Mixing
Derivation of integration property of Fourier transform
$begingroup$
For $f in mathcal{S}(mathbb R)$, where $mathcal{S}$ is a Schwartz space, denote it's Fourier transform as
$$
hat{f}(omega) := mathcal{F}f(t) := int_{-infty}^{infty} f(t) e^{-2 pi i omega t} dt
$$
and corresponding inverse Fourier transform as
$$
mathcal{F}^{-1} hat{f}(omega) := int_{-infty}^{infty} e^{2 pi i t omega} hat{f}(omega) domega
$$
I would like to know what is the formula for:
$$
mathcal{F} left[ int_{-infty}^t f(tau) dtau right]
$$
What I have so far is:
begin{alignat}{2}
f(tau) &= mathcal{F}^{-1}hat{f}(omega) qquad implies \
int_{-infty}^t f(tau) dtau &= int_{-infty}^t mathcal{F}^{-1}hat{f}(omega) dtau \
&= int_{-infty}^t left[ int_{-infty}^infty e^{2 pi i tau omega} hat{f}(omega) domega right] dtau \
&= int_{-infty}^infty hat{f}(omega) int_{-infty}^t e^{2 pi i tau omega} dtau, domega \
&= int_{-infty}^infty hat{f}(omega) (2 pi i omega)^{-1}left[ e^{2 pi i tau omega} right]_{-infty}^t domega \
end{alignat}
and then I have a problem that the limit
$$
lim_{tau to -infty}e^{2 pi i tau omega}
$$
is not well defined.
I wonder how did they got the corresponding result on this page:
integration complex-analysis analysis fourier-analysis fourier-transform
asked Feb 22 '17 at 9:39
aberdyshaberdysh
337311
$endgroup$
add a comment |
$begingroup$
For $f in mathcal{S}(mathbb R)$, where $mathcal{S}$ is a Schwartz space, denote it's Fourier transform as
$$
hat{f}(omega) := mathcal{F}f(t) := int_{-infty}^{infty} f(t) e^{-2 pi i omega t} dt
$$
and corresponding inverse Fourier transform as
$$
mathcal{F}^{-1} hat{f}(omega) := int_{-infty}^{infty} e^{2 pi i t omega} hat{f}(omega) domega
$$
I would like to know what is the formula for:
$$
mathcal{F} left[ int_{-infty}^t f(tau) dtau right]
$$
What I have so far is:
begin{alignat}{2}
f(tau) &= mathcal{F}^{-1}hat{f}(omega) qquad implies \
int_{-infty}^t f(tau) dtau &= int_{-infty}^t mathcal{F}^{-1}hat{f}(omega) dtau \
&= int_{-infty}^t left[ int_{-infty}^infty e^{2 pi i tau omega} hat{f}(omega) domega right] dtau \
&= int_{-infty}^infty hat{f}(omega) int_{-infty}^t e^{2 pi i tau omega} dtau, domega \
&= int_{-infty}^infty hat{f}(omega) (2 pi i omega)^{-1}left[ e^{2 pi i tau omega} right]_{-infty}^t domega \
end{alignat}
and then I have a problem that the limit
$$
lim_{tau to -infty}e^{2 pi i tau omega}
$$
is not well defined.
I wonder how did they got the corresponding result on this page:
integration complex-analysis analysis fourier-analysis fourier-transform
asked Feb 22 '17 at 9:39
aberdyshaberdysh
337311
$endgroup$
$begingroup$
That's not the correct definition for the Fourier transform of $L^2(mathbb R^n)$ functions, because the integral might not be convergent. Instead, you define the transform pointwise (like you did) for $mathcal S(mathbb R^n)$ and $L^1(mathbb R^n)$ functions and then, since $mathcal S(mathbb R^n)$ is dense in $L^2(mathbb R^n)$ you extend the Fourier transform operator by continuity.
$endgroup$
– rubik
Feb 22 '17 at 9:42
$begingroup$
ok, let me edit.
$endgroup$
– aberdysh
Feb 22 '17 at 9:45
add a comment |
$begingroup$
For $f in mathcal{S}(mathbb R)$, where $mathcal{S}$ is a Schwartz space, denote it's Fourier transform as
$$
hat{f}(omega) := mathcal{F}f(t) := int_{-infty}^{infty} f(t) e^{-2 pi i omega t} dt
$$
and corresponding inverse Fourier transform as
$$
mathcal{F}^{-1} hat{f}(omega) := int_{-infty}^{infty} e^{2 pi i t omega} hat{f}(omega) domega
$$
I would like to know what is the formula for:
$$
mathcal{F} left[ int_{-infty}^t f(tau) dtau right]
$$
What I have so far is:
begin{alignat}{2}
f(tau) &= mathcal{F}^{-1}hat{f}(omega) qquad implies \
int_{-infty}^t f(tau) dtau &= int_{-infty}^t mathcal{F}^{-1}hat{f}(omega) dtau \
&= int_{-infty}^t left[ int_{-infty}^infty e^{2 pi i tau omega} hat{f}(omega) domega right] dtau \
&= int_{-infty}^infty hat{f}(omega) int_{-infty}^t e^{2 pi i tau omega} dtau, domega \
&= int_{-infty}^infty hat{f}(omega) (2 pi i omega)^{-1}left[ e^{2 pi i tau omega} right]_{-infty}^t domega \
end{alignat}
and then I have a problem that the limit
$$
lim_{tau to -infty}e^{2 pi i tau omega}
$$
is not well defined.
I wonder how did they got the corresponding result on this page:
integration complex-analysis analysis fourier-analysis fourier-transform
asked Feb 22 '17 at 9:39
aberdyshaberdysh
337311
$endgroup$
For $f in mathcal{S}(mathbb R)$, where $mathcal{S}$ is a Schwartz space, denote it's Fourier transform as
$$
hat{f}(omega) := mathcal{F}f(t) := int_{-infty}^{infty} f(t) e^{-2 pi i omega t} dt
$$
and corresponding inverse Fourier transform as
$$
mathcal{F}^{-1} hat{f}(omega) := int_{-infty}^{infty} e^{2 pi i t omega} hat{f}(omega) domega
$$
I would like to know what is the formula for:
$$
mathcal{F} left[ int_{-infty}^t f(tau) dtau right]
$$
What I have so far is:
begin{alignat}{2}
f(tau) &= mathcal{F}^{-1}hat{f}(omega) qquad implies \
int_{-infty}^t f(tau) dtau &= int_{-infty}^t mathcal{F}^{-1}hat{f}(omega) dtau \
&= int_{-infty}^t left[ int_{-infty}^infty e^{2 pi i tau omega} hat{f}(omega) domega right] dtau \
&= int_{-infty}^infty hat{f}(omega) int_{-infty}^t e^{2 pi i tau omega} dtau, domega \
&= int_{-infty}^infty hat{f}(omega) (2 pi i omega)^{-1}left[ e^{2 pi i tau omega} right]_{-infty}^t domega \
end{alignat}
and then I have a problem that the limit
$$
lim_{tau to -infty}e^{2 pi i tau omega}
$$
is not well defined.
I wonder how did they got the corresponding result on this page:
integration complex-analysis analysis fourier-analysis fourier-transform
integration complex-analysis analysis fourier-analysis fourier-transform
asked Feb 22 '17 at 9:39
aberdyshaberdysh
337311
asked Feb 22 '17 at 9:39
aberdyshaberdysh
337311
edited Feb 22 '17 at 9:45
aberdysh
asked Feb 22 '17 at 9:39
aberdyshaberdysh
337311
asked Feb 22 '17 at 9:39
aberdyshaberdysh
337311
asked Feb 22 '17 at 9:39
aberdyshaberdysh
337311
337311
$begingroup$
That's not the correct definition for the Fourier transform of $L^2(mathbb R^n)$ functions, because the integral might not be convergent. Instead, you define the transform pointwise (like you did) for $mathcal S(mathbb R^n)$ and $L^1(mathbb R^n)$ functions and then, since $mathcal S(mathbb R^n)$ is dense in $L^2(mathbb R^n)$ you extend the Fourier transform operator by continuity.
$endgroup$
– rubik
Feb 22 '17 at 9:42
$begingroup$
ok, let me edit.
$endgroup$
– aberdysh
Feb 22 '17 at 9:45
add a comment |
$begingroup$
That's not the correct definition for the Fourier transform of $L^2(mathbb R^n)$ functions, because the integral might not be convergent. Instead, you define the transform pointwise (like you did) for $mathcal S(mathbb R^n)$ and $L^1(mathbb R^n)$ functions and then, since $mathcal S(mathbb R^n)$ is dense in $L^2(mathbb R^n)$ you extend the Fourier transform operator by continuity.
$endgroup$
– rubik
Feb 22 '17 at 9:42
$begingroup$
ok, let me edit.
$endgroup$
– aberdysh
Feb 22 '17 at 9:45
$begingroup$
That's not the correct definition for the Fourier transform of $L^2(mathbb R^n)$ functions, because the integral might not be convergent. Instead, you define the transform pointwise (like you did) for $mathcal S(mathbb R^n)$ and $L^1(mathbb R^n)$ functions and then, since $mathcal S(mathbb R^n)$ is dense in $L^2(mathbb R^n)$ you extend the Fourier transform operator by continuity.
$endgroup$
– rubik
Feb 22 '17 at 9:42
$begingroup$
That's not the correct definition for the Fourier transform of $L^2(mathbb R^n)$ functions, because the integral might not be convergent. Instead, you define the transform pointwise (like you did) for $mathcal S(mathbb R^n)$ and $L^1(mathbb R^n)$ functions and then, since $mathcal S(mathbb R^n)$ is dense in $L^2(mathbb R^n)$ you extend the Fourier transform operator by continuity.
$endgroup$
– rubik
Feb 22 '17 at 9:42
$begingroup$
ok, let me edit.
$endgroup$
– aberdysh
Feb 22 '17 at 9:45
$begingroup$
ok, let me edit.
$endgroup$
– aberdysh
Feb 22 '17 at 9:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Simple way is to calculate the Fourier transform of the distribution $theta(x)$ and apply convolution theorem. To get the first part, it is convenient to replace $theta (x)$ by $theta(x) e^{-epsilon x}$ and take the limit $epsilon to 0$ through positive values only after calculating the Fourier transform. This is correct because Fourier transformation is continuous in the space of distributions with weak-* topology.
answered Feb 22 '17 at 9:57
BlazejBlazej
1,485620
$endgroup$
add a comment |
$begingroup$
When $f in S(mathbb{R})$ and $int_{-infty}^infty f(t) dt = 0$ then $int_{-infty}^t f(tau)dtau in S(mathbb{R})$.
Integrating by parts $$mathcal{F}[int_{-infty}^t f(tau)dtau](xi) = int_{-infty}^inftyint_{-infty}^t f(tau) dtau e^{-2i pi xi t} dt = int_{-infty}^infty f(t) frac{e^{-2i pi xi t}}{2i pi xi} dt = frac{1}{2i pi xi} mathcal{F}[ f(t)](xi)$$
Otherwise $mathcal{F}[int_{-infty}^t f(tau)dtau]$ exists only as the Fourier transform of a tempered distribution, and the result becomes $$mathcal{F}[int_{-infty}^t f(tau)dtau](xi) = frac{1}{2i pi xi} mathcal{F}[ f(t)](xi)+ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$$
answered Feb 22 '17 at 10:25
reunsreuns
20k21148
$endgroup$
$begingroup$
I don't see where $ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$ comes from
$endgroup$
– aberdysh
Feb 22 '17 at 22:51
$begingroup$
@aberdysh it comes from the Fourier transform of distributions. Let $H(xi) = mathcal{F}[int_{-infty}^t f(tau)dtau](xi)$ then using the properties of the Fourier transform of distributions $2i pi xi H(xi)= mathcal{F}[f(t)](xi)$ i.e. $H(xi) = frac{1}{2i pi xi} mathcal{F}[f(t)](xi)+C delta(xi)$ for some $C$
$endgroup$
– reuns
Feb 22 '17 at 23:54
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
Simple way is to calculate the Fourier transform of the distribution $theta(x)$ and apply convolution theorem. To get the first part, it is convenient to replace $theta (x)$ by $theta(x) e^{-epsilon x}$ and take the limit $epsilon to 0$ through positive values only after calculating the Fourier transform. This is correct because Fourier transformation is continuous in the space of distributions with weak-* topology.
answered Feb 22 '17 at 9:57
BlazejBlazej
1,485620
$endgroup$
add a comment |
$begingroup$
Simple way is to calculate the Fourier transform of the distribution $theta(x)$ and apply convolution theorem. To get the first part, it is convenient to replace $theta (x)$ by $theta(x) e^{-epsilon x}$ and take the limit $epsilon to 0$ through positive values only after calculating the Fourier transform. This is correct because Fourier transformation is continuous in the space of distributions with weak-* topology.
answered Feb 22 '17 at 9:57
BlazejBlazej
1,485620
$endgroup$
add a comment |
$begingroup$
Simple way is to calculate the Fourier transform of the distribution $theta(x)$ and apply convolution theorem. To get the first part, it is convenient to replace $theta (x)$ by $theta(x) e^{-epsilon x}$ and take the limit $epsilon to 0$ through positive values only after calculating the Fourier transform. This is correct because Fourier transformation is continuous in the space of distributions with weak-* topology.
answered Feb 22 '17 at 9:57
BlazejBlazej
1,485620
$endgroup$
Simple way is to calculate the Fourier transform of the distribution $theta(x)$ and apply convolution theorem. To get the first part, it is convenient to replace $theta (x)$ by $theta(x) e^{-epsilon x}$ and take the limit $epsilon to 0$ through positive values only after calculating the Fourier transform. This is correct because Fourier transformation is continuous in the space of distributions with weak-* topology.
answered Feb 22 '17 at 9:57
BlazejBlazej
1,485620
answered Feb 22 '17 at 9:57
BlazejBlazej
1,485620
answered Feb 22 '17 at 9:57
BlazejBlazej
1,485620
answered Feb 22 '17 at 9:57
BlazejBlazej
1,485620
1,485620
add a comment |
add a comment |
$begingroup$
When $f in S(mathbb{R})$ and $int_{-infty}^infty f(t) dt = 0$ then $int_{-infty}^t f(tau)dtau in S(mathbb{R})$.
Integrating by parts $$mathcal{F}[int_{-infty}^t f(tau)dtau](xi) = int_{-infty}^inftyint_{-infty}^t f(tau) dtau e^{-2i pi xi t} dt = int_{-infty}^infty f(t) frac{e^{-2i pi xi t}}{2i pi xi} dt = frac{1}{2i pi xi} mathcal{F}[ f(t)](xi)$$
Otherwise $mathcal{F}[int_{-infty}^t f(tau)dtau]$ exists only as the Fourier transform of a tempered distribution, and the result becomes $$mathcal{F}[int_{-infty}^t f(tau)dtau](xi) = frac{1}{2i pi xi} mathcal{F}[ f(t)](xi)+ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$$
answered Feb 22 '17 at 10:25
reunsreuns
20k21148
$endgroup$
$begingroup$
I don't see where $ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$ comes from
$endgroup$
– aberdysh
Feb 22 '17 at 22:51
$begingroup$
@aberdysh it comes from the Fourier transform of distributions. Let $H(xi) = mathcal{F}[int_{-infty}^t f(tau)dtau](xi)$ then using the properties of the Fourier transform of distributions $2i pi xi H(xi)= mathcal{F}[f(t)](xi)$ i.e. $H(xi) = frac{1}{2i pi xi} mathcal{F}[f(t)](xi)+C delta(xi)$ for some $C$
$endgroup$
– reuns
Feb 22 '17 at 23:54
add a comment |
$begingroup$
When $f in S(mathbb{R})$ and $int_{-infty}^infty f(t) dt = 0$ then $int_{-infty}^t f(tau)dtau in S(mathbb{R})$.
Integrating by parts $$mathcal{F}[int_{-infty}^t f(tau)dtau](xi) = int_{-infty}^inftyint_{-infty}^t f(tau) dtau e^{-2i pi xi t} dt = int_{-infty}^infty f(t) frac{e^{-2i pi xi t}}{2i pi xi} dt = frac{1}{2i pi xi} mathcal{F}[ f(t)](xi)$$
Otherwise $mathcal{F}[int_{-infty}^t f(tau)dtau]$ exists only as the Fourier transform of a tempered distribution, and the result becomes $$mathcal{F}[int_{-infty}^t f(tau)dtau](xi) = frac{1}{2i pi xi} mathcal{F}[ f(t)](xi)+ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$$
answered Feb 22 '17 at 10:25
reunsreuns
20k21148
$endgroup$
$begingroup$
I don't see where $ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$ comes from
$endgroup$
– aberdysh
Feb 22 '17 at 22:51
$begingroup$
@aberdysh it comes from the Fourier transform of distributions. Let $H(xi) = mathcal{F}[int_{-infty}^t f(tau)dtau](xi)$ then using the properties of the Fourier transform of distributions $2i pi xi H(xi)= mathcal{F}[f(t)](xi)$ i.e. $H(xi) = frac{1}{2i pi xi} mathcal{F}[f(t)](xi)+C delta(xi)$ for some $C$
$endgroup$
– reuns
Feb 22 '17 at 23:54
add a comment |
$begingroup$
When $f in S(mathbb{R})$ and $int_{-infty}^infty f(t) dt = 0$ then $int_{-infty}^t f(tau)dtau in S(mathbb{R})$.
Integrating by parts $$mathcal{F}[int_{-infty}^t f(tau)dtau](xi) = int_{-infty}^inftyint_{-infty}^t f(tau) dtau e^{-2i pi xi t} dt = int_{-infty}^infty f(t) frac{e^{-2i pi xi t}}{2i pi xi} dt = frac{1}{2i pi xi} mathcal{F}[ f(t)](xi)$$
Otherwise $mathcal{F}[int_{-infty}^t f(tau)dtau]$ exists only as the Fourier transform of a tempered distribution, and the result becomes $$mathcal{F}[int_{-infty}^t f(tau)dtau](xi) = frac{1}{2i pi xi} mathcal{F}[ f(t)](xi)+ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$$
answered Feb 22 '17 at 10:25
reunsreuns
20k21148
$endgroup$
When $f in S(mathbb{R})$ and $int_{-infty}^infty f(t) dt = 0$ then $int_{-infty}^t f(tau)dtau in S(mathbb{R})$.
Integrating by parts $$mathcal{F}[int_{-infty}^t f(tau)dtau](xi) = int_{-infty}^inftyint_{-infty}^t f(tau) dtau e^{-2i pi xi t} dt = int_{-infty}^infty f(t) frac{e^{-2i pi xi t}}{2i pi xi} dt = frac{1}{2i pi xi} mathcal{F}[ f(t)](xi)$$
Otherwise $mathcal{F}[int_{-infty}^t f(tau)dtau]$ exists only as the Fourier transform of a tempered distribution, and the result becomes $$mathcal{F}[int_{-infty}^t f(tau)dtau](xi) = frac{1}{2i pi xi} mathcal{F}[ f(t)](xi)+ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$$
answered Feb 22 '17 at 10:25
reunsreuns
20k21148
edited Feb 22 '17 at 10:41
answered Feb 22 '17 at 10:25
reunsreuns
20k21148
answered Feb 22 '17 at 10:25
reunsreuns
20k21148
answered Feb 22 '17 at 10:25
reunsreuns
20k21148
20k21148
$begingroup$
I don't see where $ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$ comes from
$endgroup$
– aberdysh
Feb 22 '17 at 22:51
$begingroup$
@aberdysh it comes from the Fourier transform of distributions. Let $H(xi) = mathcal{F}[int_{-infty}^t f(tau)dtau](xi)$ then using the properties of the Fourier transform of distributions $2i pi xi H(xi)= mathcal{F}[f(t)](xi)$ i.e. $H(xi) = frac{1}{2i pi xi} mathcal{F}[f(t)](xi)+C delta(xi)$ for some $C$
$endgroup$
– reuns
Feb 22 '17 at 23:54
add a comment |
$begingroup$
I don't see where $ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$ comes from
$endgroup$
– aberdysh
Feb 22 '17 at 22:51
$begingroup$
@aberdysh it comes from the Fourier transform of distributions. Let $H(xi) = mathcal{F}[int_{-infty}^t f(tau)dtau](xi)$ then using the properties of the Fourier transform of distributions $2i pi xi H(xi)= mathcal{F}[f(t)](xi)$ i.e. $H(xi) = frac{1}{2i pi xi} mathcal{F}[f(t)](xi)+C delta(xi)$ for some $C$
$endgroup$
– reuns
Feb 22 '17 at 23:54
$begingroup$
I don't see where $ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$ comes from
$endgroup$
– aberdysh
Feb 22 '17 at 22:51
$begingroup$
I don't see where $ frac{int_{-infty}^infty f(t) dt}{2}delta(xi)$ comes from
$endgroup$
– aberdysh
Feb 22 '17 at 22:51
$begingroup$
@aberdysh it comes from the Fourier transform of distributions. Let $H(xi) = mathcal{F}[int_{-infty}^t f(tau)dtau](xi)$ then using the properties of the Fourier transform of distributions $2i pi xi H(xi)= mathcal{F}[f(t)](xi)$ i.e. $H(xi) = frac{1}{2i pi xi} mathcal{F}[f(t)](xi)+C delta(xi)$ for some $C$
$endgroup$
– reuns
Feb 22 '17 at 23:54
$begingroup$
@aberdysh it comes from the Fourier transform of distributions. Let $H(xi) = mathcal{F}[int_{-infty}^t f(tau)dtau](xi)$ then using the properties of the Fourier transform of distributions $2i pi xi H(xi)= mathcal{F}[f(t)](xi)$ i.e. $H(xi) = frac{1}{2i pi xi} mathcal{F}[f(t)](xi)+C delta(xi)$ for some $C$
$endgroup$
– reuns
Feb 22 '17 at 23:54
add a comment |
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$begingroup$
That's not the correct definition for the Fourier transform of $L^2(mathbb R^n)$ functions, because the integral might not be convergent. Instead, you define the transform pointwise (like you did) for $mathcal S(mathbb R^n)$ and $L^1(mathbb R^n)$ functions and then, since $mathcal S(mathbb R^n)$ is dense in $L^2(mathbb R^n)$ you extend the Fourier transform operator by continuity.
$endgroup$
– rubik
Feb 22 '17 at 9:42
$begingroup$
ok, let me edit.
$endgroup$
– aberdysh
Feb 22 '17 at 9:45