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How to compute the Fourier transform of $e^{-2pi {mid x mid}}$?
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I want to compute the Fourier transform of $f(x)=e^{-2pi {mid x mid}}$ where $x$ is the vector in $mathbb{R}^n$. I know that the Fourier transform of a radial function is radial but cannot see how to use the fact to calculate the following integrate
begin{equation}
hat{f}(y)=int_{mathbb{R}^n} {e^{-2pi {mid x mid}}} e^{-2pi x cdot y } dx
end{equation}
Could anyone please help me?
analysis fourier-analysis fourier-transform
edited 2 days ago
Bernard
115k637108
asked 2 days ago
Keith
1,395820
add a comment |
up vote
0
down vote
favorite
I want to compute the Fourier transform of $f(x)=e^{-2pi {mid x mid}}$ where $x$ is the vector in $mathbb{R}^n$. I know that the Fourier transform of a radial function is radial but cannot see how to use the fact to calculate the following integrate
begin{equation}
hat{f}(y)=int_{mathbb{R}^n} {e^{-2pi {mid x mid}}} e^{-2pi x cdot y } dx
end{equation}
Could anyone please help me?
analysis fourier-analysis fourier-transform
edited 2 days ago
Bernard
115k637108
asked 2 days ago
Keith
1,395820
Use polar coordinates.
– uniquesolution
2 days ago
It is general n-dimensional. I cannot see how to use the polar coordinates.
– Keith
2 days ago
You can try writing $x=rphi$ with $rinmathbb R_+$ and $phiin S^{n-1}$.
– Federico
2 days ago
Look up the Hankel Transform and its relation to the Fourier Transform for cases of circular and spherical symmetry.
– Andy Walls
2 days ago
I just had a quick glance at the excellent lecture notes {see.stanford.edu/materials/lsoftaee261/chap8.pdf} but I didn't find an answer to your question.
– Jean Marie
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to compute the Fourier transform of $f(x)=e^{-2pi {mid x mid}}$ where $x$ is the vector in $mathbb{R}^n$. I know that the Fourier transform of a radial function is radial but cannot see how to use the fact to calculate the following integrate
begin{equation}
hat{f}(y)=int_{mathbb{R}^n} {e^{-2pi {mid x mid}}} e^{-2pi x cdot y } dx
end{equation}
Could anyone please help me?
analysis fourier-analysis fourier-transform
edited 2 days ago
Bernard
115k637108
asked 2 days ago
Keith
1,395820
I want to compute the Fourier transform of $f(x)=e^{-2pi {mid x mid}}$ where $x$ is the vector in $mathbb{R}^n$. I know that the Fourier transform of a radial function is radial but cannot see how to use the fact to calculate the following integrate
begin{equation}
hat{f}(y)=int_{mathbb{R}^n} {e^{-2pi {mid x mid}}} e^{-2pi x cdot y } dx
end{equation}
Could anyone please help me?
analysis fourier-analysis fourier-transform
analysis fourier-analysis fourier-transform
edited 2 days ago
Bernard
115k637108
asked 2 days ago
Keith
1,395820
edited 2 days ago
Bernard
115k637108
asked 2 days ago
Keith
1,395820
edited 2 days ago
Bernard
115k637108
edited 2 days ago
Bernard
115k637108
edited 2 days ago
Bernard
115k637108
115k637108
asked 2 days ago
Keith
1,395820
asked 2 days ago
Keith
1,395820
asked 2 days ago
Keith
1,395820
1,395820
Use polar coordinates.
– uniquesolution
2 days ago
It is general n-dimensional. I cannot see how to use the polar coordinates.
– Keith
2 days ago
You can try writing $x=rphi$ with $rinmathbb R_+$ and $phiin S^{n-1}$.
– Federico
2 days ago
Look up the Hankel Transform and its relation to the Fourier Transform for cases of circular and spherical symmetry.
– Andy Walls
2 days ago
I just had a quick glance at the excellent lecture notes {see.stanford.edu/materials/lsoftaee261/chap8.pdf} but I didn't find an answer to your question.
– Jean Marie
2 days ago
add a comment |
Use polar coordinates.
– uniquesolution
2 days ago
It is general n-dimensional. I cannot see how to use the polar coordinates.
– Keith
2 days ago
You can try writing $x=rphi$ with $rinmathbb R_+$ and $phiin S^{n-1}$.
– Federico
2 days ago
Look up the Hankel Transform and its relation to the Fourier Transform for cases of circular and spherical symmetry.
– Andy Walls
2 days ago
I just had a quick glance at the excellent lecture notes {see.stanford.edu/materials/lsoftaee261/chap8.pdf} but I didn't find an answer to your question.
– Jean Marie
2 days ago
Use polar coordinates.
– uniquesolution
2 days ago
Use polar coordinates.
– uniquesolution
2 days ago
It is general n-dimensional. I cannot see how to use the polar coordinates.
– Keith
2 days ago
It is general n-dimensional. I cannot see how to use the polar coordinates.
– Keith
2 days ago
You can try writing $x=rphi$ with $rinmathbb R_+$ and $phiin S^{n-1}$.
– Federico
2 days ago
You can try writing $x=rphi$ with $rinmathbb R_+$ and $phiin S^{n-1}$.
– Federico
2 days ago
Look up the Hankel Transform and its relation to the Fourier Transform for cases of circular and spherical symmetry.
– Andy Walls
2 days ago
Look up the Hankel Transform and its relation to the Fourier Transform for cases of circular and spherical symmetry.
– Andy Walls
2 days ago
I just had a quick glance at the excellent lecture notes {see.stanford.edu/materials/lsoftaee261/chap8.pdf} but I didn't find an answer to your question.
– Jean Marie
2 days ago
I just had a quick glance at the excellent lecture notes {see.stanford.edu/materials/lsoftaee261/chap8.pdf} but I didn't find an answer to your question.
– Jean Marie
2 days ago
add a comment |
1 Answer
1
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oldest
votes
up vote
1
down vote
From Bracewell, the n-dimensional Fourier Transform
$$F(s_1, dots, s_n) = int_{-infty}^{infty} dots int_{-infty}^{infty} f(x_1,dots, x_n) e^{-2pi i (x_1 s_1 + dots + x_n s_n)} space dx_1 dots dx_n $$
when there is n-dimensional symmetry, can be manipulated into
$$F(q) = dfrac{2pi}{q^{frac{1}{2}n-1}} int_{0}^{infty} f(r) J_{frac{1}{2}n-1}(2pi qr)space r^{frac{1}{2}n} space dr$$
Where $r$ and $q$ are the radii from the origin in the original domain and transform domain respectively.
So that the cases of $n=1$ and $n=3$ don't seem so foreign, it is probably worth noting that
$$J_{-frac{1}{2}}(x) = sqrt{dfrac{2}{pi x}}cos(x)$$
$$J_{frac{1}{2}}(x) = sqrt{dfrac{2}{pi x}}sin(x)$$
I'm not quite sure I know what you meant by $|x|$ in your original problem statement, but if you just meant the magnitude of $x$, then the specific transform you wanted to compute looks like
$$F(q) = dfrac{2pi}{q^{frac{1}{2}n-1}} int_{0}^{infty} e^{-2pi r} J_{frac{1}{2}n-1}(2pi qr)space r^{frac{1}{2}n} space dr$$
For $n=1$
$$begin{align*}F(q) &= 2pisqrt{q} int_{0}^{infty} e^{-2pi r} sqrt{dfrac{2}{pi(2pi qr)}}cos(2pi qr)space sqrt{r} space dr\
\
&= 2 int_{0}^{infty} e^{-2pi r} cos(2pi qr)space dr \
\
&=int_{0}^{infty} e^{-2pi r} left(e^{i2pi qr} + e^{-i2pi qr}right)space dr \
\
&= left[dfrac{e^{-2pi r}e^{i2pi qr}}{-2pi (1-iq)} + dfrac{e^{-2pi r}e^{-i2pi qr}}{-2pi (1+iq)}right]Biggr|_0^infty\
\
&= dfrac{1}{2pi}left[dfrac{1}{1-iq}+dfrac{1}{1+iq}right]\
\
&= dfrac{1}{2pi} dfrac{2}{1+q^2}
end{align*}$$
which is exactly what one would expect for the 1 dimensional Fourier Transform.
For $n=2$
$$begin{align*}F(q) &= 2pi int_{0}^{infty} e^{-2pi r} J_{0}(2pi qr)space r space dr\
\
&mbox{(Hankel Transform table lookup)}\
\
&= dfrac{2pi (2pi)}{left[(2pi q)^2 + (2pi)^2right]^{frac{3}{2}}}\
\
&= dfrac{1}{2pi} dfrac{1}{left(1+q^2right)^{frac{3}{2}}}
end{align*}$$
For $n =3$
$$begin{align*}F(q) &= dfrac{2pi}{sqrt{q}} int_{0}^{infty} e^{-2pi r} sqrt{dfrac{2}{pi(2pi qr)}}sin(2pi qr)space r^{frac{3}{2}} space dr \
\
&= dfrac{2}{q} int_{0}^{infty} e^{-2pi r} sin(2pi qr)space r space dr \
\
&mbox{(Wolfram Alpha invocation in lieu of integration by parts)}
\
&= dfrac{1}{(2pi)^2}dfrac{4}{(1+q^2)^2}\
end{align*}$$
And I suppose I'll let you keep on going until you see a pattern in terms of $n$ emerge.
answered yesterday
Andy Walls
1,314126
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
From Bracewell, the n-dimensional Fourier Transform
$$F(s_1, dots, s_n) = int_{-infty}^{infty} dots int_{-infty}^{infty} f(x_1,dots, x_n) e^{-2pi i (x_1 s_1 + dots + x_n s_n)} space dx_1 dots dx_n $$
when there is n-dimensional symmetry, can be manipulated into
$$F(q) = dfrac{2pi}{q^{frac{1}{2}n-1}} int_{0}^{infty} f(r) J_{frac{1}{2}n-1}(2pi qr)space r^{frac{1}{2}n} space dr$$
Where $r$ and $q$ are the radii from the origin in the original domain and transform domain respectively.
So that the cases of $n=1$ and $n=3$ don't seem so foreign, it is probably worth noting that
$$J_{-frac{1}{2}}(x) = sqrt{dfrac{2}{pi x}}cos(x)$$
$$J_{frac{1}{2}}(x) = sqrt{dfrac{2}{pi x}}sin(x)$$
I'm not quite sure I know what you meant by $|x|$ in your original problem statement, but if you just meant the magnitude of $x$, then the specific transform you wanted to compute looks like
$$F(q) = dfrac{2pi}{q^{frac{1}{2}n-1}} int_{0}^{infty} e^{-2pi r} J_{frac{1}{2}n-1}(2pi qr)space r^{frac{1}{2}n} space dr$$
For $n=1$
$$begin{align*}F(q) &= 2pisqrt{q} int_{0}^{infty} e^{-2pi r} sqrt{dfrac{2}{pi(2pi qr)}}cos(2pi qr)space sqrt{r} space dr\
\
&= 2 int_{0}^{infty} e^{-2pi r} cos(2pi qr)space dr \
\
&=int_{0}^{infty} e^{-2pi r} left(e^{i2pi qr} + e^{-i2pi qr}right)space dr \
\
&= left[dfrac{e^{-2pi r}e^{i2pi qr}}{-2pi (1-iq)} + dfrac{e^{-2pi r}e^{-i2pi qr}}{-2pi (1+iq)}right]Biggr|_0^infty\
\
&= dfrac{1}{2pi}left[dfrac{1}{1-iq}+dfrac{1}{1+iq}right]\
\
&= dfrac{1}{2pi} dfrac{2}{1+q^2}
end{align*}$$
which is exactly what one would expect for the 1 dimensional Fourier Transform.
For $n=2$
$$begin{align*}F(q) &= 2pi int_{0}^{infty} e^{-2pi r} J_{0}(2pi qr)space r space dr\
\
&mbox{(Hankel Transform table lookup)}\
\
&= dfrac{2pi (2pi)}{left[(2pi q)^2 + (2pi)^2right]^{frac{3}{2}}}\
\
&= dfrac{1}{2pi} dfrac{1}{left(1+q^2right)^{frac{3}{2}}}
end{align*}$$
For $n =3$
$$begin{align*}F(q) &= dfrac{2pi}{sqrt{q}} int_{0}^{infty} e^{-2pi r} sqrt{dfrac{2}{pi(2pi qr)}}sin(2pi qr)space r^{frac{3}{2}} space dr \
\
&= dfrac{2}{q} int_{0}^{infty} e^{-2pi r} sin(2pi qr)space r space dr \
\
&mbox{(Wolfram Alpha invocation in lieu of integration by parts)}
\
&= dfrac{1}{(2pi)^2}dfrac{4}{(1+q^2)^2}\
end{align*}$$
And I suppose I'll let you keep on going until you see a pattern in terms of $n$ emerge.
answered yesterday
Andy Walls
1,314126
add a comment |
up vote
1
down vote
From Bracewell, the n-dimensional Fourier Transform
$$F(s_1, dots, s_n) = int_{-infty}^{infty} dots int_{-infty}^{infty} f(x_1,dots, x_n) e^{-2pi i (x_1 s_1 + dots + x_n s_n)} space dx_1 dots dx_n $$
when there is n-dimensional symmetry, can be manipulated into
$$F(q) = dfrac{2pi}{q^{frac{1}{2}n-1}} int_{0}^{infty} f(r) J_{frac{1}{2}n-1}(2pi qr)space r^{frac{1}{2}n} space dr$$
Where $r$ and $q$ are the radii from the origin in the original domain and transform domain respectively.
So that the cases of $n=1$ and $n=3$ don't seem so foreign, it is probably worth noting that
$$J_{-frac{1}{2}}(x) = sqrt{dfrac{2}{pi x}}cos(x)$$
$$J_{frac{1}{2}}(x) = sqrt{dfrac{2}{pi x}}sin(x)$$
I'm not quite sure I know what you meant by $|x|$ in your original problem statement, but if you just meant the magnitude of $x$, then the specific transform you wanted to compute looks like
$$F(q) = dfrac{2pi}{q^{frac{1}{2}n-1}} int_{0}^{infty} e^{-2pi r} J_{frac{1}{2}n-1}(2pi qr)space r^{frac{1}{2}n} space dr$$
For $n=1$
$$begin{align*}F(q) &= 2pisqrt{q} int_{0}^{infty} e^{-2pi r} sqrt{dfrac{2}{pi(2pi qr)}}cos(2pi qr)space sqrt{r} space dr\
\
&= 2 int_{0}^{infty} e^{-2pi r} cos(2pi qr)space dr \
\
&=int_{0}^{infty} e^{-2pi r} left(e^{i2pi qr} + e^{-i2pi qr}right)space dr \
\
&= left[dfrac{e^{-2pi r}e^{i2pi qr}}{-2pi (1-iq)} + dfrac{e^{-2pi r}e^{-i2pi qr}}{-2pi (1+iq)}right]Biggr|_0^infty\
\
&= dfrac{1}{2pi}left[dfrac{1}{1-iq}+dfrac{1}{1+iq}right]\
\
&= dfrac{1}{2pi} dfrac{2}{1+q^2}
end{align*}$$
which is exactly what one would expect for the 1 dimensional Fourier Transform.
For $n=2$
$$begin{align*}F(q) &= 2pi int_{0}^{infty} e^{-2pi r} J_{0}(2pi qr)space r space dr\
\
&mbox{(Hankel Transform table lookup)}\
\
&= dfrac{2pi (2pi)}{left[(2pi q)^2 + (2pi)^2right]^{frac{3}{2}}}\
\
&= dfrac{1}{2pi} dfrac{1}{left(1+q^2right)^{frac{3}{2}}}
end{align*}$$
For $n =3$
$$begin{align*}F(q) &= dfrac{2pi}{sqrt{q}} int_{0}^{infty} e^{-2pi r} sqrt{dfrac{2}{pi(2pi qr)}}sin(2pi qr)space r^{frac{3}{2}} space dr \
\
&= dfrac{2}{q} int_{0}^{infty} e^{-2pi r} sin(2pi qr)space r space dr \
\
&mbox{(Wolfram Alpha invocation in lieu of integration by parts)}
\
&= dfrac{1}{(2pi)^2}dfrac{4}{(1+q^2)^2}\
end{align*}$$
And I suppose I'll let you keep on going until you see a pattern in terms of $n$ emerge.
answered yesterday
Andy Walls
1,314126
add a comment |
up vote
1
down vote
up vote
1
down vote
From Bracewell, the n-dimensional Fourier Transform
$$F(s_1, dots, s_n) = int_{-infty}^{infty} dots int_{-infty}^{infty} f(x_1,dots, x_n) e^{-2pi i (x_1 s_1 + dots + x_n s_n)} space dx_1 dots dx_n $$
when there is n-dimensional symmetry, can be manipulated into
$$F(q) = dfrac{2pi}{q^{frac{1}{2}n-1}} int_{0}^{infty} f(r) J_{frac{1}{2}n-1}(2pi qr)space r^{frac{1}{2}n} space dr$$
Where $r$ and $q$ are the radii from the origin in the original domain and transform domain respectively.
So that the cases of $n=1$ and $n=3$ don't seem so foreign, it is probably worth noting that
$$J_{-frac{1}{2}}(x) = sqrt{dfrac{2}{pi x}}cos(x)$$
$$J_{frac{1}{2}}(x) = sqrt{dfrac{2}{pi x}}sin(x)$$
I'm not quite sure I know what you meant by $|x|$ in your original problem statement, but if you just meant the magnitude of $x$, then the specific transform you wanted to compute looks like
$$F(q) = dfrac{2pi}{q^{frac{1}{2}n-1}} int_{0}^{infty} e^{-2pi r} J_{frac{1}{2}n-1}(2pi qr)space r^{frac{1}{2}n} space dr$$
For $n=1$
$$begin{align*}F(q) &= 2pisqrt{q} int_{0}^{infty} e^{-2pi r} sqrt{dfrac{2}{pi(2pi qr)}}cos(2pi qr)space sqrt{r} space dr\
\
&= 2 int_{0}^{infty} e^{-2pi r} cos(2pi qr)space dr \
\
&=int_{0}^{infty} e^{-2pi r} left(e^{i2pi qr} + e^{-i2pi qr}right)space dr \
\
&= left[dfrac{e^{-2pi r}e^{i2pi qr}}{-2pi (1-iq)} + dfrac{e^{-2pi r}e^{-i2pi qr}}{-2pi (1+iq)}right]Biggr|_0^infty\
\
&= dfrac{1}{2pi}left[dfrac{1}{1-iq}+dfrac{1}{1+iq}right]\
\
&= dfrac{1}{2pi} dfrac{2}{1+q^2}
end{align*}$$
which is exactly what one would expect for the 1 dimensional Fourier Transform.
For $n=2$
$$begin{align*}F(q) &= 2pi int_{0}^{infty} e^{-2pi r} J_{0}(2pi qr)space r space dr\
\
&mbox{(Hankel Transform table lookup)}\
\
&= dfrac{2pi (2pi)}{left[(2pi q)^2 + (2pi)^2right]^{frac{3}{2}}}\
\
&= dfrac{1}{2pi} dfrac{1}{left(1+q^2right)^{frac{3}{2}}}
end{align*}$$
For $n =3$
$$begin{align*}F(q) &= dfrac{2pi}{sqrt{q}} int_{0}^{infty} e^{-2pi r} sqrt{dfrac{2}{pi(2pi qr)}}sin(2pi qr)space r^{frac{3}{2}} space dr \
\
&= dfrac{2}{q} int_{0}^{infty} e^{-2pi r} sin(2pi qr)space r space dr \
\
&mbox{(Wolfram Alpha invocation in lieu of integration by parts)}
\
&= dfrac{1}{(2pi)^2}dfrac{4}{(1+q^2)^2}\
end{align*}$$
And I suppose I'll let you keep on going until you see a pattern in terms of $n$ emerge.
answered yesterday
Andy Walls
1,314126
From Bracewell, the n-dimensional Fourier Transform
$$F(s_1, dots, s_n) = int_{-infty}^{infty} dots int_{-infty}^{infty} f(x_1,dots, x_n) e^{-2pi i (x_1 s_1 + dots + x_n s_n)} space dx_1 dots dx_n $$
when there is n-dimensional symmetry, can be manipulated into
$$F(q) = dfrac{2pi}{q^{frac{1}{2}n-1}} int_{0}^{infty} f(r) J_{frac{1}{2}n-1}(2pi qr)space r^{frac{1}{2}n} space dr$$
Where $r$ and $q$ are the radii from the origin in the original domain and transform domain respectively.
So that the cases of $n=1$ and $n=3$ don't seem so foreign, it is probably worth noting that
$$J_{-frac{1}{2}}(x) = sqrt{dfrac{2}{pi x}}cos(x)$$
$$J_{frac{1}{2}}(x) = sqrt{dfrac{2}{pi x}}sin(x)$$
I'm not quite sure I know what you meant by $|x|$ in your original problem statement, but if you just meant the magnitude of $x$, then the specific transform you wanted to compute looks like
$$F(q) = dfrac{2pi}{q^{frac{1}{2}n-1}} int_{0}^{infty} e^{-2pi r} J_{frac{1}{2}n-1}(2pi qr)space r^{frac{1}{2}n} space dr$$
For $n=1$
$$begin{align*}F(q) &= 2pisqrt{q} int_{0}^{infty} e^{-2pi r} sqrt{dfrac{2}{pi(2pi qr)}}cos(2pi qr)space sqrt{r} space dr\
\
&= 2 int_{0}^{infty} e^{-2pi r} cos(2pi qr)space dr \
\
&=int_{0}^{infty} e^{-2pi r} left(e^{i2pi qr} + e^{-i2pi qr}right)space dr \
\
&= left[dfrac{e^{-2pi r}e^{i2pi qr}}{-2pi (1-iq)} + dfrac{e^{-2pi r}e^{-i2pi qr}}{-2pi (1+iq)}right]Biggr|_0^infty\
\
&= dfrac{1}{2pi}left[dfrac{1}{1-iq}+dfrac{1}{1+iq}right]\
\
&= dfrac{1}{2pi} dfrac{2}{1+q^2}
end{align*}$$
which is exactly what one would expect for the 1 dimensional Fourier Transform.
For $n=2$
$$begin{align*}F(q) &= 2pi int_{0}^{infty} e^{-2pi r} J_{0}(2pi qr)space r space dr\
\
&mbox{(Hankel Transform table lookup)}\
\
&= dfrac{2pi (2pi)}{left[(2pi q)^2 + (2pi)^2right]^{frac{3}{2}}}\
\
&= dfrac{1}{2pi} dfrac{1}{left(1+q^2right)^{frac{3}{2}}}
end{align*}$$
For $n =3$
$$begin{align*}F(q) &= dfrac{2pi}{sqrt{q}} int_{0}^{infty} e^{-2pi r} sqrt{dfrac{2}{pi(2pi qr)}}sin(2pi qr)space r^{frac{3}{2}} space dr \
\
&= dfrac{2}{q} int_{0}^{infty} e^{-2pi r} sin(2pi qr)space r space dr \
\
&mbox{(Wolfram Alpha invocation in lieu of integration by parts)}
\
&= dfrac{1}{(2pi)^2}dfrac{4}{(1+q^2)^2}\
end{align*}$$
And I suppose I'll let you keep on going until you see a pattern in terms of $n$ emerge.
answered yesterday
Andy Walls
1,314126
answered yesterday
Andy Walls
1,314126
answered yesterday
Andy Walls
1,314126
answered yesterday
Andy Walls
1,314126
1,314126
add a comment |
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Use polar coordinates.
– uniquesolution
2 days ago
It is general n-dimensional. I cannot see how to use the polar coordinates.
– Keith
2 days ago
You can try writing $x=rphi$ with $rinmathbb R_+$ and $phiin S^{n-1}$.
– Federico
2 days ago
Look up the Hankel Transform and its relation to the Fourier Transform for cases of circular and spherical symmetry.
– Andy Walls
2 days ago
I just had a quick glance at the excellent lecture notes {see.stanford.edu/materials/lsoftaee261/chap8.pdf} but I didn't find an answer to your question.
– Jean Marie
2 days ago