How is this integral computed??












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I'm reading this book about electrical properties of materials where the electron is introduced as a wave. Using the equation of a wave, they bring about the "envelope" of a wave. So here is how the derivation goes:
1)Consider the equation of a wave traveling in one dimension:
$ u = ae^{-i(omega t - kz)} $ where $omega = 2 pi f$ is the frequency, $k$ is the wave number, and $a$ is the amplitude. In addition to this information, the book refers to the phase velocity defined as: ${partial z}/{partial t} = {omega}/{k}$
2)Instead of 1 single wave, let there be multiple waves that are superimposed such that $u = sum a_n e^{-i(omega_n t - k_n z)}$
3)Consider the continuous case : $$ u(z) = int_{-infty}^infty a(k)e^{-i(omega t - kz)}dk $$
4)Make two more assumptions. First assumption: the waves are zero everywhere except at a small interval $Delta k $ and here the amplitude is unity such at $a(k)=1$. This brings the equation to $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{-i(omega t - kz)}dk $$
5)Suppose that $t=0$, thus yielding the endpoint: $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk $$



When the integration is carried out, the result is $$ u(z) = Delta k e^{ik_0 z} frac {sin 1/2(Delta kz)}{1/2(Delta k z)}$$.
My question is how is the integration done here? I do notice that u is a function of z and that the integrand integrates over the wave number k. Is there some change of variables going on here? By the way I only posses knowledge of ODEs and only know how to do basic Separation of variables for PDEs.






integration definite-integrals wave-equation change-of-variable envelope







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    I'm reading this book about electrical properties of materials where the electron is introduced as a wave. Using the equation of a wave, they bring about the "envelope" of a wave. So here is how the derivation goes:
    1)Consider the equation of a wave traveling in one dimension:
    $ u = ae^{-i(omega t - kz)} $ where $omega = 2 pi f$ is the frequency, $k$ is the wave number, and $a$ is the amplitude. In addition to this information, the book refers to the phase velocity defined as: ${partial z}/{partial t} = {omega}/{k}$
    2)Instead of 1 single wave, let there be multiple waves that are superimposed such that $u = sum a_n e^{-i(omega_n t - k_n z)}$
    3)Consider the continuous case : $$ u(z) = int_{-infty}^infty a(k)e^{-i(omega t - kz)}dk $$
    4)Make two more assumptions. First assumption: the waves are zero everywhere except at a small interval $Delta k $ and here the amplitude is unity such at $a(k)=1$. This brings the equation to $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{-i(omega t - kz)}dk $$
    5)Suppose that $t=0$, thus yielding the endpoint: $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk $$



    When the integration is carried out, the result is $$ u(z) = Delta k e^{ik_0 z} frac {sin 1/2(Delta kz)}{1/2(Delta k z)}$$.
    My question is how is the integration done here? I do notice that u is a function of z and that the integrand integrates over the wave number k. Is there some change of variables going on here? By the way I only posses knowledge of ODEs and only know how to do basic Separation of variables for PDEs.






    integration definite-integrals wave-equation change-of-variable envelope







    share|cite|improve this question











    $endgroup$















      0












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      0





      $begingroup$


      I'm reading this book about electrical properties of materials where the electron is introduced as a wave. Using the equation of a wave, they bring about the "envelope" of a wave. So here is how the derivation goes:
      1)Consider the equation of a wave traveling in one dimension:
      $ u = ae^{-i(omega t - kz)} $ where $omega = 2 pi f$ is the frequency, $k$ is the wave number, and $a$ is the amplitude. In addition to this information, the book refers to the phase velocity defined as: ${partial z}/{partial t} = {omega}/{k}$
      2)Instead of 1 single wave, let there be multiple waves that are superimposed such that $u = sum a_n e^{-i(omega_n t - k_n z)}$
      3)Consider the continuous case : $$ u(z) = int_{-infty}^infty a(k)e^{-i(omega t - kz)}dk $$
      4)Make two more assumptions. First assumption: the waves are zero everywhere except at a small interval $Delta k $ and here the amplitude is unity such at $a(k)=1$. This brings the equation to $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{-i(omega t - kz)}dk $$
      5)Suppose that $t=0$, thus yielding the endpoint: $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk $$



      When the integration is carried out, the result is $$ u(z) = Delta k e^{ik_0 z} frac {sin 1/2(Delta kz)}{1/2(Delta k z)}$$.
      My question is how is the integration done here? I do notice that u is a function of z and that the integrand integrates over the wave number k. Is there some change of variables going on here? By the way I only posses knowledge of ODEs and only know how to do basic Separation of variables for PDEs.






      integration definite-integrals wave-equation change-of-variable envelope







      share|cite|improve this question











      $endgroup$




      I'm reading this book about electrical properties of materials where the electron is introduced as a wave. Using the equation of a wave, they bring about the "envelope" of a wave. So here is how the derivation goes:
      1)Consider the equation of a wave traveling in one dimension:
      $ u = ae^{-i(omega t - kz)} $ where $omega = 2 pi f$ is the frequency, $k$ is the wave number, and $a$ is the amplitude. In addition to this information, the book refers to the phase velocity defined as: ${partial z}/{partial t} = {omega}/{k}$
      2)Instead of 1 single wave, let there be multiple waves that are superimposed such that $u = sum a_n e^{-i(omega_n t - k_n z)}$
      3)Consider the continuous case : $$ u(z) = int_{-infty}^infty a(k)e^{-i(omega t - kz)}dk $$
      4)Make two more assumptions. First assumption: the waves are zero everywhere except at a small interval $Delta k $ and here the amplitude is unity such at $a(k)=1$. This brings the equation to $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{-i(omega t - kz)}dk $$
      5)Suppose that $t=0$, thus yielding the endpoint: $$ u(z) = int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk $$



      When the integration is carried out, the result is $$ u(z) = Delta k e^{ik_0 z} frac {sin 1/2(Delta kz)}{1/2(Delta k z)}$$.
      My question is how is the integration done here? I do notice that u is a function of z and that the integrand integrates over the wave number k. Is there some change of variables going on here? By the way I only posses knowledge of ODEs and only know how to do basic Separation of variables for PDEs.






      integration definite-integrals wave-equation change-of-variable envelope




      integration definite-integrals wave-equation change-of-variable envelope






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      edited Jan 16 at 21:40









      clathratus

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      asked Jan 16 at 20:04









      Jonathan AguileraJonathan Aguilera

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          The antiderivative of $e^{ikz}$ with respect to $k$ is given by $frac{e^{ikz}}{iz}$ for $z neq 0$, so if we impose this condition on $z$, we have
          begin{align}int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk &= frac{1}{iz} left[e^{i k z} right]_{k_0 - frac{Delta k}{2}}^{k_0 + frac{Delta k}{2}} \
          &= frac{1}{iz}e^{i k_0 z}left(e^{i frac{Delta k}{2}} - e^{- i frac{Delta k}{2}} right).
          end{align}



          Can you continue from here, perhaps using the fact that for any $theta in mathbb{R}$, $e^{i theta} = cos{theta} + i sin{theta}$?






          share|cite|improve this answer











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          • 1




            $begingroup$
            Thanks a lot it worked out well!
            $endgroup$
            – Jonathan Aguilera
            Jan 18 at 0:18











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          $begingroup$

          The antiderivative of $e^{ikz}$ with respect to $k$ is given by $frac{e^{ikz}}{iz}$ for $z neq 0$, so if we impose this condition on $z$, we have
          begin{align}int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk &= frac{1}{iz} left[e^{i k z} right]_{k_0 - frac{Delta k}{2}}^{k_0 + frac{Delta k}{2}} \
          &= frac{1}{iz}e^{i k_0 z}left(e^{i frac{Delta k}{2}} - e^{- i frac{Delta k}{2}} right).
          end{align}



          Can you continue from here, perhaps using the fact that for any $theta in mathbb{R}$, $e^{i theta} = cos{theta} + i sin{theta}$?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks a lot it worked out well!
            $endgroup$
            – Jonathan Aguilera
            Jan 18 at 0:18
















          1












          $begingroup$

          The antiderivative of $e^{ikz}$ with respect to $k$ is given by $frac{e^{ikz}}{iz}$ for $z neq 0$, so if we impose this condition on $z$, we have
          begin{align}int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk &= frac{1}{iz} left[e^{i k z} right]_{k_0 - frac{Delta k}{2}}^{k_0 + frac{Delta k}{2}} \
          &= frac{1}{iz}e^{i k_0 z}left(e^{i frac{Delta k}{2}} - e^{- i frac{Delta k}{2}} right).
          end{align}



          Can you continue from here, perhaps using the fact that for any $theta in mathbb{R}$, $e^{i theta} = cos{theta} + i sin{theta}$?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks a lot it worked out well!
            $endgroup$
            – Jonathan Aguilera
            Jan 18 at 0:18














          1












          1








          1





          $begingroup$

          The antiderivative of $e^{ikz}$ with respect to $k$ is given by $frac{e^{ikz}}{iz}$ for $z neq 0$, so if we impose this condition on $z$, we have
          begin{align}int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk &= frac{1}{iz} left[e^{i k z} right]_{k_0 - frac{Delta k}{2}}^{k_0 + frac{Delta k}{2}} \
          &= frac{1}{iz}e^{i k_0 z}left(e^{i frac{Delta k}{2}} - e^{- i frac{Delta k}{2}} right).
          end{align}



          Can you continue from here, perhaps using the fact that for any $theta in mathbb{R}$, $e^{i theta} = cos{theta} + i sin{theta}$?






          share|cite|improve this answer











          $endgroup$



          The antiderivative of $e^{ikz}$ with respect to $k$ is given by $frac{e^{ikz}}{iz}$ for $z neq 0$, so if we impose this condition on $z$, we have
          begin{align}int_{k_0 - Delta k /2}^{k_0 + Delta k /2} e^{i kz}dk &= frac{1}{iz} left[e^{i k z} right]_{k_0 - frac{Delta k}{2}}^{k_0 + frac{Delta k}{2}} \
          &= frac{1}{iz}e^{i k_0 z}left(e^{i frac{Delta k}{2}} - e^{- i frac{Delta k}{2}} right).
          end{align}



          Can you continue from here, perhaps using the fact that for any $theta in mathbb{R}$, $e^{i theta} = cos{theta} + i sin{theta}$?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 30 at 14:21

























          answered Jan 16 at 21:03









          E-muE-mu

          787417




          787417








          • 1




            $begingroup$
            Thanks a lot it worked out well!
            $endgroup$
            – Jonathan Aguilera
            Jan 18 at 0:18














          • 1




            $begingroup$
            Thanks a lot it worked out well!
            $endgroup$
            – Jonathan Aguilera
            Jan 18 at 0:18








          1




          1




          $begingroup$
          Thanks a lot it worked out well!
          $endgroup$
          – Jonathan Aguilera
          Jan 18 at 0:18




          $begingroup$
          Thanks a lot it worked out well!
          $endgroup$
          – Jonathan Aguilera
          Jan 18 at 0:18


















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