Integrating function with /; in its definition
$begingroup$
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}]
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
$endgroup$
add a comment |
$begingroup$
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}]
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
$endgroup$
3
$begingroup$
It's better to use ConditionalExpression, e.g.,Integrate[ConditionalExpression[x, x>0], {x, 0, 2Pi}]
$endgroup$
– Carl Woll
3 hours ago
add a comment |
$begingroup$
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}]
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
$endgroup$
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}]
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
calculus-and-analysis function-construction
edited 23 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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192838%2fintegrating-function-with-in-its-definition%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 3 hours ago
John DotyJohn Doty
7,32811124
7,32811124
add a comment |
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192838%2fintegrating-function-with-in-its-definition%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
It's better to use ConditionalExpression, e.g.,
Integrate[ConditionalExpression[x, x>0], {x, 0, 2Pi}]
$endgroup$
– Carl Woll
3 hours ago