Dirac delta integral form proof












4












$begingroup$


While reading the book "Modern Electrodynamics" by Andrew Zangwill, on page 13 I've encountered a Dirac function integral representation. The proof the book provides is the following:




begin{align*}
int_{-infty}^infty text{d}k, e^{ikx}
&= int_0^infty text{d}k, e^{ikx}
+ int_0^infty text{d}k, e^{-ikx}
\
&= lim_{epsilon to 0}
left[
int_0^infty text{d}k, e^{ik(x+iepsilon)}
+ int_0^infty text{d}k, e^{-ik(x-iepsilon)}
right].
end{align*}
The convergence factors make the integrands zero at the upper limit, so
$$
int_{-infty}^infty text{d}k, e^{ikx}
= lim_{epsilon to 0}
left[
frac{i}{x + i epsilon} - frac{i}{x - i epsilon}
right]
= lim_{epsilon to 0} frac{2 epsilon}{x^2 + epsilon^2}
= 2 pi delta(x).
$$
(Original image here.)




I don’t see how the convergence makes the first term inside the limit vanish at the upper limit. I can understand the rest from there as the the $k$-independent factor appears while integrating and the final conclussion is from one of the informal dirac delta definitions.






dirac-delta







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    While reading the book "Modern Electrodynamics" by Andrew Zangwill, on page 13 I've encountered a Dirac function integral representation. The proof the book provides is the following:




    begin{align*}
    int_{-infty}^infty text{d}k, e^{ikx}
    &= int_0^infty text{d}k, e^{ikx}
    + int_0^infty text{d}k, e^{-ikx}
    \
    &= lim_{epsilon to 0}
    left[
    int_0^infty text{d}k, e^{ik(x+iepsilon)}
    + int_0^infty text{d}k, e^{-ik(x-iepsilon)}
    right].
    end{align*}
    The convergence factors make the integrands zero at the upper limit, so
    $$
    int_{-infty}^infty text{d}k, e^{ikx}
    = lim_{epsilon to 0}
    left[
    frac{i}{x + i epsilon} - frac{i}{x - i epsilon}
    right]
    = lim_{epsilon to 0} frac{2 epsilon}{x^2 + epsilon^2}
    = 2 pi delta(x).
    $$
    (Original image here.)




    I don’t see how the convergence makes the first term inside the limit vanish at the upper limit. I can understand the rest from there as the the $k$-independent factor appears while integrating and the final conclussion is from one of the informal dirac delta definitions.






    dirac-delta







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      4



      $begingroup$


      While reading the book "Modern Electrodynamics" by Andrew Zangwill, on page 13 I've encountered a Dirac function integral representation. The proof the book provides is the following:




      begin{align*}
      int_{-infty}^infty text{d}k, e^{ikx}
      &= int_0^infty text{d}k, e^{ikx}
      + int_0^infty text{d}k, e^{-ikx}
      \
      &= lim_{epsilon to 0}
      left[
      int_0^infty text{d}k, e^{ik(x+iepsilon)}
      + int_0^infty text{d}k, e^{-ik(x-iepsilon)}
      right].
      end{align*}
      The convergence factors make the integrands zero at the upper limit, so
      $$
      int_{-infty}^infty text{d}k, e^{ikx}
      = lim_{epsilon to 0}
      left[
      frac{i}{x + i epsilon} - frac{i}{x - i epsilon}
      right]
      = lim_{epsilon to 0} frac{2 epsilon}{x^2 + epsilon^2}
      = 2 pi delta(x).
      $$
      (Original image here.)




      I don’t see how the convergence makes the first term inside the limit vanish at the upper limit. I can understand the rest from there as the the $k$-independent factor appears while integrating and the final conclussion is from one of the informal dirac delta definitions.






      dirac-delta







      share|cite|improve this question











      $endgroup$




      While reading the book "Modern Electrodynamics" by Andrew Zangwill, on page 13 I've encountered a Dirac function integral representation. The proof the book provides is the following:




      begin{align*}
      int_{-infty}^infty text{d}k, e^{ikx}
      &= int_0^infty text{d}k, e^{ikx}
      + int_0^infty text{d}k, e^{-ikx}
      \
      &= lim_{epsilon to 0}
      left[
      int_0^infty text{d}k, e^{ik(x+iepsilon)}
      + int_0^infty text{d}k, e^{-ik(x-iepsilon)}
      right].
      end{align*}
      The convergence factors make the integrands zero at the upper limit, so
      $$
      int_{-infty}^infty text{d}k, e^{ikx}
      = lim_{epsilon to 0}
      left[
      frac{i}{x + i epsilon} - frac{i}{x - i epsilon}
      right]
      = lim_{epsilon to 0} frac{2 epsilon}{x^2 + epsilon^2}
      = 2 pi delta(x).
      $$
      (Original image here.)




      I don’t see how the convergence makes the first term inside the limit vanish at the upper limit. I can understand the rest from there as the the $k$-independent factor appears while integrating and the final conclussion is from one of the informal dirac delta definitions.






      dirac-delta




      dirac-delta






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 26 '18 at 20:33









      Mefitico

      926117




      926117










      asked Feb 12 '15 at 22:04









      Mark A. RuizMark A. Ruiz

      8619




      8619






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Looking at the first one:



          $$int_0^infty dk e^{ik(x+i epsilon)} = int_0^infty dk e^{ikx} e^{-k epsilon}.$$



          Now $e^{ikx}$ is bounded in modulus while $e^{-k epsilon}$ (for fixed $epsilon$) decays exponentially in modulus. So the integral converges by comparison to $int_0^infty dk e^{-k epsilon}$.



          The second one is basically the same.



          Note that the first step is quite informal (one cannot justify it by the application of some convergence theorem which can be formulated without distribution theory.) I think you might get more insight by looking at a formal definition of the Fourier transform on tempered distributions. The formal definition is quite simple, and the desired statement (that the Fourier transform of the Dirac delta is a constant function) follows quite easily from it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I missed looking at the modolus. Can you recomend me some literature on the subject? Had very little exposition to distributions formally as a physicist.
            $endgroup$
            – Mark A. Ruiz
            Feb 12 '15 at 23:20










          • $begingroup$
            Consider volume 1 of Hormander's The analysis of partial differential equations. The notes for my current course are from there (I can check as to whether I am allowed to simply share them).
            $endgroup$
            – Ian
            Feb 12 '15 at 23:38












          • $begingroup$
            Would greatly apreaciate if so. Looking foward to that
            $endgroup$
            – Mark A. Ruiz
            Feb 12 '15 at 23:56











          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%2f1145718%2fdirac-delta-integral-form-proof%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









          0












          $begingroup$

          Looking at the first one:



          $$int_0^infty dk e^{ik(x+i epsilon)} = int_0^infty dk e^{ikx} e^{-k epsilon}.$$



          Now $e^{ikx}$ is bounded in modulus while $e^{-k epsilon}$ (for fixed $epsilon$) decays exponentially in modulus. So the integral converges by comparison to $int_0^infty dk e^{-k epsilon}$.



          The second one is basically the same.



          Note that the first step is quite informal (one cannot justify it by the application of some convergence theorem which can be formulated without distribution theory.) I think you might get more insight by looking at a formal definition of the Fourier transform on tempered distributions. The formal definition is quite simple, and the desired statement (that the Fourier transform of the Dirac delta is a constant function) follows quite easily from it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I missed looking at the modolus. Can you recomend me some literature on the subject? Had very little exposition to distributions formally as a physicist.
            $endgroup$
            – Mark A. Ruiz
            Feb 12 '15 at 23:20










          • $begingroup$
            Consider volume 1 of Hormander's The analysis of partial differential equations. The notes for my current course are from there (I can check as to whether I am allowed to simply share them).
            $endgroup$
            – Ian
            Feb 12 '15 at 23:38












          • $begingroup$
            Would greatly apreaciate if so. Looking foward to that
            $endgroup$
            – Mark A. Ruiz
            Feb 12 '15 at 23:56
















          0












          $begingroup$

          Looking at the first one:



          $$int_0^infty dk e^{ik(x+i epsilon)} = int_0^infty dk e^{ikx} e^{-k epsilon}.$$



          Now $e^{ikx}$ is bounded in modulus while $e^{-k epsilon}$ (for fixed $epsilon$) decays exponentially in modulus. So the integral converges by comparison to $int_0^infty dk e^{-k epsilon}$.



          The second one is basically the same.



          Note that the first step is quite informal (one cannot justify it by the application of some convergence theorem which can be formulated without distribution theory.) I think you might get more insight by looking at a formal definition of the Fourier transform on tempered distributions. The formal definition is quite simple, and the desired statement (that the Fourier transform of the Dirac delta is a constant function) follows quite easily from it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I missed looking at the modolus. Can you recomend me some literature on the subject? Had very little exposition to distributions formally as a physicist.
            $endgroup$
            – Mark A. Ruiz
            Feb 12 '15 at 23:20










          • $begingroup$
            Consider volume 1 of Hormander's The analysis of partial differential equations. The notes for my current course are from there (I can check as to whether I am allowed to simply share them).
            $endgroup$
            – Ian
            Feb 12 '15 at 23:38












          • $begingroup$
            Would greatly apreaciate if so. Looking foward to that
            $endgroup$
            – Mark A. Ruiz
            Feb 12 '15 at 23:56














          0












          0








          0





          $begingroup$

          Looking at the first one:



          $$int_0^infty dk e^{ik(x+i epsilon)} = int_0^infty dk e^{ikx} e^{-k epsilon}.$$



          Now $e^{ikx}$ is bounded in modulus while $e^{-k epsilon}$ (for fixed $epsilon$) decays exponentially in modulus. So the integral converges by comparison to $int_0^infty dk e^{-k epsilon}$.



          The second one is basically the same.



          Note that the first step is quite informal (one cannot justify it by the application of some convergence theorem which can be formulated without distribution theory.) I think you might get more insight by looking at a formal definition of the Fourier transform on tempered distributions. The formal definition is quite simple, and the desired statement (that the Fourier transform of the Dirac delta is a constant function) follows quite easily from it.






          share|cite|improve this answer











          $endgroup$



          Looking at the first one:



          $$int_0^infty dk e^{ik(x+i epsilon)} = int_0^infty dk e^{ikx} e^{-k epsilon}.$$



          Now $e^{ikx}$ is bounded in modulus while $e^{-k epsilon}$ (for fixed $epsilon$) decays exponentially in modulus. So the integral converges by comparison to $int_0^infty dk e^{-k epsilon}$.



          The second one is basically the same.



          Note that the first step is quite informal (one cannot justify it by the application of some convergence theorem which can be formulated without distribution theory.) I think you might get more insight by looking at a formal definition of the Fourier transform on tempered distributions. The formal definition is quite simple, and the desired statement (that the Fourier transform of the Dirac delta is a constant function) follows quite easily from it.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 12 '15 at 22:18

























          answered Feb 12 '15 at 22:11









          IanIan

          67.6k25387




          67.6k25387












          • $begingroup$
            I missed looking at the modolus. Can you recomend me some literature on the subject? Had very little exposition to distributions formally as a physicist.
            $endgroup$
            – Mark A. Ruiz
            Feb 12 '15 at 23:20










          • $begingroup$
            Consider volume 1 of Hormander's The analysis of partial differential equations. The notes for my current course are from there (I can check as to whether I am allowed to simply share them).
            $endgroup$
            – Ian
            Feb 12 '15 at 23:38












          • $begingroup$
            Would greatly apreaciate if so. Looking foward to that
            $endgroup$
            – Mark A. Ruiz
            Feb 12 '15 at 23:56


















          • $begingroup$
            I missed looking at the modolus. Can you recomend me some literature on the subject? Had very little exposition to distributions formally as a physicist.
            $endgroup$
            – Mark A. Ruiz
            Feb 12 '15 at 23:20










          • $begingroup$
            Consider volume 1 of Hormander's The analysis of partial differential equations. The notes for my current course are from there (I can check as to whether I am allowed to simply share them).
            $endgroup$
            – Ian
            Feb 12 '15 at 23:38












          • $begingroup$
            Would greatly apreaciate if so. Looking foward to that
            $endgroup$
            – Mark A. Ruiz
            Feb 12 '15 at 23:56
















          $begingroup$
          I missed looking at the modolus. Can you recomend me some literature on the subject? Had very little exposition to distributions formally as a physicist.
          $endgroup$
          – Mark A. Ruiz
          Feb 12 '15 at 23:20




          $begingroup$
          I missed looking at the modolus. Can you recomend me some literature on the subject? Had very little exposition to distributions formally as a physicist.
          $endgroup$
          – Mark A. Ruiz
          Feb 12 '15 at 23:20












          $begingroup$
          Consider volume 1 of Hormander's The analysis of partial differential equations. The notes for my current course are from there (I can check as to whether I am allowed to simply share them).
          $endgroup$
          – Ian
          Feb 12 '15 at 23:38






          $begingroup$
          Consider volume 1 of Hormander's The analysis of partial differential equations. The notes for my current course are from there (I can check as to whether I am allowed to simply share them).
          $endgroup$
          – Ian
          Feb 12 '15 at 23:38














          $begingroup$
          Would greatly apreaciate if so. Looking foward to that
          $endgroup$
          – Mark A. Ruiz
          Feb 12 '15 at 23:56




          $begingroup$
          Would greatly apreaciate if so. Looking foward to that
          $endgroup$
          – Mark A. Ruiz
          Feb 12 '15 at 23:56


















          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1145718%2fdirac-delta-integral-form-proof%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

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
          Read more