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Integrating function with /; in its definition


How to make InverseFunction work on function defined with restriction on argumentsIntegrating a periodic functionHow to Integrate trivial products of DiracDeltaIntegrating expressions with several terms and delta functionsSubstitute a function's definition for its nameMathematica `Integrate` evaluates some integrals only with some specific scaling factorPlotting a function involving Nsolve in its definitionDefinition of a periodic function doesn't work when given as the forcing function of an ODEProblem with pendulum code cant get to plot Mathematica 11.3Define Function using another Function and plot













3












$begingroup$


why



f[x_ /; x>=0]:=x;
Integrate[f[x] ,{x,0,2 Pi}]


return unevaluated? Notice that the above definition of f[x] works OK with other Mathematica functions, such as Plot



Plot[f[x], {x, 0, 2 Pi}]


Mathematica graphics



While the following works with Integrate



f[x_]:=x;
Integrate[f[x] ,{x,0,2 Pi}]


I am using version 11.3 on windows.










share|improve this question











$endgroup$








  • 3




    $begingroup$
    It's better to use ConditionalExpression, e.g., Integrate[ConditionalExpression[x, x>0], {x, 0, 2Pi}]
    $endgroup$
    – Carl Woll
    3 hours ago
















3












$begingroup$


why



f[x_ /; x>=0]:=x;
Integrate[f[x] ,{x,0,2 Pi}]


return unevaluated? Notice that the above definition of f[x] works OK with other Mathematica functions, such as Plot



Plot[f[x], {x, 0, 2 Pi}]


Mathematica graphics



While the following works with Integrate



f[x_]:=x;
Integrate[f[x] ,{x,0,2 Pi}]


I am using version 11.3 on windows.










share|improve this question











$endgroup$








  • 3




    $begingroup$
    It's better to use ConditionalExpression, e.g., Integrate[ConditionalExpression[x, x>0], {x, 0, 2Pi}]
    $endgroup$
    – Carl Woll
    3 hours ago














3












3








3





$begingroup$


why



f[x_ /; x>=0]:=x;
Integrate[f[x] ,{x,0,2 Pi}]


return unevaluated? Notice that the above definition of f[x] works OK with other Mathematica functions, such as Plot



Plot[f[x], {x, 0, 2 Pi}]


Mathematica graphics



While the following works with Integrate



f[x_]:=x;
Integrate[f[x] ,{x,0,2 Pi}]


I am using version 11.3 on windows.










share|improve this question











$endgroup$




why



f[x_ /; x>=0]:=x;
Integrate[f[x] ,{x,0,2 Pi}]


return unevaluated? Notice that the above definition of f[x] works OK with other Mathematica functions, such as Plot



Plot[f[x], {x, 0, 2 Pi}]


Mathematica graphics



While the following works with Integrate



f[x_]:=x;
Integrate[f[x] ,{x,0,2 Pi}]


I am using version 11.3 on windows.







calculus-and-analysis function-construction






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 15 mins ago









J. M. is computer-less

97.3k10303463




97.3k10303463










asked 3 hours ago









NasserNasser

58.1k489206




58.1k489206








  • 3




    $begingroup$
    It's better to use ConditionalExpression, e.g., Integrate[ConditionalExpression[x, x>0], {x, 0, 2Pi}]
    $endgroup$
    – Carl Woll
    3 hours ago














  • 3




    $begingroup$
    It's better to use ConditionalExpression, e.g., Integrate[ConditionalExpression[x, x>0], {x, 0, 2Pi}]
    $endgroup$
    – Carl Woll
    3 hours ago








3




3




$begingroup$
It's better to use ConditionalExpression, e.g., Integrate[ConditionalExpression[x, x>0], {x, 0, 2Pi}]
$endgroup$
– Carl Woll
3 hours ago




$begingroup$
It's better to use ConditionalExpression, e.g., Integrate[ConditionalExpression[x, x>0], {x, 0, 2Pi}]
$endgroup$
– Carl Woll
3 hours ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

f[x_ /; x>=0]:=x means "if whatever>=0 rewrite f[whatever] as whatever. But that doesn't apply to f[x] when x is a symbol without a numerical value. Thus, f[x] simply remains f[x]. For abstracting the notion of a function with a break like this, use Piecewise or HeavisideTheta: Integrate understands what those mean.






share|improve this answer









$endgroup$













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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    f[x_ /; x>=0]:=x means "if whatever>=0 rewrite f[whatever] as whatever. But that doesn't apply to f[x] when x is a symbol without a numerical value. Thus, f[x] simply remains f[x]. For abstracting the notion of a function with a break like this, use Piecewise or HeavisideTheta: Integrate understands what those mean.






    share|improve this answer









    $endgroup$


















      5












      $begingroup$

      f[x_ /; x>=0]:=x means "if whatever>=0 rewrite f[whatever] as whatever. But that doesn't apply to f[x] when x is a symbol without a numerical value. Thus, f[x] simply remains f[x]. For abstracting the notion of a function with a break like this, use Piecewise or HeavisideTheta: Integrate understands what those mean.






      share|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        f[x_ /; x>=0]:=x means "if whatever>=0 rewrite f[whatever] as whatever. But that doesn't apply to f[x] when x is a symbol without a numerical value. Thus, f[x] simply remains f[x]. For abstracting the notion of a function with a break like this, use Piecewise or HeavisideTheta: Integrate understands what those mean.






        share|improve this answer









        $endgroup$



        f[x_ /; x>=0]:=x means "if whatever>=0 rewrite f[whatever] as whatever. But that doesn't apply to f[x] when x is a symbol without a numerical value. Thus, f[x] simply remains f[x]. For abstracting the notion of a function with a break like this, use Piecewise or HeavisideTheta: Integrate understands what those mean.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 3 hours ago









        John DotyJohn Doty

        7,32811124




        7,32811124






























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