The Fourier transform of $frac{text{erf}(omega x)}{x}$
up vote
-1
down vote
favorite
Does anyone know the Fourier transform of
$Largefrac{text{erf}(omega x)}{x}$?
I think it should be something like $frac{4pi}{k^2}exp{(-k^2/4omega^2)}$.
Is this right? How can one go about deriving this? Any hints are much appreciated.
Thank you in advance!
fourier-transform error-function
edited Nov 18 at 8:55
Fakemistake
1,635815
asked Nov 18 at 6:48
Yang
42
add a comment |
up vote
-1
down vote
favorite
Does anyone know the Fourier transform of
$Largefrac{text{erf}(omega x)}{x}$?
I think it should be something like $frac{4pi}{k^2}exp{(-k^2/4omega^2)}$.
Is this right? How can one go about deriving this? Any hints are much appreciated.
Thank you in advance!
fourier-transform error-function
edited Nov 18 at 8:55
Fakemistake
1,635815
asked Nov 18 at 6:48
Yang
42
You mean the Fourier transform of the distribution $pv.(frac{erf(omega x)}{x})= lim_{epsilon to 0} frac{erf(omega x)}{x} 1_{|x| > epsilon}$. The method is the same as for $pv.(frac1x)$
– reuns
Nov 18 at 6:56
If $x$ is your primal variable, what is the transform variable? It surely cannot be $omega$ because you have used that in the original function.
– David G. Stork
Nov 18 at 7:03
The transform variable is just $k$, isn't it? The transform would be $int_0^infty dxfrac{text{erf}(omega x)}{x}text{e}^{-ikx}$, right?
– Yang
Nov 18 at 7:28
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Does anyone know the Fourier transform of
$Largefrac{text{erf}(omega x)}{x}$?
I think it should be something like $frac{4pi}{k^2}exp{(-k^2/4omega^2)}$.
Is this right? How can one go about deriving this? Any hints are much appreciated.
Thank you in advance!
fourier-transform error-function
edited Nov 18 at 8:55
Fakemistake
1,635815
asked Nov 18 at 6:48
Yang
42
Does anyone know the Fourier transform of
$Largefrac{text{erf}(omega x)}{x}$?
I think it should be something like $frac{4pi}{k^2}exp{(-k^2/4omega^2)}$.
Is this right? How can one go about deriving this? Any hints are much appreciated.
Thank you in advance!
fourier-transform error-function
fourier-transform error-function
edited Nov 18 at 8:55
Fakemistake
1,635815
asked Nov 18 at 6:48
Yang
42
edited Nov 18 at 8:55
Fakemistake
1,635815
asked Nov 18 at 6:48
Yang
42
edited Nov 18 at 8:55
Fakemistake
1,635815
edited Nov 18 at 8:55
Fakemistake
1,635815
edited Nov 18 at 8:55
Fakemistake
1,635815
1,635815
asked Nov 18 at 6:48
Yang
42
asked Nov 18 at 6:48
Yang
42
asked Nov 18 at 6:48
Yang
42
42
You mean the Fourier transform of the distribution $pv.(frac{erf(omega x)}{x})= lim_{epsilon to 0} frac{erf(omega x)}{x} 1_{|x| > epsilon}$. The method is the same as for $pv.(frac1x)$
– reuns
Nov 18 at 6:56
If $x$ is your primal variable, what is the transform variable? It surely cannot be $omega$ because you have used that in the original function.
– David G. Stork
Nov 18 at 7:03
The transform variable is just $k$, isn't it? The transform would be $int_0^infty dxfrac{text{erf}(omega x)}{x}text{e}^{-ikx}$, right?
– Yang
Nov 18 at 7:28
add a comment |
You mean the Fourier transform of the distribution $pv.(frac{erf(omega x)}{x})= lim_{epsilon to 0} frac{erf(omega x)}{x} 1_{|x| > epsilon}$. The method is the same as for $pv.(frac1x)$
– reuns
Nov 18 at 6:56
If $x$ is your primal variable, what is the transform variable? It surely cannot be $omega$ because you have used that in the original function.
– David G. Stork
Nov 18 at 7:03
The transform variable is just $k$, isn't it? The transform would be $int_0^infty dxfrac{text{erf}(omega x)}{x}text{e}^{-ikx}$, right?
– Yang
Nov 18 at 7:28
You mean the Fourier transform of the distribution $pv.(frac{erf(omega x)}{x})= lim_{epsilon to 0} frac{erf(omega x)}{x} 1_{|x| > epsilon}$. The method is the same as for $pv.(frac1x)$
– reuns
Nov 18 at 6:56
You mean the Fourier transform of the distribution $pv.(frac{erf(omega x)}{x})= lim_{epsilon to 0} frac{erf(omega x)}{x} 1_{|x| > epsilon}$. The method is the same as for $pv.(frac1x)$
– reuns
Nov 18 at 6:56
If $x$ is your primal variable, what is the transform variable? It surely cannot be $omega$ because you have used that in the original function.
– David G. Stork
Nov 18 at 7:03
If $x$ is your primal variable, what is the transform variable? It surely cannot be $omega$ because you have used that in the original function.
– David G. Stork
Nov 18 at 7:03
The transform variable is just $k$, isn't it? The transform would be $int_0^infty dxfrac{text{erf}(omega x)}{x}text{e}^{-ikx}$, right?
– Yang
Nov 18 at 7:28
The transform variable is just $k$, isn't it? The transform would be $int_0^infty dxfrac{text{erf}(omega x)}{x}text{e}^{-ikx}$, right?
– Yang
Nov 18 at 7:28
add a comment |
1 Answer
1
active
oldest
votes
up vote
-2
down vote
I think you have an error in your question: $omega$ is typically used for the transform variable, and hence certainly shouldn't be in the untransformed function.
Anyway, the Fourier transform of ${rm{Erf}(x) over x}$ (of your title) is:
$$frac{Gamma left(0,frac{omega ^2}{4}right)}{sqrt{2 pi }}$$
answered Nov 18 at 7:04
David G. Stork
9,33421232
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-2
down vote
I think you have an error in your question: $omega$ is typically used for the transform variable, and hence certainly shouldn't be in the untransformed function.
Anyway, the Fourier transform of ${rm{Erf}(x) over x}$ (of your title) is:
$$frac{Gamma left(0,frac{omega ^2}{4}right)}{sqrt{2 pi }}$$
answered Nov 18 at 7:04
David G. Stork
9,33421232
add a comment |
up vote
-2
down vote
I think you have an error in your question: $omega$ is typically used for the transform variable, and hence certainly shouldn't be in the untransformed function.
Anyway, the Fourier transform of ${rm{Erf}(x) over x}$ (of your title) is:
$$frac{Gamma left(0,frac{omega ^2}{4}right)}{sqrt{2 pi }}$$
answered Nov 18 at 7:04
David G. Stork
9,33421232
add a comment |
up vote
-2
down vote
up vote
-2
down vote
I think you have an error in your question: $omega$ is typically used for the transform variable, and hence certainly shouldn't be in the untransformed function.
Anyway, the Fourier transform of ${rm{Erf}(x) over x}$ (of your title) is:
$$frac{Gamma left(0,frac{omega ^2}{4}right)}{sqrt{2 pi }}$$
answered Nov 18 at 7:04
David G. Stork
9,33421232
I think you have an error in your question: $omega$ is typically used for the transform variable, and hence certainly shouldn't be in the untransformed function.
Anyway, the Fourier transform of ${rm{Erf}(x) over x}$ (of your title) is:
$$frac{Gamma left(0,frac{omega ^2}{4}right)}{sqrt{2 pi }}$$
answered Nov 18 at 7:04
David G. Stork
9,33421232
answered Nov 18 at 7:04
David G. Stork
9,33421232
answered Nov 18 at 7:04
David G. Stork
9,33421232
answered Nov 18 at 7:04
David G. Stork
9,33421232
9,33421232
add a comment |
add a comment |
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You mean the Fourier transform of the distribution $pv.(frac{erf(omega x)}{x})= lim_{epsilon to 0} frac{erf(omega x)}{x} 1_{|x| > epsilon}$. The method is the same as for $pv.(frac1x)$
– reuns
Nov 18 at 6:56
If $x$ is your primal variable, what is the transform variable? It surely cannot be $omega$ because you have used that in the original function.
– David G. Stork
Nov 18 at 7:03
The transform variable is just $k$, isn't it? The transform would be $int_0^infty dxfrac{text{erf}(omega x)}{x}text{e}^{-ikx}$, right?
– Yang
Nov 18 at 7:28