Fourier transform of product of Bessel functions











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I need help finding the Fourier transform of the function



$$
rho(vec{r}) = alpha delta_{vec{r},0} left(lambdalambda' J_1
(beta |vec{r}|)Y_1(beta |vec{r}|) - pi^2 J_0
(beta |vec{r}|)Y_0(beta |vec{r}|) right),
$$



where $alpha,beta in mathcal{R}$ and $lambda,lambda'=pm1$. The $J_i(x)$ are the $i$-th Bessel functions of the first, the $Y_i(x)$ are the $i$-th Bessel functions of the second kind. $delta_{vec{r},0}$ denotes the Kronecker delta.



The vector $vec{r} = (x,y)^T$ is discrete because I'm working on a discrete set of points. I'm a physicist, so please don't hesitate to ask for more specific information.



I tried something like



$$
rho(vec{k}) = sum_vec{r} e^{-ivec{k}cdotvec{r}}rho(vec{r}),
$$



which should in fact not be too difficult because the only vector that contributes to the sum is $vec{r} = vec{0}$ due to the Kronecker delta in $rho(vec{r})$. The problem is that $rho(vec{r})$ diverges at this point. I need some mathematical advice here. Is there anybody who can show me how to compute the Fourier transform of $rho(vec{r})$?










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  • Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
    – Andy Walls
    Aug 11 at 22:09










  • @AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
    – MeMeansMe
    Aug 13 at 7:50















up vote
1
down vote

favorite












I need help finding the Fourier transform of the function



$$
rho(vec{r}) = alpha delta_{vec{r},0} left(lambdalambda' J_1
(beta |vec{r}|)Y_1(beta |vec{r}|) - pi^2 J_0
(beta |vec{r}|)Y_0(beta |vec{r}|) right),
$$



where $alpha,beta in mathcal{R}$ and $lambda,lambda'=pm1$. The $J_i(x)$ are the $i$-th Bessel functions of the first, the $Y_i(x)$ are the $i$-th Bessel functions of the second kind. $delta_{vec{r},0}$ denotes the Kronecker delta.



The vector $vec{r} = (x,y)^T$ is discrete because I'm working on a discrete set of points. I'm a physicist, so please don't hesitate to ask for more specific information.



I tried something like



$$
rho(vec{k}) = sum_vec{r} e^{-ivec{k}cdotvec{r}}rho(vec{r}),
$$



which should in fact not be too difficult because the only vector that contributes to the sum is $vec{r} = vec{0}$ due to the Kronecker delta in $rho(vec{r})$. The problem is that $rho(vec{r})$ diverges at this point. I need some mathematical advice here. Is there anybody who can show me how to compute the Fourier transform of $rho(vec{r})$?










share|cite|improve this question
























  • Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
    – Andy Walls
    Aug 11 at 22:09










  • @AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
    – MeMeansMe
    Aug 13 at 7:50













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I need help finding the Fourier transform of the function



$$
rho(vec{r}) = alpha delta_{vec{r},0} left(lambdalambda' J_1
(beta |vec{r}|)Y_1(beta |vec{r}|) - pi^2 J_0
(beta |vec{r}|)Y_0(beta |vec{r}|) right),
$$



where $alpha,beta in mathcal{R}$ and $lambda,lambda'=pm1$. The $J_i(x)$ are the $i$-th Bessel functions of the first, the $Y_i(x)$ are the $i$-th Bessel functions of the second kind. $delta_{vec{r},0}$ denotes the Kronecker delta.



The vector $vec{r} = (x,y)^T$ is discrete because I'm working on a discrete set of points. I'm a physicist, so please don't hesitate to ask for more specific information.



I tried something like



$$
rho(vec{k}) = sum_vec{r} e^{-ivec{k}cdotvec{r}}rho(vec{r}),
$$



which should in fact not be too difficult because the only vector that contributes to the sum is $vec{r} = vec{0}$ due to the Kronecker delta in $rho(vec{r})$. The problem is that $rho(vec{r})$ diverges at this point. I need some mathematical advice here. Is there anybody who can show me how to compute the Fourier transform of $rho(vec{r})$?










share|cite|improve this question















I need help finding the Fourier transform of the function



$$
rho(vec{r}) = alpha delta_{vec{r},0} left(lambdalambda' J_1
(beta |vec{r}|)Y_1(beta |vec{r}|) - pi^2 J_0
(beta |vec{r}|)Y_0(beta |vec{r}|) right),
$$



where $alpha,beta in mathcal{R}$ and $lambda,lambda'=pm1$. The $J_i(x)$ are the $i$-th Bessel functions of the first, the $Y_i(x)$ are the $i$-th Bessel functions of the second kind. $delta_{vec{r},0}$ denotes the Kronecker delta.



The vector $vec{r} = (x,y)^T$ is discrete because I'm working on a discrete set of points. I'm a physicist, so please don't hesitate to ask for more specific information.



I tried something like



$$
rho(vec{k}) = sum_vec{r} e^{-ivec{k}cdotvec{r}}rho(vec{r}),
$$



which should in fact not be too difficult because the only vector that contributes to the sum is $vec{r} = vec{0}$ due to the Kronecker delta in $rho(vec{r})$. The problem is that $rho(vec{r})$ diverges at this point. I need some mathematical advice here. Is there anybody who can show me how to compute the Fourier transform of $rho(vec{r})$?







physics fourier-transform divergent-series bessel-functions






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edited Nov 27 at 16:54









Robert Howard

1,9181822




1,9181822










asked Aug 10 at 10:04









MeMeansMe

1354




1354












  • Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
    – Andy Walls
    Aug 11 at 22:09










  • @AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
    – MeMeansMe
    Aug 13 at 7:50


















  • Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
    – Andy Walls
    Aug 11 at 22:09










  • @AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
    – MeMeansMe
    Aug 13 at 7:50
















Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
– Andy Walls
Aug 11 at 22:09




Please confirm that this problem has strict radial symmetry. The reason I ask is that $delta^2(x,y)$ and $delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform.
– Andy Walls
Aug 11 at 22:09












@AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
– MeMeansMe
Aug 13 at 7:50




@AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :)
– MeMeansMe
Aug 13 at 7:50










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I'm almost certain the Fourier Transform doesn't exist.



I'll rewrite your radially symmetric (density?) function, $rho(vec{r})$, to be continuous, as opposed to discrete, in 2D polar coordinates:



$$d(r) = {}^2delta(r) alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$



Where ${}^2delta(r)$ is the 2D Dirac Delta function at the origin of the polar coordinate system. (For the impulse at the origin, $theta$ doesn't matter, so it doesn't appear as an argument.) In this analysis, ${}^2delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):



$${}^2delta(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{piepsilon^2} quad 0 < r < epsilon$$



so that



$$int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta = 1$$



To separate the $r$ and $theta$ portions of ${}^2delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:



$$delta_a(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{epsilon} quad 0 < r < epsilon$$



and then note
$$begin{align}\
int_0^{infty} delta_a(r) space mathrm{d}r &= 1\
&= int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta\
&= int_0^{infty} {}^2delta(r) space 2pi r space mathrm{d}r\
end{align}$$



so we have



$${}^2delta(r) = dfrac{delta_a(r)}{2pi r}$$



$d(r)$ can now be written without any dependence on $theta$ as



$$d(r) = dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$



The 2D Fourier Transform of a radially symmetric function can be expressed in terms of the Hankel Transform of order 0:



$$F(rho) = mathcal{F}left{f(r)right} = 2pi mathcal{H_0}left{f(r)right} = 2pi int_0^{infty} f(r) J_0(rho r) space r space mathrm{d}r$$



When attempting to compute the Fourier Transform



$$begin{align}mathcal{F}left{d(r)right} = D(rho) &= 2pi int_0^{infty} dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) space r space mathrm{d}r\
\
&= alpha int_0^{infty} delta_a(r) left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' int_0^{infty} delta_a(r) J_1(beta r) Y_1(beta r) J_0(rho r) mathrm{d}r - alpha pi^2 int_0^{infty} delta_a(r)J_0(beta r) Y_0(beta r) J_0(rho r) mathrm{d}r\
\
&= alphalambda lambda' lim_{r rightarrow 0+} J_1(beta r) Y_1(beta r) J_0(rho r) - alpha pi^2 lim_{r rightarrow 0+} J_0(beta r) Y_0(beta r) J_0(rho r) \
\
&= alphalambda lambda'left(-dfrac1{pi}right) - alpha pi^2 (-infty) \
end{align}$$



it diverges.



Note that if the problematic term had converged, the Fourier Transform would simply be an infinite disk of constant height above the $rhophi$ plane. That should be no surprise for the transform of a Dirac Delta function.



The problematic term would converge, if there were a factor of $1/ln(r)$:



$$lim_{r rightarrow 0+} dfrac{J_0(beta r) Y_0(beta r)}{ln(r)} J_0(rho r) = dfrac{2}{pi}$$



I don't know the physical situation you are analyzing, and whether inclusion of such a term could ever be justified.






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    up vote
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    I'm almost certain the Fourier Transform doesn't exist.



    I'll rewrite your radially symmetric (density?) function, $rho(vec{r})$, to be continuous, as opposed to discrete, in 2D polar coordinates:



    $$d(r) = {}^2delta(r) alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$



    Where ${}^2delta(r)$ is the 2D Dirac Delta function at the origin of the polar coordinate system. (For the impulse at the origin, $theta$ doesn't matter, so it doesn't appear as an argument.) In this analysis, ${}^2delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):



    $${}^2delta(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{piepsilon^2} quad 0 < r < epsilon$$



    so that



    $$int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta = 1$$



    To separate the $r$ and $theta$ portions of ${}^2delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:



    $$delta_a(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{epsilon} quad 0 < r < epsilon$$



    and then note
    $$begin{align}\
    int_0^{infty} delta_a(r) space mathrm{d}r &= 1\
    &= int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta\
    &= int_0^{infty} {}^2delta(r) space 2pi r space mathrm{d}r\
    end{align}$$



    so we have



    $${}^2delta(r) = dfrac{delta_a(r)}{2pi r}$$



    $d(r)$ can now be written without any dependence on $theta$ as



    $$d(r) = dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$



    The 2D Fourier Transform of a radially symmetric function can be expressed in terms of the Hankel Transform of order 0:



    $$F(rho) = mathcal{F}left{f(r)right} = 2pi mathcal{H_0}left{f(r)right} = 2pi int_0^{infty} f(r) J_0(rho r) space r space mathrm{d}r$$



    When attempting to compute the Fourier Transform



    $$begin{align}mathcal{F}left{d(r)right} = D(rho) &= 2pi int_0^{infty} dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) space r space mathrm{d}r\
    \
    &= alpha int_0^{infty} delta_a(r) left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) mathrm{d}r\
    \
    &= alphalambda lambda' int_0^{infty} delta_a(r) J_1(beta r) Y_1(beta r) J_0(rho r) mathrm{d}r - alpha pi^2 int_0^{infty} delta_a(r)J_0(beta r) Y_0(beta r) J_0(rho r) mathrm{d}r\
    \
    &= alphalambda lambda' lim_{r rightarrow 0+} J_1(beta r) Y_1(beta r) J_0(rho r) - alpha pi^2 lim_{r rightarrow 0+} J_0(beta r) Y_0(beta r) J_0(rho r) \
    \
    &= alphalambda lambda'left(-dfrac1{pi}right) - alpha pi^2 (-infty) \
    end{align}$$



    it diverges.



    Note that if the problematic term had converged, the Fourier Transform would simply be an infinite disk of constant height above the $rhophi$ plane. That should be no surprise for the transform of a Dirac Delta function.



    The problematic term would converge, if there were a factor of $1/ln(r)$:



    $$lim_{r rightarrow 0+} dfrac{J_0(beta r) Y_0(beta r)}{ln(r)} J_0(rho r) = dfrac{2}{pi}$$



    I don't know the physical situation you are analyzing, and whether inclusion of such a term could ever be justified.






    share|cite|improve this answer



























      up vote
      1
      down vote













      I'm almost certain the Fourier Transform doesn't exist.



      I'll rewrite your radially symmetric (density?) function, $rho(vec{r})$, to be continuous, as opposed to discrete, in 2D polar coordinates:



      $$d(r) = {}^2delta(r) alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$



      Where ${}^2delta(r)$ is the 2D Dirac Delta function at the origin of the polar coordinate system. (For the impulse at the origin, $theta$ doesn't matter, so it doesn't appear as an argument.) In this analysis, ${}^2delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):



      $${}^2delta(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{piepsilon^2} quad 0 < r < epsilon$$



      so that



      $$int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta = 1$$



      To separate the $r$ and $theta$ portions of ${}^2delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:



      $$delta_a(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{epsilon} quad 0 < r < epsilon$$



      and then note
      $$begin{align}\
      int_0^{infty} delta_a(r) space mathrm{d}r &= 1\
      &= int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta\
      &= int_0^{infty} {}^2delta(r) space 2pi r space mathrm{d}r\
      end{align}$$



      so we have



      $${}^2delta(r) = dfrac{delta_a(r)}{2pi r}$$



      $d(r)$ can now be written without any dependence on $theta$ as



      $$d(r) = dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$



      The 2D Fourier Transform of a radially symmetric function can be expressed in terms of the Hankel Transform of order 0:



      $$F(rho) = mathcal{F}left{f(r)right} = 2pi mathcal{H_0}left{f(r)right} = 2pi int_0^{infty} f(r) J_0(rho r) space r space mathrm{d}r$$



      When attempting to compute the Fourier Transform



      $$begin{align}mathcal{F}left{d(r)right} = D(rho) &= 2pi int_0^{infty} dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) space r space mathrm{d}r\
      \
      &= alpha int_0^{infty} delta_a(r) left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) mathrm{d}r\
      \
      &= alphalambda lambda' int_0^{infty} delta_a(r) J_1(beta r) Y_1(beta r) J_0(rho r) mathrm{d}r - alpha pi^2 int_0^{infty} delta_a(r)J_0(beta r) Y_0(beta r) J_0(rho r) mathrm{d}r\
      \
      &= alphalambda lambda' lim_{r rightarrow 0+} J_1(beta r) Y_1(beta r) J_0(rho r) - alpha pi^2 lim_{r rightarrow 0+} J_0(beta r) Y_0(beta r) J_0(rho r) \
      \
      &= alphalambda lambda'left(-dfrac1{pi}right) - alpha pi^2 (-infty) \
      end{align}$$



      it diverges.



      Note that if the problematic term had converged, the Fourier Transform would simply be an infinite disk of constant height above the $rhophi$ plane. That should be no surprise for the transform of a Dirac Delta function.



      The problematic term would converge, if there were a factor of $1/ln(r)$:



      $$lim_{r rightarrow 0+} dfrac{J_0(beta r) Y_0(beta r)}{ln(r)} J_0(rho r) = dfrac{2}{pi}$$



      I don't know the physical situation you are analyzing, and whether inclusion of such a term could ever be justified.






      share|cite|improve this answer

























        up vote
        1
        down vote










        up vote
        1
        down vote









        I'm almost certain the Fourier Transform doesn't exist.



        I'll rewrite your radially symmetric (density?) function, $rho(vec{r})$, to be continuous, as opposed to discrete, in 2D polar coordinates:



        $$d(r) = {}^2delta(r) alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$



        Where ${}^2delta(r)$ is the 2D Dirac Delta function at the origin of the polar coordinate system. (For the impulse at the origin, $theta$ doesn't matter, so it doesn't appear as an argument.) In this analysis, ${}^2delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):



        $${}^2delta(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{piepsilon^2} quad 0 < r < epsilon$$



        so that



        $$int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta = 1$$



        To separate the $r$ and $theta$ portions of ${}^2delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:



        $$delta_a(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{epsilon} quad 0 < r < epsilon$$



        and then note
        $$begin{align}\
        int_0^{infty} delta_a(r) space mathrm{d}r &= 1\
        &= int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta\
        &= int_0^{infty} {}^2delta(r) space 2pi r space mathrm{d}r\
        end{align}$$



        so we have



        $${}^2delta(r) = dfrac{delta_a(r)}{2pi r}$$



        $d(r)$ can now be written without any dependence on $theta$ as



        $$d(r) = dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$



        The 2D Fourier Transform of a radially symmetric function can be expressed in terms of the Hankel Transform of order 0:



        $$F(rho) = mathcal{F}left{f(r)right} = 2pi mathcal{H_0}left{f(r)right} = 2pi int_0^{infty} f(r) J_0(rho r) space r space mathrm{d}r$$



        When attempting to compute the Fourier Transform



        $$begin{align}mathcal{F}left{d(r)right} = D(rho) &= 2pi int_0^{infty} dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) space r space mathrm{d}r\
        \
        &= alpha int_0^{infty} delta_a(r) left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) mathrm{d}r\
        \
        &= alphalambda lambda' int_0^{infty} delta_a(r) J_1(beta r) Y_1(beta r) J_0(rho r) mathrm{d}r - alpha pi^2 int_0^{infty} delta_a(r)J_0(beta r) Y_0(beta r) J_0(rho r) mathrm{d}r\
        \
        &= alphalambda lambda' lim_{r rightarrow 0+} J_1(beta r) Y_1(beta r) J_0(rho r) - alpha pi^2 lim_{r rightarrow 0+} J_0(beta r) Y_0(beta r) J_0(rho r) \
        \
        &= alphalambda lambda'left(-dfrac1{pi}right) - alpha pi^2 (-infty) \
        end{align}$$



        it diverges.



        Note that if the problematic term had converged, the Fourier Transform would simply be an infinite disk of constant height above the $rhophi$ plane. That should be no surprise for the transform of a Dirac Delta function.



        The problematic term would converge, if there were a factor of $1/ln(r)$:



        $$lim_{r rightarrow 0+} dfrac{J_0(beta r) Y_0(beta r)}{ln(r)} J_0(rho r) = dfrac{2}{pi}$$



        I don't know the physical situation you are analyzing, and whether inclusion of such a term could ever be justified.






        share|cite|improve this answer














        I'm almost certain the Fourier Transform doesn't exist.



        I'll rewrite your radially symmetric (density?) function, $rho(vec{r})$, to be continuous, as opposed to discrete, in 2D polar coordinates:



        $$d(r) = {}^2delta(r) alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$



        Where ${}^2delta(r)$ is the 2D Dirac Delta function at the origin of the polar coordinate system. (For the impulse at the origin, $theta$ doesn't matter, so it doesn't appear as an argument.) In this analysis, ${}^2delta(r)$ represents the following limiting sequence of functions (that are asymmetrical about 0):



        $${}^2delta(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{piepsilon^2} quad 0 < r < epsilon$$



        so that



        $$int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta = 1$$



        To separate the $r$ and $theta$ portions of ${}^2delta(r)$, so we can use a 1D Dirac Delta function, I'll use the following limiting sequence of functions (that are asymmetric about 0) for the 1D Dirac Delta function:



        $$delta_a(r) = lim_{epsilon rightarrow 0^+} dfrac{1}{epsilon} quad 0 < r < epsilon$$



        and then note
        $$begin{align}\
        int_0^{infty} delta_a(r) space mathrm{d}r &= 1\
        &= int_0^{2pi} int_0^{infty} {}^2delta(r) space r space mathrm{d}r space mathrm{d}theta\
        &= int_0^{infty} {}^2delta(r) space 2pi r space mathrm{d}r\
        end{align}$$



        so we have



        $${}^2delta(r) = dfrac{delta_a(r)}{2pi r}$$



        $d(r)$ can now be written without any dependence on $theta$ as



        $$d(r) = dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right]$$



        The 2D Fourier Transform of a radially symmetric function can be expressed in terms of the Hankel Transform of order 0:



        $$F(rho) = mathcal{F}left{f(r)right} = 2pi mathcal{H_0}left{f(r)right} = 2pi int_0^{infty} f(r) J_0(rho r) space r space mathrm{d}r$$



        When attempting to compute the Fourier Transform



        $$begin{align}mathcal{F}left{d(r)right} = D(rho) &= 2pi int_0^{infty} dfrac{delta_a(r)}{2pi r} alpha left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) space r space mathrm{d}r\
        \
        &= alpha int_0^{infty} delta_a(r) left[lambda lambda' J_1(beta r) Y_1(beta r) - pi^2 J_0(beta r) Y_0(beta r)right] J_0(rho r) mathrm{d}r\
        \
        &= alphalambda lambda' int_0^{infty} delta_a(r) J_1(beta r) Y_1(beta r) J_0(rho r) mathrm{d}r - alpha pi^2 int_0^{infty} delta_a(r)J_0(beta r) Y_0(beta r) J_0(rho r) mathrm{d}r\
        \
        &= alphalambda lambda' lim_{r rightarrow 0+} J_1(beta r) Y_1(beta r) J_0(rho r) - alpha pi^2 lim_{r rightarrow 0+} J_0(beta r) Y_0(beta r) J_0(rho r) \
        \
        &= alphalambda lambda'left(-dfrac1{pi}right) - alpha pi^2 (-infty) \
        end{align}$$



        it diverges.



        Note that if the problematic term had converged, the Fourier Transform would simply be an infinite disk of constant height above the $rhophi$ plane. That should be no surprise for the transform of a Dirac Delta function.



        The problematic term would converge, if there were a factor of $1/ln(r)$:



        $$lim_{r rightarrow 0+} dfrac{J_0(beta r) Y_0(beta r)}{ln(r)} J_0(rho r) = dfrac{2}{pi}$$



        I don't know the physical situation you are analyzing, and whether inclusion of such a term could ever be justified.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 20 at 14:19

























        answered Aug 18 at 12:32









        Andy Walls

        1,479127




        1,479127






























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