Dirac delta integral form proof












5












$begingroup$


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











$endgroup$

















    5












    $begingroup$


    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











    $endgroup$















      5












      5








      5


      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

      928117




      928117










      asked Feb 12 '15 at 22:04









      Mark A. RuizMark A. Ruiz

      9119




      9119






















          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

          67.8k25388




          67.8k25388












          • $begingroup$
            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

          Plaza Victoria

          Image
          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
          Read more

          Puebla de Zaragoza

          Image
          Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
          Read more

          Musa

          Image
          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
          Read more