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 09/27/00 06:06am
>The following integral works in version 3 but not in version 4. The result I have given is from version 3.
>Why does it fail in version 4?
>
>In[1]:=
>Integrate[4*Pi*r^2*(Sin[\[Kappa]*r]*r*Erf[r*\[Xi]])/(\[Kappa]*r),
> {r, 0, Infinity}, Assumptions ->
> {Im[\[Kappa]] == 0, Im[\[Xi]] == 0, Re[\[Xi]] > 0, Re[\[Kappa]] > 0,
> \[Kappa] > 0, \[Xi] > 0}]
>
>Out[1]:=
>-((Pi*(\[Kappa]^4 + 2*\[Kappa]^2*\[Xi]^2 + 8*\[Xi]^4))/
> (E^(\[Kappa]^2/(4*\[Xi]^2))*\[Kappa]^4*\[Xi]^4))
>
>
>Version 4 gives the error:
>
>Integrate::''idiv'' : ''Integral of \!\(r\^2\\ \(\(Erf[\(\(r\\ \[Xi]\)\)]\)\)\\ \
>\(\(Sin[\(\(r\\ \[Kappa]\)\)]\)\)\) does not converge on \!\({0, \
>\*InterpretationBox[\''\[Infinity]\'', DirectedInfinity[1]]}\).''
>
>
>My version evaluates to:
>
>In[6]:=
>$Version
>
>Out[6]=
>''4.0 for Microsoft Windows (July 16, 1999)''
>
>In[5]:=
>$ReleaseNumber
>
>Out[5]=
>1
>
>Thank you very much.
>
>Kevin
=======
Since this integral is not convergent as an ordinary integral, the Version 4 result shows the intended behavior. The Version 3 result could be considered correct if Integrate is understood as a generalized integration operation, but that is not the intended design. The intended default behavior is for Integrate to check for basic convergence as in Version 4.
A convenient way to see that the integral is not convergent
as an ordinary integral is to look at a plot of the integrand for numerical values of the symbolic parameters. For example, after replacing both of the symbolic parameters by 1, the input
Plot[4 Pi r^2 (Sin[r] r Erf[r])/(r), {r, 0, 100}]
reveals an integrand with ever-increasing oscillations which can not be expected to converge.
You can get the Version 3 result in Version 4 by including the GenerateConditions -> False option so that Integrate does not check for convergence of the integral:
Integrate[4 Pi r^2 (Sin[\[Kappa] r] r Erf[r \[Xi]])/(\[Kappa] r), {r, 0, Infinity}, GenerateConditions -> False]
This is the intended way of getting the Version 3 result in
Version 4.
Forum Moderator
Response by David Withoff
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