$I = int_{- infty}^{infty} delta (n - ||mathbf{x}||^2) mathrm d mathbf{x} $ should not diverge












3












$begingroup$


I'm having trouble evaluating this integral:



$$I = int_{-infty}^infty delta (n - ||mathbf{x}||^2) mathrm d mathbf{x} $$



where $delta(x)$ is Dirac delta function, $mathbf x$ is the real $n$-dimensional vector $(x_1,...,x_n)$, and $||mathbf{x}||^2 = sum_{i=1}^n x_i^2$. Here I use the short-hand notation:



$$int_{-infty}^inftymathrm d mathbf{x} = int_{-infty}^inftymathrm d x_1 dots int_{-infty}^inftymathrm d x_n$$



Here is what I tried to do. Using the integral representation of Dirac's delta function $delta (x) = int_{- infty}^{infty} frac{d lambda}{2 pi} e^{i lambda x}$, I get



$$I = int_{- infty}^{infty} d mathbf{x} int_{-infty}^{infty}
frac{dlambda}{2pi} mathrm e^{ilambda(n - ||
mathbf{x}||^2)} = int_{- infty}^{infty} frac{dlambda}{2pi} e^{-i lambda n} left( int_{- infty}^{infty} e^{ilambda
x^2} dx right)^n$$



I evaluated the inner integral with Mathematica:



$$int_{- infty}^{infty} e^{ilambda
x^2} dx = sqrt{frac{pi}{2}} frac{1 +i mathrm{sgn}(lambda)}{sqrt{|lambda
|}}$$



But then when I try to substitute this I get a diverging integral in $lambda$.



What did I do wrong?



Note that the original integral is well behaved. Using polar coordinates it can be shown that $I = n^{n/2-1} pi^{n/2} / Gamma(n/2)$ (see https://doi.org/10.1093/qmath/12.1.165, or my answer below). Some step of my derivation leads to a fictitious divergence.



Update: @user587192's answer found the error in my calculation. I cannot exchange the order of integrations on $x$ and $lambda$. My next question is whether there is a way to rescue this calculation? Manipulations with the Fourier transform of the Dirac delta are very convenient in more complicated situations. This is evident from the algebraic complexity of the spherical coordinates appraoch. Is there way to exploit the convenience of the Fourier representation of the Dirac delta here?










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$endgroup$












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo
    Nov 30 '18 at 21:58
















3












$begingroup$


I'm having trouble evaluating this integral:



$$I = int_{-infty}^infty delta (n - ||mathbf{x}||^2) mathrm d mathbf{x} $$



where $delta(x)$ is Dirac delta function, $mathbf x$ is the real $n$-dimensional vector $(x_1,...,x_n)$, and $||mathbf{x}||^2 = sum_{i=1}^n x_i^2$. Here I use the short-hand notation:



$$int_{-infty}^inftymathrm d mathbf{x} = int_{-infty}^inftymathrm d x_1 dots int_{-infty}^inftymathrm d x_n$$



Here is what I tried to do. Using the integral representation of Dirac's delta function $delta (x) = int_{- infty}^{infty} frac{d lambda}{2 pi} e^{i lambda x}$, I get



$$I = int_{- infty}^{infty} d mathbf{x} int_{-infty}^{infty}
frac{dlambda}{2pi} mathrm e^{ilambda(n - ||
mathbf{x}||^2)} = int_{- infty}^{infty} frac{dlambda}{2pi} e^{-i lambda n} left( int_{- infty}^{infty} e^{ilambda
x^2} dx right)^n$$



I evaluated the inner integral with Mathematica:



$$int_{- infty}^{infty} e^{ilambda
x^2} dx = sqrt{frac{pi}{2}} frac{1 +i mathrm{sgn}(lambda)}{sqrt{|lambda
|}}$$



But then when I try to substitute this I get a diverging integral in $lambda$.



What did I do wrong?



Note that the original integral is well behaved. Using polar coordinates it can be shown that $I = n^{n/2-1} pi^{n/2} / Gamma(n/2)$ (see https://doi.org/10.1093/qmath/12.1.165, or my answer below). Some step of my derivation leads to a fictitious divergence.



Update: @user587192's answer found the error in my calculation. I cannot exchange the order of integrations on $x$ and $lambda$. My next question is whether there is a way to rescue this calculation? Manipulations with the Fourier transform of the Dirac delta are very convenient in more complicated situations. This is evident from the algebraic complexity of the spherical coordinates appraoch. Is there way to exploit the convenience of the Fourier representation of the Dirac delta here?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo
    Nov 30 '18 at 21:58














3












3








3


1



$begingroup$


I'm having trouble evaluating this integral:



$$I = int_{-infty}^infty delta (n - ||mathbf{x}||^2) mathrm d mathbf{x} $$



where $delta(x)$ is Dirac delta function, $mathbf x$ is the real $n$-dimensional vector $(x_1,...,x_n)$, and $||mathbf{x}||^2 = sum_{i=1}^n x_i^2$. Here I use the short-hand notation:



$$int_{-infty}^inftymathrm d mathbf{x} = int_{-infty}^inftymathrm d x_1 dots int_{-infty}^inftymathrm d x_n$$



Here is what I tried to do. Using the integral representation of Dirac's delta function $delta (x) = int_{- infty}^{infty} frac{d lambda}{2 pi} e^{i lambda x}$, I get



$$I = int_{- infty}^{infty} d mathbf{x} int_{-infty}^{infty}
frac{dlambda}{2pi} mathrm e^{ilambda(n - ||
mathbf{x}||^2)} = int_{- infty}^{infty} frac{dlambda}{2pi} e^{-i lambda n} left( int_{- infty}^{infty} e^{ilambda
x^2} dx right)^n$$



I evaluated the inner integral with Mathematica:



$$int_{- infty}^{infty} e^{ilambda
x^2} dx = sqrt{frac{pi}{2}} frac{1 +i mathrm{sgn}(lambda)}{sqrt{|lambda
|}}$$



But then when I try to substitute this I get a diverging integral in $lambda$.



What did I do wrong?



Note that the original integral is well behaved. Using polar coordinates it can be shown that $I = n^{n/2-1} pi^{n/2} / Gamma(n/2)$ (see https://doi.org/10.1093/qmath/12.1.165, or my answer below). Some step of my derivation leads to a fictitious divergence.



Update: @user587192's answer found the error in my calculation. I cannot exchange the order of integrations on $x$ and $lambda$. My next question is whether there is a way to rescue this calculation? Manipulations with the Fourier transform of the Dirac delta are very convenient in more complicated situations. This is evident from the algebraic complexity of the spherical coordinates appraoch. Is there way to exploit the convenience of the Fourier representation of the Dirac delta here?










share|cite|improve this question











$endgroup$




I'm having trouble evaluating this integral:



$$I = int_{-infty}^infty delta (n - ||mathbf{x}||^2) mathrm d mathbf{x} $$



where $delta(x)$ is Dirac delta function, $mathbf x$ is the real $n$-dimensional vector $(x_1,...,x_n)$, and $||mathbf{x}||^2 = sum_{i=1}^n x_i^2$. Here I use the short-hand notation:



$$int_{-infty}^inftymathrm d mathbf{x} = int_{-infty}^inftymathrm d x_1 dots int_{-infty}^inftymathrm d x_n$$



Here is what I tried to do. Using the integral representation of Dirac's delta function $delta (x) = int_{- infty}^{infty} frac{d lambda}{2 pi} e^{i lambda x}$, I get



$$I = int_{- infty}^{infty} d mathbf{x} int_{-infty}^{infty}
frac{dlambda}{2pi} mathrm e^{ilambda(n - ||
mathbf{x}||^2)} = int_{- infty}^{infty} frac{dlambda}{2pi} e^{-i lambda n} left( int_{- infty}^{infty} e^{ilambda
x^2} dx right)^n$$



I evaluated the inner integral with Mathematica:



$$int_{- infty}^{infty} e^{ilambda
x^2} dx = sqrt{frac{pi}{2}} frac{1 +i mathrm{sgn}(lambda)}{sqrt{|lambda
|}}$$



But then when I try to substitute this I get a diverging integral in $lambda$.



What did I do wrong?



Note that the original integral is well behaved. Using polar coordinates it can be shown that $I = n^{n/2-1} pi^{n/2} / Gamma(n/2)$ (see https://doi.org/10.1093/qmath/12.1.165, or my answer below). Some step of my derivation leads to a fictitious divergence.



Update: @user587192's answer found the error in my calculation. I cannot exchange the order of integrations on $x$ and $lambda$. My next question is whether there is a way to rescue this calculation? Manipulations with the Fourier transform of the Dirac delta are very convenient in more complicated situations. This is evident from the algebraic complexity of the spherical coordinates appraoch. Is there way to exploit the convenience of the Fourier representation of the Dirac delta here?







definite-integrals area surfaces dirac-delta spheres






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share|cite|improve this question








edited Nov 30 '18 at 16:00







becko

















asked Nov 29 '18 at 23:16









beckobecko

2,35931942




2,35931942












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo
    Nov 30 '18 at 21:58


















  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo
    Nov 30 '18 at 21:58
















$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo
Nov 30 '18 at 21:58




$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo
Nov 30 '18 at 21:58










2 Answers
2






active

oldest

votes


















1












$begingroup$

Here is the solution in (hyper)spherical coordinates, proving that $I$ should be well behaved. The spherical coordinates are defined by:



$$x_1 = r cos(phi_1),quad x_2 = r sin(phi_1)cos(phi_2),dots,quad x_n = r sin(phi_1)dotssin(phi_{n-2})sin(theta)$$



where $0lephi_ilepi$, $0lethetale2pi$ and $0le r<infty$. The volume element is then:



$$mathrm d x_1 dots mathrm d x_n = r^{n-1}sin^{n-2}(phi_1)dotssin(phi_{n-2}) mathrm dr mathrm dphi_1dotsmathrm dphi_{n-2}mathrm dtheta$$



We have:



$$begin{aligned}
I & = int_{mathbb{R}^n} delta (n - | mathbf{x} |^2) mathrm d
mathbf{x}\
& = int_0^{pi} mathrm d phi_1 ldots int_0^{pi}mathrm d phi_{n - 2} int_0^{2 pi}
mathrm d theta int_0^{infty} mathrm d r , delta (n - r^2) r^{n - 1} sin^{n -
2} (phi_1) ldots sin (phi_{n - 2})\
& = left( int_0^{infty} delta (n - r^2) r^{n - 1} mathrm d r right)
left( int_0^{2 pi} mathrm d theta mathrm d theta right) left(
int_0^{pi} sin^{n - 2} (phi_1) mathrm d phi_1 right) ldots left(
int_0^{pi} sin (phi_{n - 2}) mathrm d phi_{n - 2} right)\
& = frac{pi^{n / 2} n^{n / 2 - 1}}{Gamma (n / 2)}
end{aligned}$$



This is not an answer to my question. My question is: What did I do wrong to that leads to a divergent integral?






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This is not an answer to your question. I think you'd better add it to your post instead of here.
    $endgroup$
    – user587192
    Nov 30 '18 at 14:03






  • 1




    $begingroup$
    @user587192 It is not an answer. But it is customary on this site to sometimes post solution attempts as (incomplete) answers.
    $endgroup$
    – becko
    Nov 30 '18 at 14:08



















0












$begingroup$

There is a trick one can do to use the Fourier transform and still have converging integrals. The trick is to multiply the integrand by $mathrm e^{(1-||mathbf x||^2)} = 1$. Then



$$begin{aligned}
int_{- infty}^{infty} delta (1 - ||mathbf x||^2) mathrm d
mathbf x & = int_{- infty}^{infty} mathrm d mathbf x
int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{(1
+mathfrak{i} lambda) (1 - ||mathbf x||^2)}\
& = int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{1
+mathfrak{i} lambda} left( int_{- infty}^{infty} mathrm e^{- (1
+mathfrak{i} lambda) x^2} mathrm dx right)^N\
& = int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{1
+mathfrak{i} lambda} left( frac{pi}{1 +mathfrak{i} lambda} right)^{N
/ 2}\
& = pi^{N / 2} / Gamma (N / 2)
end{aligned}$$



which is the correct result. Here all intermediate integrals converge and the integration order can be exchanged.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Here is the solution in (hyper)spherical coordinates, proving that $I$ should be well behaved. The spherical coordinates are defined by:



    $$x_1 = r cos(phi_1),quad x_2 = r sin(phi_1)cos(phi_2),dots,quad x_n = r sin(phi_1)dotssin(phi_{n-2})sin(theta)$$



    where $0lephi_ilepi$, $0lethetale2pi$ and $0le r<infty$. The volume element is then:



    $$mathrm d x_1 dots mathrm d x_n = r^{n-1}sin^{n-2}(phi_1)dotssin(phi_{n-2}) mathrm dr mathrm dphi_1dotsmathrm dphi_{n-2}mathrm dtheta$$



    We have:



    $$begin{aligned}
    I & = int_{mathbb{R}^n} delta (n - | mathbf{x} |^2) mathrm d
    mathbf{x}\
    & = int_0^{pi} mathrm d phi_1 ldots int_0^{pi}mathrm d phi_{n - 2} int_0^{2 pi}
    mathrm d theta int_0^{infty} mathrm d r , delta (n - r^2) r^{n - 1} sin^{n -
    2} (phi_1) ldots sin (phi_{n - 2})\
    & = left( int_0^{infty} delta (n - r^2) r^{n - 1} mathrm d r right)
    left( int_0^{2 pi} mathrm d theta mathrm d theta right) left(
    int_0^{pi} sin^{n - 2} (phi_1) mathrm d phi_1 right) ldots left(
    int_0^{pi} sin (phi_{n - 2}) mathrm d phi_{n - 2} right)\
    & = frac{pi^{n / 2} n^{n / 2 - 1}}{Gamma (n / 2)}
    end{aligned}$$



    This is not an answer to my question. My question is: What did I do wrong to that leads to a divergent integral?






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      This is not an answer to your question. I think you'd better add it to your post instead of here.
      $endgroup$
      – user587192
      Nov 30 '18 at 14:03






    • 1




      $begingroup$
      @user587192 It is not an answer. But it is customary on this site to sometimes post solution attempts as (incomplete) answers.
      $endgroup$
      – becko
      Nov 30 '18 at 14:08
















    1












    $begingroup$

    Here is the solution in (hyper)spherical coordinates, proving that $I$ should be well behaved. The spherical coordinates are defined by:



    $$x_1 = r cos(phi_1),quad x_2 = r sin(phi_1)cos(phi_2),dots,quad x_n = r sin(phi_1)dotssin(phi_{n-2})sin(theta)$$



    where $0lephi_ilepi$, $0lethetale2pi$ and $0le r<infty$. The volume element is then:



    $$mathrm d x_1 dots mathrm d x_n = r^{n-1}sin^{n-2}(phi_1)dotssin(phi_{n-2}) mathrm dr mathrm dphi_1dotsmathrm dphi_{n-2}mathrm dtheta$$



    We have:



    $$begin{aligned}
    I & = int_{mathbb{R}^n} delta (n - | mathbf{x} |^2) mathrm d
    mathbf{x}\
    & = int_0^{pi} mathrm d phi_1 ldots int_0^{pi}mathrm d phi_{n - 2} int_0^{2 pi}
    mathrm d theta int_0^{infty} mathrm d r , delta (n - r^2) r^{n - 1} sin^{n -
    2} (phi_1) ldots sin (phi_{n - 2})\
    & = left( int_0^{infty} delta (n - r^2) r^{n - 1} mathrm d r right)
    left( int_0^{2 pi} mathrm d theta mathrm d theta right) left(
    int_0^{pi} sin^{n - 2} (phi_1) mathrm d phi_1 right) ldots left(
    int_0^{pi} sin (phi_{n - 2}) mathrm d phi_{n - 2} right)\
    & = frac{pi^{n / 2} n^{n / 2 - 1}}{Gamma (n / 2)}
    end{aligned}$$



    This is not an answer to my question. My question is: What did I do wrong to that leads to a divergent integral?






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      This is not an answer to your question. I think you'd better add it to your post instead of here.
      $endgroup$
      – user587192
      Nov 30 '18 at 14:03






    • 1




      $begingroup$
      @user587192 It is not an answer. But it is customary on this site to sometimes post solution attempts as (incomplete) answers.
      $endgroup$
      – becko
      Nov 30 '18 at 14:08














    1












    1








    1





    $begingroup$

    Here is the solution in (hyper)spherical coordinates, proving that $I$ should be well behaved. The spherical coordinates are defined by:



    $$x_1 = r cos(phi_1),quad x_2 = r sin(phi_1)cos(phi_2),dots,quad x_n = r sin(phi_1)dotssin(phi_{n-2})sin(theta)$$



    where $0lephi_ilepi$, $0lethetale2pi$ and $0le r<infty$. The volume element is then:



    $$mathrm d x_1 dots mathrm d x_n = r^{n-1}sin^{n-2}(phi_1)dotssin(phi_{n-2}) mathrm dr mathrm dphi_1dotsmathrm dphi_{n-2}mathrm dtheta$$



    We have:



    $$begin{aligned}
    I & = int_{mathbb{R}^n} delta (n - | mathbf{x} |^2) mathrm d
    mathbf{x}\
    & = int_0^{pi} mathrm d phi_1 ldots int_0^{pi}mathrm d phi_{n - 2} int_0^{2 pi}
    mathrm d theta int_0^{infty} mathrm d r , delta (n - r^2) r^{n - 1} sin^{n -
    2} (phi_1) ldots sin (phi_{n - 2})\
    & = left( int_0^{infty} delta (n - r^2) r^{n - 1} mathrm d r right)
    left( int_0^{2 pi} mathrm d theta mathrm d theta right) left(
    int_0^{pi} sin^{n - 2} (phi_1) mathrm d phi_1 right) ldots left(
    int_0^{pi} sin (phi_{n - 2}) mathrm d phi_{n - 2} right)\
    & = frac{pi^{n / 2} n^{n / 2 - 1}}{Gamma (n / 2)}
    end{aligned}$$



    This is not an answer to my question. My question is: What did I do wrong to that leads to a divergent integral?






    share|cite|improve this answer











    $endgroup$



    Here is the solution in (hyper)spherical coordinates, proving that $I$ should be well behaved. The spherical coordinates are defined by:



    $$x_1 = r cos(phi_1),quad x_2 = r sin(phi_1)cos(phi_2),dots,quad x_n = r sin(phi_1)dotssin(phi_{n-2})sin(theta)$$



    where $0lephi_ilepi$, $0lethetale2pi$ and $0le r<infty$. The volume element is then:



    $$mathrm d x_1 dots mathrm d x_n = r^{n-1}sin^{n-2}(phi_1)dotssin(phi_{n-2}) mathrm dr mathrm dphi_1dotsmathrm dphi_{n-2}mathrm dtheta$$



    We have:



    $$begin{aligned}
    I & = int_{mathbb{R}^n} delta (n - | mathbf{x} |^2) mathrm d
    mathbf{x}\
    & = int_0^{pi} mathrm d phi_1 ldots int_0^{pi}mathrm d phi_{n - 2} int_0^{2 pi}
    mathrm d theta int_0^{infty} mathrm d r , delta (n - r^2) r^{n - 1} sin^{n -
    2} (phi_1) ldots sin (phi_{n - 2})\
    & = left( int_0^{infty} delta (n - r^2) r^{n - 1} mathrm d r right)
    left( int_0^{2 pi} mathrm d theta mathrm d theta right) left(
    int_0^{pi} sin^{n - 2} (phi_1) mathrm d phi_1 right) ldots left(
    int_0^{pi} sin (phi_{n - 2}) mathrm d phi_{n - 2} right)\
    & = frac{pi^{n / 2} n^{n / 2 - 1}}{Gamma (n / 2)}
    end{aligned}$$



    This is not an answer to my question. My question is: What did I do wrong to that leads to a divergent integral?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 30 '18 at 14:53

























    answered Nov 30 '18 at 14:00









    beckobecko

    2,35931942




    2,35931942








    • 1




      $begingroup$
      This is not an answer to your question. I think you'd better add it to your post instead of here.
      $endgroup$
      – user587192
      Nov 30 '18 at 14:03






    • 1




      $begingroup$
      @user587192 It is not an answer. But it is customary on this site to sometimes post solution attempts as (incomplete) answers.
      $endgroup$
      – becko
      Nov 30 '18 at 14:08














    • 1




      $begingroup$
      This is not an answer to your question. I think you'd better add it to your post instead of here.
      $endgroup$
      – user587192
      Nov 30 '18 at 14:03






    • 1




      $begingroup$
      @user587192 It is not an answer. But it is customary on this site to sometimes post solution attempts as (incomplete) answers.
      $endgroup$
      – becko
      Nov 30 '18 at 14:08








    1




    1




    $begingroup$
    This is not an answer to your question. I think you'd better add it to your post instead of here.
    $endgroup$
    – user587192
    Nov 30 '18 at 14:03




    $begingroup$
    This is not an answer to your question. I think you'd better add it to your post instead of here.
    $endgroup$
    – user587192
    Nov 30 '18 at 14:03




    1




    1




    $begingroup$
    @user587192 It is not an answer. But it is customary on this site to sometimes post solution attempts as (incomplete) answers.
    $endgroup$
    – becko
    Nov 30 '18 at 14:08




    $begingroup$
    @user587192 It is not an answer. But it is customary on this site to sometimes post solution attempts as (incomplete) answers.
    $endgroup$
    – becko
    Nov 30 '18 at 14:08











    0












    $begingroup$

    There is a trick one can do to use the Fourier transform and still have converging integrals. The trick is to multiply the integrand by $mathrm e^{(1-||mathbf x||^2)} = 1$. Then



    $$begin{aligned}
    int_{- infty}^{infty} delta (1 - ||mathbf x||^2) mathrm d
    mathbf x & = int_{- infty}^{infty} mathrm d mathbf x
    int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{(1
    +mathfrak{i} lambda) (1 - ||mathbf x||^2)}\
    & = int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{1
    +mathfrak{i} lambda} left( int_{- infty}^{infty} mathrm e^{- (1
    +mathfrak{i} lambda) x^2} mathrm dx right)^N\
    & = int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{1
    +mathfrak{i} lambda} left( frac{pi}{1 +mathfrak{i} lambda} right)^{N
    / 2}\
    & = pi^{N / 2} / Gamma (N / 2)
    end{aligned}$$



    which is the correct result. Here all intermediate integrals converge and the integration order can be exchanged.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      There is a trick one can do to use the Fourier transform and still have converging integrals. The trick is to multiply the integrand by $mathrm e^{(1-||mathbf x||^2)} = 1$. Then



      $$begin{aligned}
      int_{- infty}^{infty} delta (1 - ||mathbf x||^2) mathrm d
      mathbf x & = int_{- infty}^{infty} mathrm d mathbf x
      int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{(1
      +mathfrak{i} lambda) (1 - ||mathbf x||^2)}\
      & = int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{1
      +mathfrak{i} lambda} left( int_{- infty}^{infty} mathrm e^{- (1
      +mathfrak{i} lambda) x^2} mathrm dx right)^N\
      & = int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{1
      +mathfrak{i} lambda} left( frac{pi}{1 +mathfrak{i} lambda} right)^{N
      / 2}\
      & = pi^{N / 2} / Gamma (N / 2)
      end{aligned}$$



      which is the correct result. Here all intermediate integrals converge and the integration order can be exchanged.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        There is a trick one can do to use the Fourier transform and still have converging integrals. The trick is to multiply the integrand by $mathrm e^{(1-||mathbf x||^2)} = 1$. Then



        $$begin{aligned}
        int_{- infty}^{infty} delta (1 - ||mathbf x||^2) mathrm d
        mathbf x & = int_{- infty}^{infty} mathrm d mathbf x
        int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{(1
        +mathfrak{i} lambda) (1 - ||mathbf x||^2)}\
        & = int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{1
        +mathfrak{i} lambda} left( int_{- infty}^{infty} mathrm e^{- (1
        +mathfrak{i} lambda) x^2} mathrm dx right)^N\
        & = int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{1
        +mathfrak{i} lambda} left( frac{pi}{1 +mathfrak{i} lambda} right)^{N
        / 2}\
        & = pi^{N / 2} / Gamma (N / 2)
        end{aligned}$$



        which is the correct result. Here all intermediate integrals converge and the integration order can be exchanged.






        share|cite|improve this answer











        $endgroup$



        There is a trick one can do to use the Fourier transform and still have converging integrals. The trick is to multiply the integrand by $mathrm e^{(1-||mathbf x||^2)} = 1$. Then



        $$begin{aligned}
        int_{- infty}^{infty} delta (1 - ||mathbf x||^2) mathrm d
        mathbf x & = int_{- infty}^{infty} mathrm d mathbf x
        int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{(1
        +mathfrak{i} lambda) (1 - ||mathbf x||^2)}\
        & = int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{1
        +mathfrak{i} lambda} left( int_{- infty}^{infty} mathrm e^{- (1
        +mathfrak{i} lambda) x^2} mathrm dx right)^N\
        & = int_{- infty}^{infty} frac{mathrm d lambda}{2 pi} mathrm e^{1
        +mathfrak{i} lambda} left( frac{pi}{1 +mathfrak{i} lambda} right)^{N
        / 2}\
        & = pi^{N / 2} / Gamma (N / 2)
        end{aligned}$$



        which is the correct result. Here all intermediate integrals converge and the integration order can be exchanged.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 18:46

























        answered Nov 30 '18 at 17:44









        beckobecko

        2,35931942




        2,35931942






























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