A Triple integral
$begingroup$
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?
fourier-transform greens-function
$endgroup$
add a comment |
$begingroup$
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?
fourier-transform greens-function
$endgroup$
add a comment |
$begingroup$
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?
fourier-transform greens-function
$endgroup$
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
fourier-transform greens-function
edited Dec 25 '18 at 11:18
asked Dec 23 '18 at 17:40
user628607
add a comment |
add a comment |
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.
$endgroup$
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1 Answer
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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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Dec 23 '18 at 23:37
answered Dec 23 '18 at 23:31
Andy WallsAndy Walls
1,754139
1,754139
add a comment |
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