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





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          1 Answer
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          1 Answer
          1






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          active

          oldest

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          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


















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