Inverse Fourier Transform of the Hdamard's finite part - 2 dimensions

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I'm trying to compute the Fourier transform of the Hadamard's finite part distribution in two dimensions, and it's giving me a little trouble.



In two dimensions, we defined the finite part $textPFleft(1/k^2right)$, where $kequiv|veck|=sqrtk_x^2+k_y^2$, as the distribution such that for each test function $f$
beginequation
leftlangle textPFleft(frac1k^2right),frightrangle =textPFiint!d^2veckfracf(veck)k^2 =iint_k<1!!!d^2veckfracf(veck)-f(0)k^2+iint_k>1!!!d^2veckfracf(veck)k^2
endequation



Quoting my professor:




Sloppily then we can compute the inverse Fourier transform as
beginequation
frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac1(2pi)^2textPFiint!d^2veckfrace^iveckcdotvecxk^2 = frac1(2pi)^2textPFint_0^infty!dk frackk^2int_0^2pidvartheta,e^ikrcosvartheta
endequation
Where we used spherical coordinates in two dimensions, and we called $requiv |vecx|=sqrtx^2+y^2$. Using Bessel $J_n$ functions, one eventually get
beginequation
frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac1(2pi)^2textPFint_0^infty!dk, (2pi)fracJ_0(kr)k tag1
endequation
Using now the definition of finite part, and changing variable $uequiv rk$, it can be showed that
beginequation
frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac12pileft-log(r)+int_0^1!dufracJ_0(u)-J_0(0)u+int_1^infty!dufracJ_0(u)uright tag2
endequation




First of all, I can't understand very well how do we get to equation (1), in the sense that it's seems that going in spherical coordinates allows us to "delete" one dimension, in the sense that equation (1) seems the Hadamard's finite part of the one dimensional variable $k$. I really don't understand the meaning of this equation.



Then, if we accept that, I cannot understand how to get to equation (2). Indeed if we rename $u=kr$ we get
beginequation
frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac12piint_0^1/r!du,fracJ_0(u)-J_0(0)u +frac12piint_1/r^infty!du,fracJ_0(u)u
endequation
And i really can't understand how to go on, and get the $log(r)$ term.




distribution-theory fourier-transform




share|cite|improve this question

























    up vote
    1
    down vote

    favorite












    I'm trying to compute the Fourier transform of the Hadamard's finite part distribution in two dimensions, and it's giving me a little trouble.



    In two dimensions, we defined the finite part $textPFleft(1/k^2right)$, where $kequiv|veck|=sqrtk_x^2+k_y^2$, as the distribution such that for each test function $f$
    beginequation
    leftlangle textPFleft(frac1k^2right),frightrangle =textPFiint!d^2veckfracf(veck)k^2 =iint_k<1!!!d^2veckfracf(veck)-f(0)k^2+iint_k>1!!!d^2veckfracf(veck)k^2
    endequation



    Quoting my professor:




    Sloppily then we can compute the inverse Fourier transform as
    beginequation
    frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac1(2pi)^2textPFiint!d^2veckfrace^iveckcdotvecxk^2 = frac1(2pi)^2textPFint_0^infty!dk frackk^2int_0^2pidvartheta,e^ikrcosvartheta
    endequation
    Where we used spherical coordinates in two dimensions, and we called $requiv |vecx|=sqrtx^2+y^2$. Using Bessel $J_n$ functions, one eventually get
    beginequation
    frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac1(2pi)^2textPFint_0^infty!dk, (2pi)fracJ_0(kr)k tag1
    endequation
    Using now the definition of finite part, and changing variable $uequiv rk$, it can be showed that
    beginequation
    frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac12pileft-log(r)+int_0^1!dufracJ_0(u)-J_0(0)u+int_1^infty!dufracJ_0(u)uright tag2
    endequation




    First of all, I can't understand very well how do we get to equation (1), in the sense that it's seems that going in spherical coordinates allows us to "delete" one dimension, in the sense that equation (1) seems the Hadamard's finite part of the one dimensional variable $k$. I really don't understand the meaning of this equation.



    Then, if we accept that, I cannot understand how to get to equation (2). Indeed if we rename $u=kr$ we get
    beginequation
    frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac12piint_0^1/r!du,fracJ_0(u)-J_0(0)u +frac12piint_1/r^infty!du,fracJ_0(u)u
    endequation
    And i really can't understand how to go on, and get the $log(r)$ term.




    distribution-theory fourier-transform




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I'm trying to compute the Fourier transform of the Hadamard's finite part distribution in two dimensions, and it's giving me a little trouble.



      In two dimensions, we defined the finite part $textPFleft(1/k^2right)$, where $kequiv|veck|=sqrtk_x^2+k_y^2$, as the distribution such that for each test function $f$
      beginequation
      leftlangle textPFleft(frac1k^2right),frightrangle =textPFiint!d^2veckfracf(veck)k^2 =iint_k<1!!!d^2veckfracf(veck)-f(0)k^2+iint_k>1!!!d^2veckfracf(veck)k^2
      endequation



      Quoting my professor:




      Sloppily then we can compute the inverse Fourier transform as
      beginequation
      frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac1(2pi)^2textPFiint!d^2veckfrace^iveckcdotvecxk^2 = frac1(2pi)^2textPFint_0^infty!dk frackk^2int_0^2pidvartheta,e^ikrcosvartheta
      endequation
      Where we used spherical coordinates in two dimensions, and we called $requiv |vecx|=sqrtx^2+y^2$. Using Bessel $J_n$ functions, one eventually get
      beginequation
      frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac1(2pi)^2textPFint_0^infty!dk, (2pi)fracJ_0(kr)k tag1
      endequation
      Using now the definition of finite part, and changing variable $uequiv rk$, it can be showed that
      beginequation
      frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac12pileft-log(r)+int_0^1!dufracJ_0(u)-J_0(0)u+int_1^infty!dufracJ_0(u)uright tag2
      endequation




      First of all, I can't understand very well how do we get to equation (1), in the sense that it's seems that going in spherical coordinates allows us to "delete" one dimension, in the sense that equation (1) seems the Hadamard's finite part of the one dimensional variable $k$. I really don't understand the meaning of this equation.



      Then, if we accept that, I cannot understand how to get to equation (2). Indeed if we rename $u=kr$ we get
      beginequation
      frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac12piint_0^1/r!du,fracJ_0(u)-J_0(0)u +frac12piint_1/r^infty!du,fracJ_0(u)u
      endequation
      And i really can't understand how to go on, and get the $log(r)$ term.




      distribution-theory fourier-transform




      share|cite|improve this question













      I'm trying to compute the Fourier transform of the Hadamard's finite part distribution in two dimensions, and it's giving me a little trouble.



      In two dimensions, we defined the finite part $textPFleft(1/k^2right)$, where $kequiv|veck|=sqrtk_x^2+k_y^2$, as the distribution such that for each test function $f$
      beginequation
      leftlangle textPFleft(frac1k^2right),frightrangle =textPFiint!d^2veckfracf(veck)k^2 =iint_k<1!!!d^2veckfracf(veck)-f(0)k^2+iint_k>1!!!d^2veckfracf(veck)k^2
      endequation



      Quoting my professor:




      Sloppily then we can compute the inverse Fourier transform as
      beginequation
      frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac1(2pi)^2textPFiint!d^2veckfrace^iveckcdotvecxk^2 = frac1(2pi)^2textPFint_0^infty!dk frackk^2int_0^2pidvartheta,e^ikrcosvartheta
      endequation
      Where we used spherical coordinates in two dimensions, and we called $requiv |vecx|=sqrtx^2+y^2$. Using Bessel $J_n$ functions, one eventually get
      beginequation
      frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac1(2pi)^2textPFint_0^infty!dk, (2pi)fracJ_0(kr)k tag1
      endequation
      Using now the definition of finite part, and changing variable $uequiv rk$, it can be showed that
      beginequation
      frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac12pileft-log(r)+int_0^1!dufracJ_0(u)-J_0(0)u+int_1^infty!dufracJ_0(u)uright tag2
      endequation




      First of all, I can't understand very well how do we get to equation (1), in the sense that it's seems that going in spherical coordinates allows us to "delete" one dimension, in the sense that equation (1) seems the Hadamard's finite part of the one dimensional variable $k$. I really don't understand the meaning of this equation.



      Then, if we accept that, I cannot understand how to get to equation (2). Indeed if we rename $u=kr$ we get
      beginequation
      frac1(2pi)^2leftlangle textPFleft(frac1k^2right),e^iveckcdotvecxrightrangle = frac12piint_0^1/r!du,fracJ_0(u)-J_0(0)u +frac12piint_1/r^infty!du,fracJ_0(u)u
      endequation
      And i really can't understand how to go on, and get the $log(r)$ term.





      distribution-theory fourier-transform





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      edited Jul 14 at 14:51
























      asked Jul 14 at 14:41









      M. M. R.

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          1 Answer
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          up vote
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          I'll change the notation slightly and use $k$ and $w$ for the vectors. Write the action of $1/|k|^2$ on $e^i k cdot w$ by definition as if $e^i k cdot w$ were a test function. In the resulting integral, make the change of variables $k_x = rho cos phi, k_y = rho sin phi$:
          $$left< frac 1 , e^i k cdot w right> =
          iint_ frac e^i k cdot w - 1 dk_x dk_y +
          iint_ frac e^i k cdot w dk_x dk_y = \
          int_0^1 int_0^2pi
          frac exp(i rho (w_x cos phi + w_y sin phi)) - 1
          rho dphi drho + \
          int_1^infty int_0^2pi
          frac exp(i rho (w_x cos phi + w_y sin phi)) rho dphi drho.$$
          We know that
          $$w_x cos phi + w_y sin phi =
          r left( frac w_x r cos phi + frac w_y r sin phi right) =
          r cos(phi + phi_0)$$
          for some $phi_0$, where $r = |w|$; since we're integrating over a complete period, the result does not depend on $phi_0$, and
          $$int_0^2pi e^i rho r cos phi dphi =
          2 pi J_0(rho r) Rightarrow \
          frac 1 2pi left< frac 1 , e^i k cdot w right> =
          int_0^1 frac J_0(rho r) - 1 rho drho +
          int_1^infty frac J_0(rho r) rho drho = \
          int_0^r frac J_0(u) - 1 u du +
          int_r^infty frac J_0(u) u du = \
          left( int_0^1 frac J_0(u) - 1 u du +
          int_1^r frac J_0(u) - 1 u du right) + \
          left( int_1^infty frac J_0(u) u du +
          int_r^1 frac J_0(u) u du right) = \
          -ln r + int_0^1 frac J_0(u) - 1 u du +
          int_1^infty frac J_0(u) u du = \
          -ln r - gamma + ln 2.$$
          The last step is for extra points.






          share|cite|improve this answer





















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            1 Answer
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            active

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            up vote
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            accepted










            I'll change the notation slightly and use $k$ and $w$ for the vectors. Write the action of $1/|k|^2$ on $e^i k cdot w$ by definition as if $e^i k cdot w$ were a test function. In the resulting integral, make the change of variables $k_x = rho cos phi, k_y = rho sin phi$:
            $$left< frac 1 , e^i k cdot w right> =
            iint_ frac e^i k cdot w - 1 dk_x dk_y +
            iint_ frac e^i k cdot w dk_x dk_y = \
            int_0^1 int_0^2pi
            frac exp(i rho (w_x cos phi + w_y sin phi)) - 1
            rho dphi drho + \
            int_1^infty int_0^2pi
            frac exp(i rho (w_x cos phi + w_y sin phi)) rho dphi drho.$$
            We know that
            $$w_x cos phi + w_y sin phi =
            r left( frac w_x r cos phi + frac w_y r sin phi right) =
            r cos(phi + phi_0)$$
            for some $phi_0$, where $r = |w|$; since we're integrating over a complete period, the result does not depend on $phi_0$, and
            $$int_0^2pi e^i rho r cos phi dphi =
            2 pi J_0(rho r) Rightarrow \
            frac 1 2pi left< frac 1 , e^i k cdot w right> =
            int_0^1 frac J_0(rho r) - 1 rho drho +
            int_1^infty frac J_0(rho r) rho drho = \
            int_0^r frac J_0(u) - 1 u du +
            int_r^infty frac J_0(u) u du = \
            left( int_0^1 frac J_0(u) - 1 u du +
            int_1^r frac J_0(u) - 1 u du right) + \
            left( int_1^infty frac J_0(u) u du +
            int_r^1 frac J_0(u) u du right) = \
            -ln r + int_0^1 frac J_0(u) - 1 u du +
            int_1^infty frac J_0(u) u du = \
            -ln r - gamma + ln 2.$$
            The last step is for extra points.






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              I'll change the notation slightly and use $k$ and $w$ for the vectors. Write the action of $1/|k|^2$ on $e^i k cdot w$ by definition as if $e^i k cdot w$ were a test function. In the resulting integral, make the change of variables $k_x = rho cos phi, k_y = rho sin phi$:
              $$left< frac 1 , e^i k cdot w right> =
              iint_ frac e^i k cdot w - 1 dk_x dk_y +
              iint_ frac e^i k cdot w dk_x dk_y = \
              int_0^1 int_0^2pi
              frac exp(i rho (w_x cos phi + w_y sin phi)) - 1
              rho dphi drho + \
              int_1^infty int_0^2pi
              frac exp(i rho (w_x cos phi + w_y sin phi)) rho dphi drho.$$
              We know that
              $$w_x cos phi + w_y sin phi =
              r left( frac w_x r cos phi + frac w_y r sin phi right) =
              r cos(phi + phi_0)$$
              for some $phi_0$, where $r = |w|$; since we're integrating over a complete period, the result does not depend on $phi_0$, and
              $$int_0^2pi e^i rho r cos phi dphi =
              2 pi J_0(rho r) Rightarrow \
              frac 1 2pi left< frac 1 , e^i k cdot w right> =
              int_0^1 frac J_0(rho r) - 1 rho drho +
              int_1^infty frac J_0(rho r) rho drho = \
              int_0^r frac J_0(u) - 1 u du +
              int_r^infty frac J_0(u) u du = \
              left( int_0^1 frac J_0(u) - 1 u du +
              int_1^r frac J_0(u) - 1 u du right) + \
              left( int_1^infty frac J_0(u) u du +
              int_r^1 frac J_0(u) u du right) = \
              -ln r + int_0^1 frac J_0(u) - 1 u du +
              int_1^infty frac J_0(u) u du = \
              -ln r - gamma + ln 2.$$
              The last step is for extra points.






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                I'll change the notation slightly and use $k$ and $w$ for the vectors. Write the action of $1/|k|^2$ on $e^i k cdot w$ by definition as if $e^i k cdot w$ were a test function. In the resulting integral, make the change of variables $k_x = rho cos phi, k_y = rho sin phi$:
                $$left< frac 1 , e^i k cdot w right> =
                iint_ frac e^i k cdot w - 1 dk_x dk_y +
                iint_ frac e^i k cdot w dk_x dk_y = \
                int_0^1 int_0^2pi
                frac exp(i rho (w_x cos phi + w_y sin phi)) - 1
                rho dphi drho + \
                int_1^infty int_0^2pi
                frac exp(i rho (w_x cos phi + w_y sin phi)) rho dphi drho.$$
                We know that
                $$w_x cos phi + w_y sin phi =
                r left( frac w_x r cos phi + frac w_y r sin phi right) =
                r cos(phi + phi_0)$$
                for some $phi_0$, where $r = |w|$; since we're integrating over a complete period, the result does not depend on $phi_0$, and
                $$int_0^2pi e^i rho r cos phi dphi =
                2 pi J_0(rho r) Rightarrow \
                frac 1 2pi left< frac 1 , e^i k cdot w right> =
                int_0^1 frac J_0(rho r) - 1 rho drho +
                int_1^infty frac J_0(rho r) rho drho = \
                int_0^r frac J_0(u) - 1 u du +
                int_r^infty frac J_0(u) u du = \
                left( int_0^1 frac J_0(u) - 1 u du +
                int_1^r frac J_0(u) - 1 u du right) + \
                left( int_1^infty frac J_0(u) u du +
                int_r^1 frac J_0(u) u du right) = \
                -ln r + int_0^1 frac J_0(u) - 1 u du +
                int_1^infty frac J_0(u) u du = \
                -ln r - gamma + ln 2.$$
                The last step is for extra points.






                share|cite|improve this answer













                I'll change the notation slightly and use $k$ and $w$ for the vectors. Write the action of $1/|k|^2$ on $e^i k cdot w$ by definition as if $e^i k cdot w$ were a test function. In the resulting integral, make the change of variables $k_x = rho cos phi, k_y = rho sin phi$:
                $$left< frac 1 , e^i k cdot w right> =
                iint_ frac e^i k cdot w - 1 dk_x dk_y +
                iint_ frac e^i k cdot w dk_x dk_y = \
                int_0^1 int_0^2pi
                frac exp(i rho (w_x cos phi + w_y sin phi)) - 1
                rho dphi drho + \
                int_1^infty int_0^2pi
                frac exp(i rho (w_x cos phi + w_y sin phi)) rho dphi drho.$$
                We know that
                $$w_x cos phi + w_y sin phi =
                r left( frac w_x r cos phi + frac w_y r sin phi right) =
                r cos(phi + phi_0)$$
                for some $phi_0$, where $r = |w|$; since we're integrating over a complete period, the result does not depend on $phi_0$, and
                $$int_0^2pi e^i rho r cos phi dphi =
                2 pi J_0(rho r) Rightarrow \
                frac 1 2pi left< frac 1 , e^i k cdot w right> =
                int_0^1 frac J_0(rho r) - 1 rho drho +
                int_1^infty frac J_0(rho r) rho drho = \
                int_0^r frac J_0(u) - 1 u du +
                int_r^infty frac J_0(u) u du = \
                left( int_0^1 frac J_0(u) - 1 u du +
                int_1^r frac J_0(u) - 1 u du right) + \
                left( int_1^infty frac J_0(u) u du +
                int_r^1 frac J_0(u) u du right) = \
                -ln r + int_0^1 frac J_0(u) - 1 u du +
                int_1^infty frac J_0(u) u du = \
                -ln r - gamma + ln 2.$$
                The last step is for extra points.







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                answered Jul 15 at 2:06









                Maxim

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                    Color the edges and diagonals of a regular polygon

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                    Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
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                    Relationship between determinant of matrix and determinant of adjoint?

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                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
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