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MathGroup Archive 2007

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convergence conditions correction

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74916] convergence conditions correction
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Wed, 11 Apr 2007 01:56:51 -0400 (EDT)

(Part of Mathematica code was ommited in previous post. The present
version is
to be considered and ignore the first one.)

Hello.

Consider the integral

In[642]:=
f = HoldForm[Integrate[1/(x^2 + 2*a*x + 1), {x, 0, Infinity}]]

The following provides enough evidence that the integral converges
for a>-1.

In[679]:=
1/(x^2 + 2*a*x + 1) + O[x, 0]^4
1/(x^2 + 2*a*x + 1) + O[x, Infinity]^4
Reduce[x^2 + 2*a*x + 1 == 0 && x > 0 && a <= -1, x, Reals]
Reduce[x^2 + 2*a*x + 1 == 0 && x > 0 && a > -1, x, Reals]

Out[679]=
SeriesData[x, 0, {1, -2*a, -1 + 4*a^2, 4*a - 8*a^3}, 0, 4, 1]
Out[680]=
SeriesData[x, Infinity, {1, -2*a}, 2, 4, 1]
Out[681]=
(a < -1 && (x == -a - Sqrt[-1 + a^2] || x == -a + Sqrt[-1 + a^2])) =
||
(a == -1 && x == 1)
Out[682]=
False


However Mathematica returns

In[684]:=
ReleaseHold[f /. Integrate[x___] :> Integrate[x, Assumptions -> a >
-1]]
Out[684]=
If[a >= Re[Sqrt[-1 + a^2]] || Im[Sqrt[-1 + a^2]] != 0, (Pi -
4*ArcTan[a/Sqrt[1 - a^2]] + 2*I*Log[-(I/Sqrt[1 - a])] +
    I*Log[1 - a])/(4*Sqrt[1 - a^2]), Integrate[1/(1 + 2*a*x + x^2),
{x, 0, Infinity},
   Assumptions -> Im[Sqrt[-1 + a^2]] == 0 && a < Re[Sqrt[-1 + a^2]]]]

which I do consider quite complicated.

However

In[689]:=
ReleaseHold[f /. Integrate[x___] :> Integrate[x, Assumptions -> a >
1]]
ReleaseHold[f /. Integrate[x___] :> Integrate[x, Assumptions -> a ==
1]]
ReleaseHold[f /. Integrate[x___] :> Integrate[x, Assumptions -> -1 < a
< 1]]

Out[689]=
Log[-1 + 2*a*(a + Sqrt[-1 + a^2])]/(2*Sqrt[-1 + a^2])
Out[690]=
1
Out[691]=
ArcCos[a]/Sqrt[1 - a^2]

In[697]:=
ReleaseHold[f /. Integrate[x___] :> Integrate[x, Assumptions -> a <=
-1]]
Integrate::gener: Unable to check convergence.
Integrate::idiv: Integral of 1/(1 + 2*a*x + x^2) does not converge on
{0,=C3=A2=CB=86=C5=BE}.
Out[697]=
Integrate[1/(1 + 2*a*x + x^2), {x, 0, Infinity}, Assumptions -> a <
-1]

So, in view of these results (both are correct) would be better for
Integrate
to return something like the following for the setting Assumptions-
>a>-1?

ReleaseHold[f /. Integrate[x___] :> Integrate[x, Assumptions -> a
>-1]]
If[-1<a<1, ArcCos[a]/Sqrt[1 - a^2], Log[-1 + 2*a*(a + Sqrt[-1 +
a^2])]/
(2*Sqrt[-1 + a^2])]


Dimitris



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