Mixing
Fourier transform and lebesgue integral
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Let $f:mathbb{R}^d rightarrow mathbb{R}^d$ be the function with $f(x) = exp(-frac{1}{2}mid x mid^2)$. Show that the fourier transform of $f$ is given by $hat f = (sqrt{2 pi})^d f$.
The fourier transform is given by $hat f (xi ) := displaystyleint_{mathbb{R}^d} f(x) exp(-ixi cdot x) mu(dx)$.
I would like to start with looking at the case $d = 1$ but I dont know how to proceed.
calculus lebesgue-integral fourier-transform
asked yesterday
Arjihad
356111
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down vote
favorite
Let $f:mathbb{R}^d rightarrow mathbb{R}^d$ be the function with $f(x) = exp(-frac{1}{2}mid x mid^2)$. Show that the fourier transform of $f$ is given by $hat f = (sqrt{2 pi})^d f$.
The fourier transform is given by $hat f (xi ) := displaystyleint_{mathbb{R}^d} f(x) exp(-ixi cdot x) mu(dx)$.
I would like to start with looking at the case $d = 1$ but I dont know how to proceed.
calculus lebesgue-integral fourier-transform
asked yesterday
Arjihad
356111
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f:mathbb{R}^d rightarrow mathbb{R}^d$ be the function with $f(x) = exp(-frac{1}{2}mid x mid^2)$. Show that the fourier transform of $f$ is given by $hat f = (sqrt{2 pi})^d f$.
The fourier transform is given by $hat f (xi ) := displaystyleint_{mathbb{R}^d} f(x) exp(-ixi cdot x) mu(dx)$.
I would like to start with looking at the case $d = 1$ but I dont know how to proceed.
calculus lebesgue-integral fourier-transform
asked yesterday
Arjihad
356111
Let $f:mathbb{R}^d rightarrow mathbb{R}^d$ be the function with $f(x) = exp(-frac{1}{2}mid x mid^2)$. Show that the fourier transform of $f$ is given by $hat f = (sqrt{2 pi})^d f$.
The fourier transform is given by $hat f (xi ) := displaystyleint_{mathbb{R}^d} f(x) exp(-ixi cdot x) mu(dx)$.
I would like to start with looking at the case $d = 1$ but I dont know how to proceed.
calculus lebesgue-integral fourier-transform
calculus lebesgue-integral fourier-transform
asked yesterday
Arjihad
356111
asked yesterday
Arjihad
356111
asked yesterday
Arjihad
356111
asked yesterday
Arjihad
356111
asked yesterday
Arjihad
356111
356111
add a comment |
add a comment |
2 Answers
2
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0
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It's a Gaussian integral---complete the square up in the exponential to get $(2pi)^{d/2} e^{-|xi|^2/2}$. Standard result on multivariate Gaussians (and no need to worry about Lebesgue).
answered yesterday
Richard Martin
1,3438
1
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
add a comment |
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0
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For $d=1$, we have $f(x)=e^{-x^2/2}$ and therefore (remember that $(ixi+x)^2=-xi^2+2ixi x+x^2$)
begin{equation*}
begin{split}
hat{f}(xi) & = int_{-infty}^{infty} e^{-1/2(2ixi x + x^2)} {rm d}mu(x) \
& = int_{-infty}^{infty} e^{-1/2((ixi+x)^2+xi^2)} {rm d}mu(x) \
& = f(xi) int_{-infty}^{infty} e^{-(ixi+x)^2/2} {rm d}mu(x) \
& =f(xi) int_{-infty}^{infty} e^{-y^2/2} {rm d}mu(y) \
& = sqrt{2pi} f(xi).
end{split}
end{equation*}
It's pretty similar for $d>1$.
answered yesterday
rldias
2,9301522
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
Change variable and the contribution from the ends of the contour (which gets shifted up or down) is negligible.
– Richard Martin
yesterday
How do I continue for $d=2$ and so on?
– Arjihad
12 hours ago
For any $dgeq 1$ you will get $$hat{f}(xi)=f(xi)int_{mathbb{R}^d} e^{-|y|^2/2} {rm d}mu(y)$$ which is just $hat{f}(xi)=(2pi)^{d/2} f(xi)$.
– rldias
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It's a Gaussian integral---complete the square up in the exponential to get $(2pi)^{d/2} e^{-|xi|^2/2}$. Standard result on multivariate Gaussians (and no need to worry about Lebesgue).
answered yesterday
Richard Martin
1,3438
1
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
add a comment |
up vote
0
down vote
It's a Gaussian integral---complete the square up in the exponential to get $(2pi)^{d/2} e^{-|xi|^2/2}$. Standard result on multivariate Gaussians (and no need to worry about Lebesgue).
answered yesterday
Richard Martin
1,3438
1
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
It's a Gaussian integral---complete the square up in the exponential to get $(2pi)^{d/2} e^{-|xi|^2/2}$. Standard result on multivariate Gaussians (and no need to worry about Lebesgue).
answered yesterday
Richard Martin
1,3438
It's a Gaussian integral---complete the square up in the exponential to get $(2pi)^{d/2} e^{-|xi|^2/2}$. Standard result on multivariate Gaussians (and no need to worry about Lebesgue).
answered yesterday
Richard Martin
1,3438
answered yesterday
Richard Martin
1,3438
answered yesterday
Richard Martin
1,3438
answered yesterday
Richard Martin
1,3438
1,3438
1
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
add a comment |
1
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
1
1
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
add a comment |
up vote
0
down vote
For $d=1$, we have $f(x)=e^{-x^2/2}$ and therefore (remember that $(ixi+x)^2=-xi^2+2ixi x+x^2$)
begin{equation*}
begin{split}
hat{f}(xi) & = int_{-infty}^{infty} e^{-1/2(2ixi x + x^2)} {rm d}mu(x) \
& = int_{-infty}^{infty} e^{-1/2((ixi+x)^2+xi^2)} {rm d}mu(x) \
& = f(xi) int_{-infty}^{infty} e^{-(ixi+x)^2/2} {rm d}mu(x) \
& =f(xi) int_{-infty}^{infty} e^{-y^2/2} {rm d}mu(y) \
& = sqrt{2pi} f(xi).
end{split}
end{equation*}
It's pretty similar for $d>1$.
answered yesterday
rldias
2,9301522
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
Change variable and the contribution from the ends of the contour (which gets shifted up or down) is negligible.
– Richard Martin
yesterday
How do I continue for $d=2$ and so on?
– Arjihad
12 hours ago
For any $dgeq 1$ you will get $$hat{f}(xi)=f(xi)int_{mathbb{R}^d} e^{-|y|^2/2} {rm d}mu(y)$$ which is just $hat{f}(xi)=(2pi)^{d/2} f(xi)$.
– rldias
4 hours ago
add a comment |
up vote
0
down vote
For $d=1$, we have $f(x)=e^{-x^2/2}$ and therefore (remember that $(ixi+x)^2=-xi^2+2ixi x+x^2$)
begin{equation*}
begin{split}
hat{f}(xi) & = int_{-infty}^{infty} e^{-1/2(2ixi x + x^2)} {rm d}mu(x) \
& = int_{-infty}^{infty} e^{-1/2((ixi+x)^2+xi^2)} {rm d}mu(x) \
& = f(xi) int_{-infty}^{infty} e^{-(ixi+x)^2/2} {rm d}mu(x) \
& =f(xi) int_{-infty}^{infty} e^{-y^2/2} {rm d}mu(y) \
& = sqrt{2pi} f(xi).
end{split}
end{equation*}
It's pretty similar for $d>1$.
answered yesterday
rldias
2,9301522
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
Change variable and the contribution from the ends of the contour (which gets shifted up or down) is negligible.
– Richard Martin
yesterday
How do I continue for $d=2$ and so on?
– Arjihad
12 hours ago
For any $dgeq 1$ you will get $$hat{f}(xi)=f(xi)int_{mathbb{R}^d} e^{-|y|^2/2} {rm d}mu(y)$$ which is just $hat{f}(xi)=(2pi)^{d/2} f(xi)$.
– rldias
4 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
For $d=1$, we have $f(x)=e^{-x^2/2}$ and therefore (remember that $(ixi+x)^2=-xi^2+2ixi x+x^2$)
begin{equation*}
begin{split}
hat{f}(xi) & = int_{-infty}^{infty} e^{-1/2(2ixi x + x^2)} {rm d}mu(x) \
& = int_{-infty}^{infty} e^{-1/2((ixi+x)^2+xi^2)} {rm d}mu(x) \
& = f(xi) int_{-infty}^{infty} e^{-(ixi+x)^2/2} {rm d}mu(x) \
& =f(xi) int_{-infty}^{infty} e^{-y^2/2} {rm d}mu(y) \
& = sqrt{2pi} f(xi).
end{split}
end{equation*}
It's pretty similar for $d>1$.
answered yesterday
rldias
2,9301522
For $d=1$, we have $f(x)=e^{-x^2/2}$ and therefore (remember that $(ixi+x)^2=-xi^2+2ixi x+x^2$)
begin{equation*}
begin{split}
hat{f}(xi) & = int_{-infty}^{infty} e^{-1/2(2ixi x + x^2)} {rm d}mu(x) \
& = int_{-infty}^{infty} e^{-1/2((ixi+x)^2+xi^2)} {rm d}mu(x) \
& = f(xi) int_{-infty}^{infty} e^{-(ixi+x)^2/2} {rm d}mu(x) \
& =f(xi) int_{-infty}^{infty} e^{-y^2/2} {rm d}mu(y) \
& = sqrt{2pi} f(xi).
end{split}
end{equation*}
It's pretty similar for $d>1$.
answered yesterday
rldias
2,9301522
answered yesterday
rldias
2,9301522
answered yesterday
rldias
2,9301522
answered yesterday
rldias
2,9301522
2,9301522
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
Change variable and the contribution from the ends of the contour (which gets shifted up or down) is negligible.
– Richard Martin
yesterday
How do I continue for $d=2$ and so on?
– Arjihad
12 hours ago
For any $dgeq 1$ you will get $$hat{f}(xi)=f(xi)int_{mathbb{R}^d} e^{-|y|^2/2} {rm d}mu(y)$$ which is just $hat{f}(xi)=(2pi)^{d/2} f(xi)$.
– rldias
4 hours ago
add a comment |
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
Change variable and the contribution from the ends of the contour (which gets shifted up or down) is negligible.
– Richard Martin
yesterday
How do I continue for $d=2$ and so on?
– Arjihad
12 hours ago
For any $dgeq 1$ you will get $$hat{f}(xi)=f(xi)int_{mathbb{R}^d} e^{-|y|^2/2} {rm d}mu(y)$$ which is just $hat{f}(xi)=(2pi)^{d/2} f(xi)$.
– rldias
4 hours ago
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
How do you know that $int_{-infty}^infty e^{- (x-i xi)^2/2}dx =int_{-infty}^infty e^{- x^2/2}dx$
– reuns
yesterday
Change variable and the contribution from the ends of the contour (which gets shifted up or down) is negligible.
– Richard Martin
yesterday
Change variable and the contribution from the ends of the contour (which gets shifted up or down) is negligible.
– Richard Martin
yesterday
How do I continue for $d=2$ and so on?
– Arjihad
12 hours ago
How do I continue for $d=2$ and so on?
– Arjihad
12 hours ago
For any $dgeq 1$ you will get $$hat{f}(xi)=f(xi)int_{mathbb{R}^d} e^{-|y|^2/2} {rm d}mu(y)$$ which is just $hat{f}(xi)=(2pi)^{d/2} f(xi)$.
– rldias
4 hours ago
For any $dgeq 1$ you will get $$hat{f}(xi)=f(xi)int_{mathbb{R}^d} e^{-|y|^2/2} {rm d}mu(y)$$ which is just $hat{f}(xi)=(2pi)^{d/2} f(xi)$.
– rldias
4 hours ago
add a comment |
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