Evaluating $int_{-infty}^{infty} xf(x)delta(x-a) dx$.












1















Evaluate



$$int_{-infty}^{infty} xf(x)delta(x-a) dx.$$




I suspect that I could create another function, let's say $g(x)=xf(x)$ and perform the integral which would just give $g(a)=af(a)$ but I'm not too sure. Thanks!










share|cite|improve this question
























  • Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
    – Eevee Trainer
    Nov 27 '18 at 2:33








  • 1




    take a look here for a similar question. Yes, the answer is $a f(a)$
    – Masacroso
    Nov 27 '18 at 2:35












  • Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
    – John L Winters
    Nov 27 '18 at 2:41












  • Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
    – Shaun
    Nov 27 '18 at 2:46


















1















Evaluate



$$int_{-infty}^{infty} xf(x)delta(x-a) dx.$$




I suspect that I could create another function, let's say $g(x)=xf(x)$ and perform the integral which would just give $g(a)=af(a)$ but I'm not too sure. Thanks!










share|cite|improve this question
























  • Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
    – Eevee Trainer
    Nov 27 '18 at 2:33








  • 1




    take a look here for a similar question. Yes, the answer is $a f(a)$
    – Masacroso
    Nov 27 '18 at 2:35












  • Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
    – John L Winters
    Nov 27 '18 at 2:41












  • Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
    – Shaun
    Nov 27 '18 at 2:46
















1












1








1








Evaluate



$$int_{-infty}^{infty} xf(x)delta(x-a) dx.$$




I suspect that I could create another function, let's say $g(x)=xf(x)$ and perform the integral which would just give $g(a)=af(a)$ but I'm not too sure. Thanks!










share|cite|improve this question
















Evaluate



$$int_{-infty}^{infty} xf(x)delta(x-a) dx.$$




I suspect that I could create another function, let's say $g(x)=xf(x)$ and perform the integral which would just give $g(a)=af(a)$ but I'm not too sure. Thanks!







integration special-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 2:42









Shaun

8,810113680




8,810113680










asked Nov 27 '18 at 2:32









killerownage2006

61




61












  • Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
    – Eevee Trainer
    Nov 27 '18 at 2:33








  • 1




    take a look here for a similar question. Yes, the answer is $a f(a)$
    – Masacroso
    Nov 27 '18 at 2:35












  • Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
    – John L Winters
    Nov 27 '18 at 2:41












  • Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
    – Shaun
    Nov 27 '18 at 2:46




















  • Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
    – Eevee Trainer
    Nov 27 '18 at 2:33








  • 1




    take a look here for a similar question. Yes, the answer is $a f(a)$
    – Masacroso
    Nov 27 '18 at 2:35












  • Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
    – John L Winters
    Nov 27 '18 at 2:41












  • Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
    – Shaun
    Nov 27 '18 at 2:46


















Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
– Eevee Trainer
Nov 27 '18 at 2:33






Is this function $delta(x-a)$ any type of special function? I rarely see $delta$ used as a name for a function if not, and such context would be absolutely important. (Granted names are arbitrary but I generally just tend to see $f, g, h$ etc.)
– Eevee Trainer
Nov 27 '18 at 2:33






1




1




take a look here for a similar question. Yes, the answer is $a f(a)$
– Masacroso
Nov 27 '18 at 2:35






take a look here for a similar question. Yes, the answer is $a f(a)$
– Masacroso
Nov 27 '18 at 2:35














Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
– John L Winters
Nov 27 '18 at 2:41






Yes it is the special function area 1 width zero. So it just selects the value at a point, like a sampler. It is used in study of signals and systems with LaPlace transforms. Commonly known as Delta Function. It is first derivative of step function. Such a trivial integral would not be online.
– John L Winters
Nov 27 '18 at 2:41














Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
– Shaun
Nov 27 '18 at 2:46






Subtle point: One has to assume that $f(a)$ exists (because, for instance, $f$ might be defined a.e. except at $a$, otherwise).
– Shaun
Nov 27 '18 at 2:46












1 Answer
1






active

oldest

votes


















3














The general rule in integrating over a delta function is:



$$intlimits_{-infty}^infty g(x) delta(x-a) dx = g(a)$$



Here your $g(x)$ is $x f(x)$.






share|cite|improve this answer





















  • Okay thank you, I just wanted to confirm if I could do this.
    – killerownage2006
    Nov 27 '18 at 3:06











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: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015252%2fevaluating-int-infty-infty-xfx-deltax-a-dx%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









3














The general rule in integrating over a delta function is:



$$intlimits_{-infty}^infty g(x) delta(x-a) dx = g(a)$$



Here your $g(x)$ is $x f(x)$.






share|cite|improve this answer





















  • Okay thank you, I just wanted to confirm if I could do this.
    – killerownage2006
    Nov 27 '18 at 3:06
















3














The general rule in integrating over a delta function is:



$$intlimits_{-infty}^infty g(x) delta(x-a) dx = g(a)$$



Here your $g(x)$ is $x f(x)$.






share|cite|improve this answer





















  • Okay thank you, I just wanted to confirm if I could do this.
    – killerownage2006
    Nov 27 '18 at 3:06














3












3








3






The general rule in integrating over a delta function is:



$$intlimits_{-infty}^infty g(x) delta(x-a) dx = g(a)$$



Here your $g(x)$ is $x f(x)$.






share|cite|improve this answer












The general rule in integrating over a delta function is:



$$intlimits_{-infty}^infty g(x) delta(x-a) dx = g(a)$$



Here your $g(x)$ is $x f(x)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 '18 at 2:44









David G. Stork

9,96021232




9,96021232












  • Okay thank you, I just wanted to confirm if I could do this.
    – killerownage2006
    Nov 27 '18 at 3:06


















  • Okay thank you, I just wanted to confirm if I could do this.
    – killerownage2006
    Nov 27 '18 at 3:06
















Okay thank you, I just wanted to confirm if I could do this.
– killerownage2006
Nov 27 '18 at 3:06




Okay thank you, I just wanted to confirm if I could do this.
– killerownage2006
Nov 27 '18 at 3:06


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics 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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015252%2fevaluating-int-infty-infty-xfx-deltax-a-dx%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Plaza Victoria

Puebla de Zaragoza

Musa