqrt[3]) +
(9 - Sqrt[69])^(1/3)*(-9 - 3*I*Sqrt[3] + I*Sqrt[23] +
Sqrt[69])))) -
(2*(8*(9 - Sqrt[69])^(2/3) + 2^(1/3)*3^(1/6)*(2*I*2^(1/3)*3^(1/6)*(I
+ Sqrt[3]) +
(9 - Sqrt[69])^(1/3)*(-9*I - 3*Sqrt[3] + Sqrt[23] +
I*Sqrt[69])))*
Log[1 + (-1 - I*Sqrt[3])/(2^(2/3)*(27 - 3*Sqrt[69])^(1/3)) + (I*(I
+ Sqrt[3])*((1/2)*(9 - Sqrt[69]))^(1/3))/(2*3^(2/3))])/
(3*(4*(9 - Sqrt[69])^(2/3) +
2^(1/3)*3^(1/6)*(2*I*2^(1/3)*3^(1/6)*(I + Sqrt[3]) +
(9 - Sqrt[69])^(1/3)*(-9*I - 3*Sqrt[3] + Sqrt[23] +
I*Sqrt[69]))))
0.3712169752602468
Am I missing something simpler than this setting?
Is it a built -in function that RootSum and this function acts like
the pair {RootReduce,ToRadicals} below?
Reduce[1 + 2*x + 3*x^2 + x^3 == 0, Reals]
ToRadicals[%]
RootReduce[%]
x == Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 1]
x == -1 - (2/(3*(9 - Sqrt[69])))^(1/3) - ((1/2)*(9 - Sqrt[69]))^(1/3)/
3^(2/3)
x == Root[1 + 2*#1 + 3*#1^2 + #1^3 & , 1]
Thanks!
Dimitris
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