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convergence conditions
*To*: mathgroup at smc.vnet.net
*Subject*: [mg74915] convergence conditions
*From*: "dimitris" <dimmechan at yahoo.com>
*Date*: Wed, 11 Apr 2007 01:56:20 -0400 (EDT)
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.
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,=E2=88=9E}.
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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