Mathematica Asked on January 20, 2021
I’ m writing a function with the following line:
f[α_,x_, y_] := f[α,x,y] =
Assuming[α > 0,Normalize[ E^(-α √(x^2 + y^2)),
Integrate[#,{x,-∞,∞},{y,-∞,∞}]&]];
The output I have is
ConditionalExpression[(E^(-Sqrt[x^2 + y^2] α) α^2)/(
2 π), Re[α] > 0]
I don’ t understand why it does ignore a constraint I gave in the assumptions.
Thanks
Integrate[
E^(-[Alpha] [Sqrt](x^2 +
y^2)), {x, -[Infinity], [Infinity]}, {y, -[Infinity],
[Infinity]}, Assumptions -> a > 0]
(ConditionalExpression[(2 [Pi])/[Alpha]^2, Re[[Alpha]] > 0]
)
There is an internal parameter to the notebooks called $Assumption
.
There is the documentation page for Assumptions
for the built-ins Simplify
, Refine
and Integrate
.
And there is Assuming
. Assuming
is the mightiest of them but seldom in need. As my solution shows the internal option is well suited with the default settings of $Assumption
for solving this appropriate.
By default Mathematica assumes that parameters as Complexes
.
So the output does take Your assumption in the form of the constrain in the ConditionalExpression: α > 0 -> Re[α] > 0
Mightier is than
Assuming[[Alpha] > 0 && Element[[Alpha], Reals],
Normalize[E^(-[Alpha] [Sqrt](x^2 + y^2)),
Integrate[#, {x, -[Infinity], [Infinity]}, {y, -[Infinity],
[Infinity]}] &]];
If the intention is to maintain the ConditionalExpression the Normalize has to go outmost and the internal Assumptions have to be used.
Normalize[E^(-[Alpha] [Sqrt](x^2 + y^2)),
Integrate[#, {x, -[Infinity], [Infinity]}, {y, -[Infinity],
[Infinity]}, Assumptions -> [Alpha] > 0] &]
(* ConditionalExpression[(E^(-Sqrt[x^2 + y^2] [Alpha]) [Alpha]^2)/( 2 [Pi]), Re[[Alpha]] > 0] *)
Normalize[E^(-[Alpha] [Sqrt](x^2 + y^2)),
Integrate[#, {x, -[Infinity], [Infinity]}, {y, -[Infinity],
[Infinity]}] &]
(* ConditionalExpression[(E^(-Sqrt[x^2 + y^2] [Alpha]) [Alpha]^2)/( 2 [Pi]), Re[[Alpha]] > 0] *)
Excuse. The concept is either the assumption is given or computized by internals. Now flaw, no error, nothing forgotten.
Normalize[E^(-[Alpha] [Sqrt](x^2 + y^2)),
Integrate[#, {x, -[Infinity], [Infinity]}, {y, -[Infinity],
[Infinity]},
Assumptions -> ([Alpha] > 0 && Element[[Alpha], Reals]) &]]
E^(-Sqrt[x^2 +
y^2] [Alpha])/(Assumptions -> [Alpha] >
0 && ([Alpha] [Element] Reals) & (!(
*SubsuperscriptBox[([Integral]), (-[Infinity]),
([Infinity])](
*SubsuperscriptBox[([Integral]), (-[Infinity]),
([Infinity])]1 [DifferentialD]y [DifferentialD]x)))
#1)[E^(-Sqrt[x^2 + y^2] [Alpha])]
Which is not nicely evaluable. With Rules
this works.
Correct answer by Steffen Jaeschke on January 20, 2021
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