A Triple integral












1












$begingroup$


While trying to obtain a Green's function for a PDE,I stumbled upon this integral



$displaystyleint_{-infty}^{infty}int_{-infty}^{infty}int_{-infty}^{infty}frac{e^{i(s_{x}x+s_{y}y+s_{z}z)}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{x}ds_{y}ds_{z}$



How can I evaluate this triple integral?










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    While trying to obtain a Green's function for a PDE,I stumbled upon this integral



    $displaystyleint_{-infty}^{infty}int_{-infty}^{infty}int_{-infty}^{infty}frac{e^{i(s_{x}x+s_{y}y+s_{z}z)}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{x}ds_{y}ds_{z}$



    How can I evaluate this triple integral?










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      While trying to obtain a Green's function for a PDE,I stumbled upon this integral



      $displaystyleint_{-infty}^{infty}int_{-infty}^{infty}int_{-infty}^{infty}frac{e^{i(s_{x}x+s_{y}y+s_{z}z)}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{x}ds_{y}ds_{z}$



      How can I evaluate this triple integral?










      share|cite|improve this question











      $endgroup$




      While trying to obtain a Green's function for a PDE,I stumbled upon this integral



      $displaystyleint_{-infty}^{infty}int_{-infty}^{infty}int_{-infty}^{infty}frac{e^{i(s_{x}x+s_{y}y+s_{z}z)}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{x}ds_{y}ds_{z}$



      How can I evaluate this triple integral?







      fourier-transform greens-function






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 25 '18 at 11:18

























      asked Dec 23 '18 at 17:40







      user628607





























          1 Answer
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          $begingroup$

          You can use the Inverse Fourier Transform to evaluate the 3 integrals. In the derivation below, I have only used the properties and transform pairs listed on this page: https://en.wikipedia.org/wiki/Fourier_transform



          $$begin{align*}&displaystyleint_{-infty}^{infty}int_{-infty}^{infty}int_{-infty}^{infty}frac{e^{i(s_{x}x+s_{y}y+s_{z}z)}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{x}ds_{y}ds_{z} \
          \
          &= int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} int_{-infty}^{infty}frac{e^{is_{z}z}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{z}ds_{y}ds_{x} \
          \
          &= int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} space 2pidfrac{i}{2a}dfrac{1}{2}spacemathscr{F}^{-1}left{frac{2}{ileft(s_{z} -left[a - frac{s_{x}^{2}+s_{y}^{2}}{2a}right] right)}right}ds_{y}ds_{x}\
          \
          &= dfrac{ipi}{2a} int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} space mathrm{sgn}(z) space e^{izleft(a - frac{s_{x}^{2}+s_{y}^{2}}{2a}right) } ds_{y}ds_{x}\
          \
          &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{y}^{2}} space e^{i s_{y}y} space ds_{y}ds_{x}\
          \
          &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x} int_{-infty}^{infty} 2pi e^{-ifrac{pi}{4}}sqrt{frac{a/2z}{pi}}sqrt{frac{pi}{a/2z}}e^{-ileft(frac{s_{y}^{2}}{4(a/2z)}-frac{pi}{4}right)} space frac{1}{2pi}e^{i s_{y}y} space ds_{y}ds_{x}\
          \
          &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a/2z}{pi}}mathscr{F}^{-1}left{sqrt{frac{pi}{a/2z}}e^{-ileft(frac{s_{y}^{2}}{4(a/2z)}-frac{pi}{4}right)} right}ds_{x}\
          \
          &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2}ds_{x}\
          \
          &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space ds_{x}\
          \
          &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}x^2} \
          \
          &= pi^2dfrac{mathrm{sgn}(z)}{z} e^{iaz} space e^{ifrac{a}{2z}left(x^2+y^2right)}\
          \
          end{align*}$$



          if I didn't make an error.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            You can use the Inverse Fourier Transform to evaluate the 3 integrals. In the derivation below, I have only used the properties and transform pairs listed on this page: https://en.wikipedia.org/wiki/Fourier_transform



            $$begin{align*}&displaystyleint_{-infty}^{infty}int_{-infty}^{infty}int_{-infty}^{infty}frac{e^{i(s_{x}x+s_{y}y+s_{z}z)}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{x}ds_{y}ds_{z} \
            \
            &= int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} int_{-infty}^{infty}frac{e^{is_{z}z}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{z}ds_{y}ds_{x} \
            \
            &= int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} space 2pidfrac{i}{2a}dfrac{1}{2}spacemathscr{F}^{-1}left{frac{2}{ileft(s_{z} -left[a - frac{s_{x}^{2}+s_{y}^{2}}{2a}right] right)}right}ds_{y}ds_{x}\
            \
            &= dfrac{ipi}{2a} int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} space mathrm{sgn}(z) space e^{izleft(a - frac{s_{x}^{2}+s_{y}^{2}}{2a}right) } ds_{y}ds_{x}\
            \
            &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{y}^{2}} space e^{i s_{y}y} space ds_{y}ds_{x}\
            \
            &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x} int_{-infty}^{infty} 2pi e^{-ifrac{pi}{4}}sqrt{frac{a/2z}{pi}}sqrt{frac{pi}{a/2z}}e^{-ileft(frac{s_{y}^{2}}{4(a/2z)}-frac{pi}{4}right)} space frac{1}{2pi}e^{i s_{y}y} space ds_{y}ds_{x}\
            \
            &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a/2z}{pi}}mathscr{F}^{-1}left{sqrt{frac{pi}{a/2z}}e^{-ileft(frac{s_{y}^{2}}{4(a/2z)}-frac{pi}{4}right)} right}ds_{x}\
            \
            &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2}ds_{x}\
            \
            &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space ds_{x}\
            \
            &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}x^2} \
            \
            &= pi^2dfrac{mathrm{sgn}(z)}{z} e^{iaz} space e^{ifrac{a}{2z}left(x^2+y^2right)}\
            \
            end{align*}$$



            if I didn't make an error.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              You can use the Inverse Fourier Transform to evaluate the 3 integrals. In the derivation below, I have only used the properties and transform pairs listed on this page: https://en.wikipedia.org/wiki/Fourier_transform



              $$begin{align*}&displaystyleint_{-infty}^{infty}int_{-infty}^{infty}int_{-infty}^{infty}frac{e^{i(s_{x}x+s_{y}y+s_{z}z)}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{x}ds_{y}ds_{z} \
              \
              &= int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} int_{-infty}^{infty}frac{e^{is_{z}z}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{z}ds_{y}ds_{x} \
              \
              &= int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} space 2pidfrac{i}{2a}dfrac{1}{2}spacemathscr{F}^{-1}left{frac{2}{ileft(s_{z} -left[a - frac{s_{x}^{2}+s_{y}^{2}}{2a}right] right)}right}ds_{y}ds_{x}\
              \
              &= dfrac{ipi}{2a} int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} space mathrm{sgn}(z) space e^{izleft(a - frac{s_{x}^{2}+s_{y}^{2}}{2a}right) } ds_{y}ds_{x}\
              \
              &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{y}^{2}} space e^{i s_{y}y} space ds_{y}ds_{x}\
              \
              &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x} int_{-infty}^{infty} 2pi e^{-ifrac{pi}{4}}sqrt{frac{a/2z}{pi}}sqrt{frac{pi}{a/2z}}e^{-ileft(frac{s_{y}^{2}}{4(a/2z)}-frac{pi}{4}right)} space frac{1}{2pi}e^{i s_{y}y} space ds_{y}ds_{x}\
              \
              &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a/2z}{pi}}mathscr{F}^{-1}left{sqrt{frac{pi}{a/2z}}e^{-ileft(frac{s_{y}^{2}}{4(a/2z)}-frac{pi}{4}right)} right}ds_{x}\
              \
              &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2}ds_{x}\
              \
              &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space ds_{x}\
              \
              &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}x^2} \
              \
              &= pi^2dfrac{mathrm{sgn}(z)}{z} e^{iaz} space e^{ifrac{a}{2z}left(x^2+y^2right)}\
              \
              end{align*}$$



              if I didn't make an error.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                You can use the Inverse Fourier Transform to evaluate the 3 integrals. In the derivation below, I have only used the properties and transform pairs listed on this page: https://en.wikipedia.org/wiki/Fourier_transform



                $$begin{align*}&displaystyleint_{-infty}^{infty}int_{-infty}^{infty}int_{-infty}^{infty}frac{e^{i(s_{x}x+s_{y}y+s_{z}z)}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{x}ds_{y}ds_{z} \
                \
                &= int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} int_{-infty}^{infty}frac{e^{is_{z}z}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{z}ds_{y}ds_{x} \
                \
                &= int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} space 2pidfrac{i}{2a}dfrac{1}{2}spacemathscr{F}^{-1}left{frac{2}{ileft(s_{z} -left[a - frac{s_{x}^{2}+s_{y}^{2}}{2a}right] right)}right}ds_{y}ds_{x}\
                \
                &= dfrac{ipi}{2a} int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} space mathrm{sgn}(z) space e^{izleft(a - frac{s_{x}^{2}+s_{y}^{2}}{2a}right) } ds_{y}ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{y}^{2}} space e^{i s_{y}y} space ds_{y}ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x} int_{-infty}^{infty} 2pi e^{-ifrac{pi}{4}}sqrt{frac{a/2z}{pi}}sqrt{frac{pi}{a/2z}}e^{-ileft(frac{s_{y}^{2}}{4(a/2z)}-frac{pi}{4}right)} space frac{1}{2pi}e^{i s_{y}y} space ds_{y}ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a/2z}{pi}}mathscr{F}^{-1}left{sqrt{frac{pi}{a/2z}}e^{-ileft(frac{s_{y}^{2}}{4(a/2z)}-frac{pi}{4}right)} right}ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2}ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}x^2} \
                \
                &= pi^2dfrac{mathrm{sgn}(z)}{z} e^{iaz} space e^{ifrac{a}{2z}left(x^2+y^2right)}\
                \
                end{align*}$$



                if I didn't make an error.






                share|cite|improve this answer











                $endgroup$



                You can use the Inverse Fourier Transform to evaluate the 3 integrals. In the derivation below, I have only used the properties and transform pairs listed on this page: https://en.wikipedia.org/wiki/Fourier_transform



                $$begin{align*}&displaystyleint_{-infty}^{infty}int_{-infty}^{infty}int_{-infty}^{infty}frac{e^{i(s_{x}x+s_{y}y+s_{z}z)}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{x}ds_{y}ds_{z} \
                \
                &= int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} int_{-infty}^{infty}frac{e^{is_{z}z}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{z}ds_{y}ds_{x} \
                \
                &= int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} space 2pidfrac{i}{2a}dfrac{1}{2}spacemathscr{F}^{-1}left{frac{2}{ileft(s_{z} -left[a - frac{s_{x}^{2}+s_{y}^{2}}{2a}right] right)}right}ds_{y}ds_{x}\
                \
                &= dfrac{ipi}{2a} int_{-infty}^{infty} e^{i s_{x}x} int_{-infty}^{infty} e^{i s_{y}y} space mathrm{sgn}(z) space e^{izleft(a - frac{s_{x}^{2}+s_{y}^{2}}{2a}right) } ds_{y}ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{y}^{2}} space e^{i s_{y}y} space ds_{y}ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x} int_{-infty}^{infty} 2pi e^{-ifrac{pi}{4}}sqrt{frac{a/2z}{pi}}sqrt{frac{pi}{a/2z}}e^{-ileft(frac{s_{y}^{2}}{4(a/2z)}-frac{pi}{4}right)} space frac{1}{2pi}e^{i s_{y}y} space ds_{y}ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a/2z}{pi}}mathscr{F}^{-1}left{sqrt{frac{pi}{a/2z}}e^{-ileft(frac{s_{y}^{2}}{4(a/2z)}-frac{pi}{4}right)} right}ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2}ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2} int_{-infty}^{infty} e^{-ifrac{z}{2a}s_{x}^{2}} space e^{i s_{x}x}space ds_{x}\
                \
                &= dfrac{ipi}{2a} mathrm{sgn}(z) space e^{iaz} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}y^2} space 2pi e^{-ifrac{pi}{4}}sqrt{frac{a}{2pi z}} e^{ifrac{a}{2z}x^2} \
                \
                &= pi^2dfrac{mathrm{sgn}(z)}{z} e^{iaz} space e^{ifrac{a}{2z}left(x^2+y^2right)}\
                \
                end{align*}$$



                if I didn't make an error.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 23 '18 at 23:37

























                answered Dec 23 '18 at 23:31









                Andy WallsAndy Walls

                1,754139




                1,754139






























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