How do I perform the following variable change:
IntegrateChangeVariables[Inactive[Integrate][2x E^-x^2,{x,1,\[Infinity]}],u,u==E^-x^2]
Returns unevaluated. Am I missing some condition?
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1 Answer
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I think it is because you have two branches and it does not know which one to use
sol=Solve[u == Exp[-x^2], x, Reals]
When you pick the right one, then it works
sol2 = x /. Last[sol] // Normal (*to remove conditional part*)
IntegrateChangeVariables[Inactive[Integrate][2 x E^-x^2, {x, 1, ∞}], u, x == sol2]
Update
The issue of sign wrong is known and was asked to be reported it before on similar problem with sign.
IntegrateChangeVariables producing incorrect result
(I just noticed it is same OP who asked this question as the above linked to question).
Also note that this function is marked as EXPERIMENTAL so bugs are to be expected? It is better not to use EXPERIMENTAL functions for production code.
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2$\begingroup$ Your result is a negative number
-1/E, whereas the original integral is a positive number1/E. $\endgroup$ Jan 29 at 17:14 -
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Normalwill remove the condition from aConditionalExpression, e.g.,sol2 = x /. Last[sol] // Normal$\endgroup$ Jan 29 at 18:40 -
$\begingroup$ @BobHanlon good suggestion,. I did not know that. Will update, thanks. $\endgroup$– NasserJan 29 at 21:19
Integrate[2 x E^-x^2, {x, 1, \[Infinity]}]which evaluates to1/E$\endgroup$Assuming[x >= 1, IntegrateChangeVariables[ Inactive[Integrate][2 x E^-x^2, {x, 1, \[Infinity]}], u, u == E^-x^2]]However, as with @Nasser answer, the sign of the result is wrong. $\endgroup$