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
real-analysis analysis fourier-analysis harmonic-analysis spherical-harmonics
add a comment |
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
real-analysis analysis fourier-analysis harmonic-analysis spherical-harmonics
add a comment |
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
real-analysis analysis fourier-analysis harmonic-analysis spherical-harmonics
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
real-analysis analysis fourier-analysis harmonic-analysis spherical-harmonics
edited Nov 15 at 16:31
asked Nov 14 at 18:00
Medo
612213
612213
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1 Answer
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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.
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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
add a comment |
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.
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
add a comment |
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.
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.
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
add a comment |
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
add a comment |
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