Does this oscillatory integral exist?











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Let $ngeq 2$ and consider the improper integral
$$I:=int_{mathbb{R}^{n}}F(x)dx$$ where $F$ is a continuous function.



If $I$ exists then



$$I=lim_{Rrightarrow +infty}int_{B_{R}}F(x)dx,$$
where $B_{R}$ is a ball with radius $R$.
So if this limit does not exist we know that the integral does not exist.
Does the existence of this limit imply the existence of the integral ?



Motivation:



I am interested in the existence of the integral
$$int_{mathbb{R}^{3}}frac{e^{dot{imath}|x-y|^2}}{1+|y|}dy.$$



Using spherical coordinates (I do not even know if we are allowed to change variables here. Are we ? )
$$int_{mathbb{R}^{3}}frac{e^{dot{imath}|x-y|^2}}{1+|y|}dy=
int_{mathbb{S}^{2}}int_{0}^{infty}
frac{e^{dot{imath}|rhoomega-x|^2}rho^2}{1+rho}drho domega\
=e^{i|x|^{2}}int_{mathbb{S}^{2}}int_{0}^{infty}
frac{e^{dot{imath} (rho^2-2xcdot omega,rho)}rho^2}{1+rho}drho domega.$$



Observations:



1-The inner integral does not exist for any $x$ and $omega$.



2-We can not change order of integration



3-The limit



$$lim_{Rrightarrow infty}int_{mathbb{S}^{2}}int_{0}^{R}
frac{e^{dot{imath} (rho^2-2xcdot omega,rho)}rho^2}{1+rho}drho domega$$

exists. Simply apply the very nice formula [Grafakos, classical Fourier analysis-Appendix D]:
$$int_{mathbb{S}^{n-1}}
F(x.omega)domega=c int_{-1}^{1}(sqrt{1-s^2})^{n-3}
F(s|x|)ds.$$

then benefit from the oscillation in both variables $rho$ and $omega$ and integrate by parts in both variables.



Any ideas how to handle this ?



Thank you so much










share|cite|improve this question




























    up vote
    3
    down vote

    favorite












    Let $ngeq 2$ and consider the improper integral
    $$I:=int_{mathbb{R}^{n}}F(x)dx$$ where $F$ is a continuous function.



    If $I$ exists then



    $$I=lim_{Rrightarrow +infty}int_{B_{R}}F(x)dx,$$
    where $B_{R}$ is a ball with radius $R$.
    So if this limit does not exist we know that the integral does not exist.
    Does the existence of this limit imply the existence of the integral ?



    Motivation:



    I am interested in the existence of the integral
    $$int_{mathbb{R}^{3}}frac{e^{dot{imath}|x-y|^2}}{1+|y|}dy.$$



    Using spherical coordinates (I do not even know if we are allowed to change variables here. Are we ? )
    $$int_{mathbb{R}^{3}}frac{e^{dot{imath}|x-y|^2}}{1+|y|}dy=
    int_{mathbb{S}^{2}}int_{0}^{infty}
    frac{e^{dot{imath}|rhoomega-x|^2}rho^2}{1+rho}drho domega\
    =e^{i|x|^{2}}int_{mathbb{S}^{2}}int_{0}^{infty}
    frac{e^{dot{imath} (rho^2-2xcdot omega,rho)}rho^2}{1+rho}drho domega.$$



    Observations:



    1-The inner integral does not exist for any $x$ and $omega$.



    2-We can not change order of integration



    3-The limit



    $$lim_{Rrightarrow infty}int_{mathbb{S}^{2}}int_{0}^{R}
    frac{e^{dot{imath} (rho^2-2xcdot omega,rho)}rho^2}{1+rho}drho domega$$

    exists. Simply apply the very nice formula [Grafakos, classical Fourier analysis-Appendix D]:
    $$int_{mathbb{S}^{n-1}}
    F(x.omega)domega=c int_{-1}^{1}(sqrt{1-s^2})^{n-3}
    F(s|x|)ds.$$

    then benefit from the oscillation in both variables $rho$ and $omega$ and integrate by parts in both variables.



    Any ideas how to handle this ?



    Thank you so much










    share|cite|improve this question


























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      Let $ngeq 2$ and consider the improper integral
      $$I:=int_{mathbb{R}^{n}}F(x)dx$$ where $F$ is a continuous function.



      If $I$ exists then



      $$I=lim_{Rrightarrow +infty}int_{B_{R}}F(x)dx,$$
      where $B_{R}$ is a ball with radius $R$.
      So if this limit does not exist we know that the integral does not exist.
      Does the existence of this limit imply the existence of the integral ?



      Motivation:



      I am interested in the existence of the integral
      $$int_{mathbb{R}^{3}}frac{e^{dot{imath}|x-y|^2}}{1+|y|}dy.$$



      Using spherical coordinates (I do not even know if we are allowed to change variables here. Are we ? )
      $$int_{mathbb{R}^{3}}frac{e^{dot{imath}|x-y|^2}}{1+|y|}dy=
      int_{mathbb{S}^{2}}int_{0}^{infty}
      frac{e^{dot{imath}|rhoomega-x|^2}rho^2}{1+rho}drho domega\
      =e^{i|x|^{2}}int_{mathbb{S}^{2}}int_{0}^{infty}
      frac{e^{dot{imath} (rho^2-2xcdot omega,rho)}rho^2}{1+rho}drho domega.$$



      Observations:



      1-The inner integral does not exist for any $x$ and $omega$.



      2-We can not change order of integration



      3-The limit



      $$lim_{Rrightarrow infty}int_{mathbb{S}^{2}}int_{0}^{R}
      frac{e^{dot{imath} (rho^2-2xcdot omega,rho)}rho^2}{1+rho}drho domega$$

      exists. Simply apply the very nice formula [Grafakos, classical Fourier analysis-Appendix D]:
      $$int_{mathbb{S}^{n-1}}
      F(x.omega)domega=c int_{-1}^{1}(sqrt{1-s^2})^{n-3}
      F(s|x|)ds.$$

      then benefit from the oscillation in both variables $rho$ and $omega$ and integrate by parts in both variables.



      Any ideas how to handle this ?



      Thank you so much










      share|cite|improve this question















      Let $ngeq 2$ and consider the improper integral
      $$I:=int_{mathbb{R}^{n}}F(x)dx$$ where $F$ is a continuous function.



      If $I$ exists then



      $$I=lim_{Rrightarrow +infty}int_{B_{R}}F(x)dx,$$
      where $B_{R}$ is a ball with radius $R$.
      So if this limit does not exist we know that the integral does not exist.
      Does the existence of this limit imply the existence of the integral ?



      Motivation:



      I am interested in the existence of the integral
      $$int_{mathbb{R}^{3}}frac{e^{dot{imath}|x-y|^2}}{1+|y|}dy.$$



      Using spherical coordinates (I do not even know if we are allowed to change variables here. Are we ? )
      $$int_{mathbb{R}^{3}}frac{e^{dot{imath}|x-y|^2}}{1+|y|}dy=
      int_{mathbb{S}^{2}}int_{0}^{infty}
      frac{e^{dot{imath}|rhoomega-x|^2}rho^2}{1+rho}drho domega\
      =e^{i|x|^{2}}int_{mathbb{S}^{2}}int_{0}^{infty}
      frac{e^{dot{imath} (rho^2-2xcdot omega,rho)}rho^2}{1+rho}drho domega.$$



      Observations:



      1-The inner integral does not exist for any $x$ and $omega$.



      2-We can not change order of integration



      3-The limit



      $$lim_{Rrightarrow infty}int_{mathbb{S}^{2}}int_{0}^{R}
      frac{e^{dot{imath} (rho^2-2xcdot omega,rho)}rho^2}{1+rho}drho domega$$

      exists. Simply apply the very nice formula [Grafakos, classical Fourier analysis-Appendix D]:
      $$int_{mathbb{S}^{n-1}}
      F(x.omega)domega=c int_{-1}^{1}(sqrt{1-s^2})^{n-3}
      F(s|x|)ds.$$

      then benefit from the oscillation in both variables $rho$ and $omega$ and integrate by parts in both variables.



      Any ideas how to handle this ?



      Thank you so much







      real-analysis analysis fourier-analysis harmonic-analysis spherical-harmonics






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      edited Nov 15 at 16:31

























      asked Nov 14 at 18:00









      Medo

      612213




      612213






















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          Does the existence of this limit imply the existence of the integral ?




          No. Take $n=1$. Then $$lim_{Rtoinfty}int_{-R}^{R}x,dx$$ exists and equals $0$ because postive parts are exactly canceling negative parts as $R$ grows. But I would not say $int_{mathbb{R}}x,dx$ exists.





          Maybe there is a question yet to answer if you require the function to be positive.






          share|cite|improve this answer





















          • I think the most common definition of $int_0^{infty}f(x),dx$ is $lim_{Rtoinfty}int_{0}^{R}f(x),dx$. In that case, you of course cannot find such an $f$. So what definition of $int_0^{infty}f(x),dx$ are you using?
            – alex.jordan
            Nov 14 at 23:01










          • Exactly my question: How define the improper integral $int_{mathbb{R}^n} F(x)dx$ when $ngeq2$ and $F$ is a continuous function.
            – Medo
            Nov 14 at 23:04











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          up vote
          0
          down vote














          Does the existence of this limit imply the existence of the integral ?




          No. Take $n=1$. Then $$lim_{Rtoinfty}int_{-R}^{R}x,dx$$ exists and equals $0$ because postive parts are exactly canceling negative parts as $R$ grows. But I would not say $int_{mathbb{R}}x,dx$ exists.





          Maybe there is a question yet to answer if you require the function to be positive.






          share|cite|improve this answer





















          • I think the most common definition of $int_0^{infty}f(x),dx$ is $lim_{Rtoinfty}int_{0}^{R}f(x),dx$. In that case, you of course cannot find such an $f$. So what definition of $int_0^{infty}f(x),dx$ are you using?
            – alex.jordan
            Nov 14 at 23:01










          • Exactly my question: How define the improper integral $int_{mathbb{R}^n} F(x)dx$ when $ngeq2$ and $F$ is a continuous function.
            – Medo
            Nov 14 at 23:04















          up vote
          0
          down vote














          Does the existence of this limit imply the existence of the integral ?




          No. Take $n=1$. Then $$lim_{Rtoinfty}int_{-R}^{R}x,dx$$ exists and equals $0$ because postive parts are exactly canceling negative parts as $R$ grows. But I would not say $int_{mathbb{R}}x,dx$ exists.





          Maybe there is a question yet to answer if you require the function to be positive.






          share|cite|improve this answer





















          • I think the most common definition of $int_0^{infty}f(x),dx$ is $lim_{Rtoinfty}int_{0}^{R}f(x),dx$. In that case, you of course cannot find such an $f$. So what definition of $int_0^{infty}f(x),dx$ are you using?
            – alex.jordan
            Nov 14 at 23:01










          • Exactly my question: How define the improper integral $int_{mathbb{R}^n} F(x)dx$ when $ngeq2$ and $F$ is a continuous function.
            – Medo
            Nov 14 at 23:04













          up vote
          0
          down vote










          up vote
          0
          down vote










          Does the existence of this limit imply the existence of the integral ?




          No. Take $n=1$. Then $$lim_{Rtoinfty}int_{-R}^{R}x,dx$$ exists and equals $0$ because postive parts are exactly canceling negative parts as $R$ grows. But I would not say $int_{mathbb{R}}x,dx$ exists.





          Maybe there is a question yet to answer if you require the function to be positive.






          share|cite|improve this answer













          Does the existence of this limit imply the existence of the integral ?




          No. Take $n=1$. Then $$lim_{Rtoinfty}int_{-R}^{R}x,dx$$ exists and equals $0$ because postive parts are exactly canceling negative parts as $R$ grows. But I would not say $int_{mathbb{R}}x,dx$ exists.





          Maybe there is a question yet to answer if you require the function to be positive.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 14 at 22:53









          alex.jordan

          37.9k559118




          37.9k559118












          • I think the most common definition of $int_0^{infty}f(x),dx$ is $lim_{Rtoinfty}int_{0}^{R}f(x),dx$. In that case, you of course cannot find such an $f$. So what definition of $int_0^{infty}f(x),dx$ are you using?
            – alex.jordan
            Nov 14 at 23:01










          • Exactly my question: How define the improper integral $int_{mathbb{R}^n} F(x)dx$ when $ngeq2$ and $F$ is a continuous function.
            – Medo
            Nov 14 at 23:04


















          • I think the most common definition of $int_0^{infty}f(x),dx$ is $lim_{Rtoinfty}int_{0}^{R}f(x),dx$. In that case, you of course cannot find such an $f$. So what definition of $int_0^{infty}f(x),dx$ are you using?
            – alex.jordan
            Nov 14 at 23:01










          • Exactly my question: How define the improper integral $int_{mathbb{R}^n} F(x)dx$ when $ngeq2$ and $F$ is a continuous function.
            – Medo
            Nov 14 at 23:04
















          I think the most common definition of $int_0^{infty}f(x),dx$ is $lim_{Rtoinfty}int_{0}^{R}f(x),dx$. In that case, you of course cannot find such an $f$. So what definition of $int_0^{infty}f(x),dx$ are you using?
          – alex.jordan
          Nov 14 at 23:01




          I think the most common definition of $int_0^{infty}f(x),dx$ is $lim_{Rtoinfty}int_{0}^{R}f(x),dx$. In that case, you of course cannot find such an $f$. So what definition of $int_0^{infty}f(x),dx$ are you using?
          – alex.jordan
          Nov 14 at 23:01












          Exactly my question: How define the improper integral $int_{mathbb{R}^n} F(x)dx$ when $ngeq2$ and $F$ is a continuous function.
          – Medo
          Nov 14 at 23:04




          Exactly my question: How define the improper integral $int_{mathbb{R}^n} F(x)dx$ when $ngeq2$ and $F$ is a continuous function.
          – Medo
          Nov 14 at 23:04


















           

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