Gaussians as good kernels












0












$begingroup$


Theorem. If $delta>0$ and
$$K_{delta}(x) = delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}},$$
then
$$widehat{K_{delta}}(xi) = e^{- pi delta xi^2}.$$
Proof.

I started the proof in a very expected way.
$$widehat{K_{delta}}(xi) = int limits_{-infty}^{infty} delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}} e^{-2 pi ix xi} mbox{d}x. tag{1}$$
Unfortunately I don't know how to evaluate $(1)$. I think using the following equality might help
$$int limits_{-infty}^{infty} e^{- pi x} mbox{d}x =1$$
but I don't know how to use it. I would appreciate any hints or tips.






real-analysis fourier-transform







share|cite|improve this question









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    0












    $begingroup$


    Theorem. If $delta>0$ and
    $$K_{delta}(x) = delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}},$$
    then
    $$widehat{K_{delta}}(xi) = e^{- pi delta xi^2}.$$
    Proof.

    I started the proof in a very expected way.
    $$widehat{K_{delta}}(xi) = int limits_{-infty}^{infty} delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}} e^{-2 pi ix xi} mbox{d}x. tag{1}$$
    Unfortunately I don't know how to evaluate $(1)$. I think using the following equality might help
    $$int limits_{-infty}^{infty} e^{- pi x} mbox{d}x =1$$
    but I don't know how to use it. I would appreciate any hints or tips.






    real-analysis fourier-transform







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Theorem. If $delta>0$ and
      $$K_{delta}(x) = delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}},$$
      then
      $$widehat{K_{delta}}(xi) = e^{- pi delta xi^2}.$$
      Proof.

      I started the proof in a very expected way.
      $$widehat{K_{delta}}(xi) = int limits_{-infty}^{infty} delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}} e^{-2 pi ix xi} mbox{d}x. tag{1}$$
      Unfortunately I don't know how to evaluate $(1)$. I think using the following equality might help
      $$int limits_{-infty}^{infty} e^{- pi x} mbox{d}x =1$$
      but I don't know how to use it. I would appreciate any hints or tips.






      real-analysis fourier-transform







      share|cite|improve this question









      $endgroup$




      Theorem. If $delta>0$ and
      $$K_{delta}(x) = delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}},$$
      then
      $$widehat{K_{delta}}(xi) = e^{- pi delta xi^2}.$$
      Proof.

      I started the proof in a very expected way.
      $$widehat{K_{delta}}(xi) = int limits_{-infty}^{infty} delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}} e^{-2 pi ix xi} mbox{d}x. tag{1}$$
      Unfortunately I don't know how to evaluate $(1)$. I think using the following equality might help
      $$int limits_{-infty}^{infty} e^{- pi x} mbox{d}x =1$$
      but I don't know how to use it. I would appreciate any hints or tips.






      real-analysis fourier-transform




      real-analysis fourier-transform






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 19 '18 at 14:54









      HendrraHendrra

      1,200516




      1,200516






















          1 Answer
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          $begingroup$

          The integral is equal to
          $$
          int _{-infty}^{infty} delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}} e^{-2 pi ix xi}dx = int _{-infty}^{infty}e^{- pi x^2} e^{-2 pi isqrt{delta} x xi}dx= e^{-pideltaxi^2}int _{-infty}^{infty}e^{- pi (x+isqrt{delta}xi)^2} dx.
          $$
          Note that
          $$
          int _{-infty}^{infty}e^{- pi (x+isqrt{delta}xi)^2} dx=int_{Im(z)=sqrt{delta}xi}e^{- pi z^2} dz.
          $$           By Cauchy's integral theorem, we can show that
          $$
          int_{Im(z)=sqrt{delta}xi}e^{- pi z^2} dz =int_{Im(z)=0}e^{- pi z^2} dz=int_{-infty}^infty e^{- pi z^2} dz=1.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you explain a bit more the second line? I can't see how did you get $Im{z}$ in the integration limit.
            $endgroup$
            – Hendrra
            Dec 19 '18 at 15:09






          • 1




            $begingroup$
            The left hand side is just the line integral of $zmapsto e^{-pi z^2}$ over $Im(z) = sqrt{delta}xi$ parametrized by $gamma(t) = t+isqrt{delta}xi, quad -infty <t<infty$.
            $endgroup$
            – Song
            Dec 19 '18 at 15:12










          • $begingroup$
            Thank you! I appreciate your help and knowledge!
            $endgroup$
            – Hendrra
            Dec 19 '18 at 15:16










          • $begingroup$
            I'm glad it helped :)
            $endgroup$
            – Song
            Dec 19 '18 at 15:19











          Your Answer





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          $begingroup$

          The integral is equal to
          $$
          int _{-infty}^{infty} delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}} e^{-2 pi ix xi}dx = int _{-infty}^{infty}e^{- pi x^2} e^{-2 pi isqrt{delta} x xi}dx= e^{-pideltaxi^2}int _{-infty}^{infty}e^{- pi (x+isqrt{delta}xi)^2} dx.
          $$
          Note that
          $$
          int _{-infty}^{infty}e^{- pi (x+isqrt{delta}xi)^2} dx=int_{Im(z)=sqrt{delta}xi}e^{- pi z^2} dz.
          $$           By Cauchy's integral theorem, we can show that
          $$
          int_{Im(z)=sqrt{delta}xi}e^{- pi z^2} dz =int_{Im(z)=0}e^{- pi z^2} dz=int_{-infty}^infty e^{- pi z^2} dz=1.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you explain a bit more the second line? I can't see how did you get $Im{z}$ in the integration limit.
            $endgroup$
            – Hendrra
            Dec 19 '18 at 15:09






          • 1




            $begingroup$
            The left hand side is just the line integral of $zmapsto e^{-pi z^2}$ over $Im(z) = sqrt{delta}xi$ parametrized by $gamma(t) = t+isqrt{delta}xi, quad -infty <t<infty$.
            $endgroup$
            – Song
            Dec 19 '18 at 15:12










          • $begingroup$
            Thank you! I appreciate your help and knowledge!
            $endgroup$
            – Hendrra
            Dec 19 '18 at 15:16










          • $begingroup$
            I'm glad it helped :)
            $endgroup$
            – Song
            Dec 19 '18 at 15:19
















          1












          $begingroup$

          The integral is equal to
          $$
          int _{-infty}^{infty} delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}} e^{-2 pi ix xi}dx = int _{-infty}^{infty}e^{- pi x^2} e^{-2 pi isqrt{delta} x xi}dx= e^{-pideltaxi^2}int _{-infty}^{infty}e^{- pi (x+isqrt{delta}xi)^2} dx.
          $$
          Note that
          $$
          int _{-infty}^{infty}e^{- pi (x+isqrt{delta}xi)^2} dx=int_{Im(z)=sqrt{delta}xi}e^{- pi z^2} dz.
          $$           By Cauchy's integral theorem, we can show that
          $$
          int_{Im(z)=sqrt{delta}xi}e^{- pi z^2} dz =int_{Im(z)=0}e^{- pi z^2} dz=int_{-infty}^infty e^{- pi z^2} dz=1.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you explain a bit more the second line? I can't see how did you get $Im{z}$ in the integration limit.
            $endgroup$
            – Hendrra
            Dec 19 '18 at 15:09






          • 1




            $begingroup$
            The left hand side is just the line integral of $zmapsto e^{-pi z^2}$ over $Im(z) = sqrt{delta}xi$ parametrized by $gamma(t) = t+isqrt{delta}xi, quad -infty <t<infty$.
            $endgroup$
            – Song
            Dec 19 '18 at 15:12










          • $begingroup$
            Thank you! I appreciate your help and knowledge!
            $endgroup$
            – Hendrra
            Dec 19 '18 at 15:16










          • $begingroup$
            I'm glad it helped :)
            $endgroup$
            – Song
            Dec 19 '18 at 15:19














          1












          1








          1





          $begingroup$

          The integral is equal to
          $$
          int _{-infty}^{infty} delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}} e^{-2 pi ix xi}dx = int _{-infty}^{infty}e^{- pi x^2} e^{-2 pi isqrt{delta} x xi}dx= e^{-pideltaxi^2}int _{-infty}^{infty}e^{- pi (x+isqrt{delta}xi)^2} dx.
          $$
          Note that
          $$
          int _{-infty}^{infty}e^{- pi (x+isqrt{delta}xi)^2} dx=int_{Im(z)=sqrt{delta}xi}e^{- pi z^2} dz.
          $$           By Cauchy's integral theorem, we can show that
          $$
          int_{Im(z)=sqrt{delta}xi}e^{- pi z^2} dz =int_{Im(z)=0}e^{- pi z^2} dz=int_{-infty}^infty e^{- pi z^2} dz=1.
          $$






          share|cite|improve this answer









          $endgroup$



          The integral is equal to
          $$
          int _{-infty}^{infty} delta^{-frac{1}{2}}e^{- pi frac{x^2}{delta}} e^{-2 pi ix xi}dx = int _{-infty}^{infty}e^{- pi x^2} e^{-2 pi isqrt{delta} x xi}dx= e^{-pideltaxi^2}int _{-infty}^{infty}e^{- pi (x+isqrt{delta}xi)^2} dx.
          $$
          Note that
          $$
          int _{-infty}^{infty}e^{- pi (x+isqrt{delta}xi)^2} dx=int_{Im(z)=sqrt{delta}xi}e^{- pi z^2} dz.
          $$           By Cauchy's integral theorem, we can show that
          $$
          int_{Im(z)=sqrt{delta}xi}e^{- pi z^2} dz =int_{Im(z)=0}e^{- pi z^2} dz=int_{-infty}^infty e^{- pi z^2} dz=1.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 15:07









          SongSong

          14.8k1635




          14.8k1635












          • $begingroup$
            Can you explain a bit more the second line? I can't see how did you get $Im{z}$ in the integration limit.
            $endgroup$
            – Hendrra
            Dec 19 '18 at 15:09






          • 1




            $begingroup$
            The left hand side is just the line integral of $zmapsto e^{-pi z^2}$ over $Im(z) = sqrt{delta}xi$ parametrized by $gamma(t) = t+isqrt{delta}xi, quad -infty <t<infty$.
            $endgroup$
            – Song
            Dec 19 '18 at 15:12










          • $begingroup$
            Thank you! I appreciate your help and knowledge!
            $endgroup$
            – Hendrra
            Dec 19 '18 at 15:16










          • $begingroup$
            I'm glad it helped :)
            $endgroup$
            – Song
            Dec 19 '18 at 15:19


















          • $begingroup$
            Can you explain a bit more the second line? I can't see how did you get $Im{z}$ in the integration limit.
            $endgroup$
            – Hendrra
            Dec 19 '18 at 15:09






          • 1




            $begingroup$
            The left hand side is just the line integral of $zmapsto e^{-pi z^2}$ over $Im(z) = sqrt{delta}xi$ parametrized by $gamma(t) = t+isqrt{delta}xi, quad -infty <t<infty$.
            $endgroup$
            – Song
            Dec 19 '18 at 15:12










          • $begingroup$
            Thank you! I appreciate your help and knowledge!
            $endgroup$
            – Hendrra
            Dec 19 '18 at 15:16










          • $begingroup$
            I'm glad it helped :)
            $endgroup$
            – Song
            Dec 19 '18 at 15:19
















          $begingroup$
          Can you explain a bit more the second line? I can't see how did you get $Im{z}$ in the integration limit.
          $endgroup$
          – Hendrra
          Dec 19 '18 at 15:09




          $begingroup$
          Can you explain a bit more the second line? I can't see how did you get $Im{z}$ in the integration limit.
          $endgroup$
          – Hendrra
          Dec 19 '18 at 15:09




          1




          1




          $begingroup$
          The left hand side is just the line integral of $zmapsto e^{-pi z^2}$ over $Im(z) = sqrt{delta}xi$ parametrized by $gamma(t) = t+isqrt{delta}xi, quad -infty <t<infty$.
          $endgroup$
          – Song
          Dec 19 '18 at 15:12




          $begingroup$
          The left hand side is just the line integral of $zmapsto e^{-pi z^2}$ over $Im(z) = sqrt{delta}xi$ parametrized by $gamma(t) = t+isqrt{delta}xi, quad -infty <t<infty$.
          $endgroup$
          – Song
          Dec 19 '18 at 15:12












          $begingroup$
          Thank you! I appreciate your help and knowledge!
          $endgroup$
          – Hendrra
          Dec 19 '18 at 15:16




          $begingroup$
          Thank you! I appreciate your help and knowledge!
          $endgroup$
          – Hendrra
          Dec 19 '18 at 15:16












          $begingroup$
          I'm glad it helped :)
          $endgroup$
          – Song
          Dec 19 '18 at 15:19




          $begingroup$
          I'm glad it helped :)
          $endgroup$
          – Song
          Dec 19 '18 at 15:19


















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