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
9,28521232
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
9,28521232
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
9,28521232
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
9,28521232
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
9,28521232
answered Nov 18 at 7:04
David G. Stork
9,28521232
answered Nov 18 at 7:04
David G. Stork
9,28521232
answered Nov 18 at 7:04
David G. Stork
9,28521232
9,28521232
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003215%2fthe-fourier-transform-of-frac-texterf-omega-xx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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