Integrating function with /; in its definitionHow to make InverseFunction work on function defined with...
The need of reserving one's ability in job interviews
Do natural melee weapons (from racial traits) trigger Improved Divine Smite?
Was it really inappropriate to write a pull request for the company I interviewed with?
Can inspiration allow the Rogue to make a Sneak Attack?
Preparing as much as possible of a cake in advance
Can you run a ground wire from stove directly to ground pole in the ground
Problems with rounding giving too many digits
Dukha vs legitimate need
When to use the term transposed instead of modulation?
What can I do if someone tampers with my SSH public key?
Is there a way to find out the age of climbing ropes?
Can a Mimic (container form) actually hold loot?
Error in TransformedField
Should I use HTTPS on a domain that will only be used for redirection?
Naming Characters after Friends/Family
Can a Mexican citizen living in US under DACA drive to Canada?
Called into a meeting and told we are being made redundant (laid off) and "not to share outside". Can I tell my partner?
ESPP--any reason not to go all in?
Does the US political system, in principle, allow for a no-party system?
Can a space-faring robot still function over a billion years?
I can't die. Who am I?
Is "cogitate" an appropriate word for this?
Why aren't there more gauls like Obelix?
Should we avoid writing fiction about historical events without extensive research?
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
$begingroup$
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 13 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