Fourier Transform Of $h(x,y)=Acdot e^{-2pi^2(x^2+y^2)}$












2












$begingroup$



Prove: $$h(x,y)=Acdot e^{-2pi^2(x^2+y^2)}$$ is $$H(u,v)=frac{A}{sqrt{2}}cdot e^{-frac{u^2+v^2}{2}}$$
After fourier transform




$$F(u,v)=int_{-infty}^{infty}int_{-infty}^{infty}f(x,y)e^{-i2pi (ux+vy)}dxdy$$



$$H(u,v)=int_{-infty}^{infty}int_{-infty}^{infty}Acdot e^{-2pi^2(x^2+y^2)}e^{-i2pi (ux+vy)}dxdy$$



$$H(u,v)=Aint_{-infty}^{infty}int_{-infty}^{infty}e^{-2pi^2(x^2+y^2)}e^{-i2pi (ux+vy)}dxdy$$



$$H(u,v)=Aint_{-infty}^{infty}int_{-infty}^{infty}e^{-2pi^2x^2-2pi^2y^2-i2pi ux -i2pi vy}dxdy$$



How should I group $-2pi^2x^2-2pi^2y^2-i2pi ux -i2pi vy$ in order to integrate?






fourier-analysis fourier-transform







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    2












    $begingroup$



    Prove: $$h(x,y)=Acdot e^{-2pi^2(x^2+y^2)}$$ is $$H(u,v)=frac{A}{sqrt{2}}cdot e^{-frac{u^2+v^2}{2}}$$
    After fourier transform




    $$F(u,v)=int_{-infty}^{infty}int_{-infty}^{infty}f(x,y)e^{-i2pi (ux+vy)}dxdy$$



    $$H(u,v)=int_{-infty}^{infty}int_{-infty}^{infty}Acdot e^{-2pi^2(x^2+y^2)}e^{-i2pi (ux+vy)}dxdy$$



    $$H(u,v)=Aint_{-infty}^{infty}int_{-infty}^{infty}e^{-2pi^2(x^2+y^2)}e^{-i2pi (ux+vy)}dxdy$$



    $$H(u,v)=Aint_{-infty}^{infty}int_{-infty}^{infty}e^{-2pi^2x^2-2pi^2y^2-i2pi ux -i2pi vy}dxdy$$



    How should I group $-2pi^2x^2-2pi^2y^2-i2pi ux -i2pi vy$ in order to integrate?






    fourier-analysis fourier-transform







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$



      Prove: $$h(x,y)=Acdot e^{-2pi^2(x^2+y^2)}$$ is $$H(u,v)=frac{A}{sqrt{2}}cdot e^{-frac{u^2+v^2}{2}}$$
      After fourier transform




      $$F(u,v)=int_{-infty}^{infty}int_{-infty}^{infty}f(x,y)e^{-i2pi (ux+vy)}dxdy$$



      $$H(u,v)=int_{-infty}^{infty}int_{-infty}^{infty}Acdot e^{-2pi^2(x^2+y^2)}e^{-i2pi (ux+vy)}dxdy$$



      $$H(u,v)=Aint_{-infty}^{infty}int_{-infty}^{infty}e^{-2pi^2(x^2+y^2)}e^{-i2pi (ux+vy)}dxdy$$



      $$H(u,v)=Aint_{-infty}^{infty}int_{-infty}^{infty}e^{-2pi^2x^2-2pi^2y^2-i2pi ux -i2pi vy}dxdy$$



      How should I group $-2pi^2x^2-2pi^2y^2-i2pi ux -i2pi vy$ in order to integrate?






      fourier-analysis fourier-transform







      share|cite|improve this question









      $endgroup$





      Prove: $$h(x,y)=Acdot e^{-2pi^2(x^2+y^2)}$$ is $$H(u,v)=frac{A}{sqrt{2}}cdot e^{-frac{u^2+v^2}{2}}$$
      After fourier transform




      $$F(u,v)=int_{-infty}^{infty}int_{-infty}^{infty}f(x,y)e^{-i2pi (ux+vy)}dxdy$$



      $$H(u,v)=int_{-infty}^{infty}int_{-infty}^{infty}Acdot e^{-2pi^2(x^2+y^2)}e^{-i2pi (ux+vy)}dxdy$$



      $$H(u,v)=Aint_{-infty}^{infty}int_{-infty}^{infty}e^{-2pi^2(x^2+y^2)}e^{-i2pi (ux+vy)}dxdy$$



      $$H(u,v)=Aint_{-infty}^{infty}int_{-infty}^{infty}e^{-2pi^2x^2-2pi^2y^2-i2pi ux -i2pi vy}dxdy$$



      How should I group $-2pi^2x^2-2pi^2y^2-i2pi ux -i2pi vy$ in order to integrate?






      fourier-analysis fourier-transform




      fourier-analysis fourier-transform






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      asked Jan 24 at 16:40









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          $begingroup$

          Complete squares



          begin{align}
          -2pi^2 x^2 - 2pi^2y^2 - i2pi u x - i 2pi v y &=
          -2pi^2left(x + frac{i u}{pi}right)^2 - frac{u^2}{2}
          -2pi^2left(y + frac{i u}{pi}right)^2 - frac{v^2}{2}
          end{align}



          So the integral becomes



          $$
          H(u, v) = A e^{-u^2/2 - v^2/2}left(int_{-infty}^{+infty} e^{-2pi^2(x + iu/pi)^2}{rm d}xright)left(int_{-infty}^{+infty} e^{-2pi^2(y + iv/pi)^2}{rm d}yright)
          $$



          The integrals are fairly easy to calculate



          $$
          int_{-infty}^{+infty} e^{-2pi^2(x + iu/pi)^2}{rm d}x = frac{1}{sqrt{2pi}}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So we will end up with $frac{A}{2}e^{-frac{u^2+v^2}{2}}$?
            $endgroup$
            – gbox
            Jan 24 at 17:18








          • 1




            $begingroup$
            @gbox $(A/2pi)e^{-(u^2 + v^2)/2}$
            $endgroup$
            – caverac
            Jan 24 at 17:20










          • $begingroup$
            yes sorry, so we still miss a sqrt in $2pi$
            $endgroup$
            – gbox
            Jan 24 at 17:53






          • 1




            $begingroup$
            @gbox That could be the definition of Fourier transform you are applying. There's always a factor of $2pi$ in there, authors use different conventions. You need to figure out which one is the one you're using math.stackexchange.com/questions/301860/…
            $endgroup$
            – caverac
            Jan 24 at 18:12











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          1












          $begingroup$

          Complete squares



          begin{align}
          -2pi^2 x^2 - 2pi^2y^2 - i2pi u x - i 2pi v y &=
          -2pi^2left(x + frac{i u}{pi}right)^2 - frac{u^2}{2}
          -2pi^2left(y + frac{i u}{pi}right)^2 - frac{v^2}{2}
          end{align}



          So the integral becomes



          $$
          H(u, v) = A e^{-u^2/2 - v^2/2}left(int_{-infty}^{+infty} e^{-2pi^2(x + iu/pi)^2}{rm d}xright)left(int_{-infty}^{+infty} e^{-2pi^2(y + iv/pi)^2}{rm d}yright)
          $$



          The integrals are fairly easy to calculate



          $$
          int_{-infty}^{+infty} e^{-2pi^2(x + iu/pi)^2}{rm d}x = frac{1}{sqrt{2pi}}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So we will end up with $frac{A}{2}e^{-frac{u^2+v^2}{2}}$?
            $endgroup$
            – gbox
            Jan 24 at 17:18








          • 1




            $begingroup$
            @gbox $(A/2pi)e^{-(u^2 + v^2)/2}$
            $endgroup$
            – caverac
            Jan 24 at 17:20










          • $begingroup$
            yes sorry, so we still miss a sqrt in $2pi$
            $endgroup$
            – gbox
            Jan 24 at 17:53






          • 1




            $begingroup$
            @gbox That could be the definition of Fourier transform you are applying. There's always a factor of $2pi$ in there, authors use different conventions. You need to figure out which one is the one you're using math.stackexchange.com/questions/301860/…
            $endgroup$
            – caverac
            Jan 24 at 18:12
















          1












          $begingroup$

          Complete squares



          begin{align}
          -2pi^2 x^2 - 2pi^2y^2 - i2pi u x - i 2pi v y &=
          -2pi^2left(x + frac{i u}{pi}right)^2 - frac{u^2}{2}
          -2pi^2left(y + frac{i u}{pi}right)^2 - frac{v^2}{2}
          end{align}



          So the integral becomes



          $$
          H(u, v) = A e^{-u^2/2 - v^2/2}left(int_{-infty}^{+infty} e^{-2pi^2(x + iu/pi)^2}{rm d}xright)left(int_{-infty}^{+infty} e^{-2pi^2(y + iv/pi)^2}{rm d}yright)
          $$



          The integrals are fairly easy to calculate



          $$
          int_{-infty}^{+infty} e^{-2pi^2(x + iu/pi)^2}{rm d}x = frac{1}{sqrt{2pi}}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So we will end up with $frac{A}{2}e^{-frac{u^2+v^2}{2}}$?
            $endgroup$
            – gbox
            Jan 24 at 17:18








          • 1




            $begingroup$
            @gbox $(A/2pi)e^{-(u^2 + v^2)/2}$
            $endgroup$
            – caverac
            Jan 24 at 17:20










          • $begingroup$
            yes sorry, so we still miss a sqrt in $2pi$
            $endgroup$
            – gbox
            Jan 24 at 17:53






          • 1




            $begingroup$
            @gbox That could be the definition of Fourier transform you are applying. There's always a factor of $2pi$ in there, authors use different conventions. You need to figure out which one is the one you're using math.stackexchange.com/questions/301860/…
            $endgroup$
            – caverac
            Jan 24 at 18:12














          1












          1








          1





          $begingroup$

          Complete squares



          begin{align}
          -2pi^2 x^2 - 2pi^2y^2 - i2pi u x - i 2pi v y &=
          -2pi^2left(x + frac{i u}{pi}right)^2 - frac{u^2}{2}
          -2pi^2left(y + frac{i u}{pi}right)^2 - frac{v^2}{2}
          end{align}



          So the integral becomes



          $$
          H(u, v) = A e^{-u^2/2 - v^2/2}left(int_{-infty}^{+infty} e^{-2pi^2(x + iu/pi)^2}{rm d}xright)left(int_{-infty}^{+infty} e^{-2pi^2(y + iv/pi)^2}{rm d}yright)
          $$



          The integrals are fairly easy to calculate



          $$
          int_{-infty}^{+infty} e^{-2pi^2(x + iu/pi)^2}{rm d}x = frac{1}{sqrt{2pi}}
          $$






          share|cite|improve this answer









          $endgroup$



          Complete squares



          begin{align}
          -2pi^2 x^2 - 2pi^2y^2 - i2pi u x - i 2pi v y &=
          -2pi^2left(x + frac{i u}{pi}right)^2 - frac{u^2}{2}
          -2pi^2left(y + frac{i u}{pi}right)^2 - frac{v^2}{2}
          end{align}



          So the integral becomes



          $$
          H(u, v) = A e^{-u^2/2 - v^2/2}left(int_{-infty}^{+infty} e^{-2pi^2(x + iu/pi)^2}{rm d}xright)left(int_{-infty}^{+infty} e^{-2pi^2(y + iv/pi)^2}{rm d}yright)
          $$



          The integrals are fairly easy to calculate



          $$
          int_{-infty}^{+infty} e^{-2pi^2(x + iu/pi)^2}{rm d}x = frac{1}{sqrt{2pi}}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 24 at 16:58









          caveraccaverac

          14.8k31130




          14.8k31130












          • $begingroup$
            So we will end up with $frac{A}{2}e^{-frac{u^2+v^2}{2}}$?
            $endgroup$
            – gbox
            Jan 24 at 17:18








          • 1




            $begingroup$
            @gbox $(A/2pi)e^{-(u^2 + v^2)/2}$
            $endgroup$
            – caverac
            Jan 24 at 17:20










          • $begingroup$
            yes sorry, so we still miss a sqrt in $2pi$
            $endgroup$
            – gbox
            Jan 24 at 17:53






          • 1




            $begingroup$
            @gbox That could be the definition of Fourier transform you are applying. There's always a factor of $2pi$ in there, authors use different conventions. You need to figure out which one is the one you're using math.stackexchange.com/questions/301860/…
            $endgroup$
            – caverac
            Jan 24 at 18:12


















          • $begingroup$
            So we will end up with $frac{A}{2}e^{-frac{u^2+v^2}{2}}$?
            $endgroup$
            – gbox
            Jan 24 at 17:18








          • 1




            $begingroup$
            @gbox $(A/2pi)e^{-(u^2 + v^2)/2}$
            $endgroup$
            – caverac
            Jan 24 at 17:20










          • $begingroup$
            yes sorry, so we still miss a sqrt in $2pi$
            $endgroup$
            – gbox
            Jan 24 at 17:53






          • 1




            $begingroup$
            @gbox That could be the definition of Fourier transform you are applying. There's always a factor of $2pi$ in there, authors use different conventions. You need to figure out which one is the one you're using math.stackexchange.com/questions/301860/…
            $endgroup$
            – caverac
            Jan 24 at 18:12
















          $begingroup$
          So we will end up with $frac{A}{2}e^{-frac{u^2+v^2}{2}}$?
          $endgroup$
          – gbox
          Jan 24 at 17:18






          $begingroup$
          So we will end up with $frac{A}{2}e^{-frac{u^2+v^2}{2}}$?
          $endgroup$
          – gbox
          Jan 24 at 17:18






          1




          1




          $begingroup$
          @gbox $(A/2pi)e^{-(u^2 + v^2)/2}$
          $endgroup$
          – caverac
          Jan 24 at 17:20




          $begingroup$
          @gbox $(A/2pi)e^{-(u^2 + v^2)/2}$
          $endgroup$
          – caverac
          Jan 24 at 17:20












          $begingroup$
          yes sorry, so we still miss a sqrt in $2pi$
          $endgroup$
          – gbox
          Jan 24 at 17:53




          $begingroup$
          yes sorry, so we still miss a sqrt in $2pi$
          $endgroup$
          – gbox
          Jan 24 at 17:53




          1




          1




          $begingroup$
          @gbox That could be the definition of Fourier transform you are applying. There's always a factor of $2pi$ in there, authors use different conventions. You need to figure out which one is the one you're using math.stackexchange.com/questions/301860/…
          $endgroup$
          – caverac
          Jan 24 at 18:12




          $begingroup$
          @gbox That could be the definition of Fourier transform you are applying. There's always a factor of $2pi$ in there, authors use different conventions. You need to figure out which one is the one you're using math.stackexchange.com/questions/301860/…
          $endgroup$
          – caverac
          Jan 24 at 18:12


















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