Fourier transform of $e^{-ax^2}$ in $mathbb{R}^2$












1












$begingroup$


I'm trying to solve the following Fourier transform, or at least I'm trying to simplify it a bit



$$ mathcal{F}(e^{-ax^2})(xi,eta) = int_{mathbb{R^2}} e^{-i(xxi+yeta)}e^{-ax^2}dxdy $$



for $a > 0$.
One of the problems I'm finding is that the integrand function is not summable in $mathbb{R}^2$, hence integrating is not that immediate. Thank you for any help.






integration fourier-transform







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try polar coordinates.
    $endgroup$
    – lightxbulb
    Jan 23 at 11:42
















1












$begingroup$


I'm trying to solve the following Fourier transform, or at least I'm trying to simplify it a bit



$$ mathcal{F}(e^{-ax^2})(xi,eta) = int_{mathbb{R^2}} e^{-i(xxi+yeta)}e^{-ax^2}dxdy $$



for $a > 0$.
One of the problems I'm finding is that the integrand function is not summable in $mathbb{R}^2$, hence integrating is not that immediate. Thank you for any help.






integration fourier-transform







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try polar coordinates.
    $endgroup$
    – lightxbulb
    Jan 23 at 11:42














1












1








1





$begingroup$


I'm trying to solve the following Fourier transform, or at least I'm trying to simplify it a bit



$$ mathcal{F}(e^{-ax^2})(xi,eta) = int_{mathbb{R^2}} e^{-i(xxi+yeta)}e^{-ax^2}dxdy $$



for $a > 0$.
One of the problems I'm finding is that the integrand function is not summable in $mathbb{R}^2$, hence integrating is not that immediate. Thank you for any help.






integration fourier-transform







share|cite|improve this question









$endgroup$




I'm trying to solve the following Fourier transform, or at least I'm trying to simplify it a bit



$$ mathcal{F}(e^{-ax^2})(xi,eta) = int_{mathbb{R^2}} e^{-i(xxi+yeta)}e^{-ax^2}dxdy $$



for $a > 0$.
One of the problems I'm finding is that the integrand function is not summable in $mathbb{R}^2$, hence integrating is not that immediate. Thank you for any help.






integration fourier-transform




integration fourier-transform






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 23 at 11:13









AlakaeiAlakaei

83




83












  • $begingroup$
    Try polar coordinates.
    $endgroup$
    – lightxbulb
    Jan 23 at 11:42


















  • $begingroup$
    Try polar coordinates.
    $endgroup$
    – lightxbulb
    Jan 23 at 11:42
















$begingroup$
Try polar coordinates.
$endgroup$
– lightxbulb
Jan 23 at 11:42




$begingroup$
Try polar coordinates.
$endgroup$
– lightxbulb
Jan 23 at 11:42










1 Answer
1






active

oldest

votes


















1












$begingroup$

As you pointed out, the integral you wrote is not well defined. You have to think of $f(x,y) = e^{-ax^2}$ as a tempered distribution. Let $f_1(x) = e^{-ax^2}$ and $f_2(y) = 1$. The function $f$ can be written as $f_1 otimes f_2$. Therefore we have
$$mathcal F(f)(xi, eta) = mathcal F(f_1 otimes f_2)(xi, eta) = mathcal F(f_1)otimesmathcal F(f_2)(xi, eta).$$



The Fourier transform of $f_1(x) = e^{-ax^2}$ is $sqrt{frac{pi}{a}}e^{-frac{xi^2}{4a}}$ and the Fourier transform of $f_2(y) = 1$ is $2pi delta(eta)$.



So, we obtain the formula
$$mathcal F(f) =2pi sqrt{frac{pi}{a}}e^{-frac{xi^2}{4a}}delta(eta).$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Awesome thank you. Just a thing: I believe it's $check{delta}(eta) = delta( - eta)$. Is this correct?
    $endgroup$
    – Alakaei
    Jan 23 at 14:31












  • $begingroup$
    You are welcome. I'm not sure what do you mean by $check delta$. I just used that the Fouruer transform of $1$ is $2pi delta$, where $delta$ is the Dirac delta.
    $endgroup$
    – Hugocito
    Jan 23 at 19:59






  • 1




    $begingroup$
    Okay I said something correct, but trivial, and you are correct. I got confused because by the inversion formula is easy provable that if $u$ is a tempered distribution, then $mathcal{F}mathcal{F}(u) = 2pi check{u}$, where if $f$ is a function then $check{f}(x) = f(-x)$, if $u$ a distribution $check{u}(f) = u(check{f})$. But in Dirac's delta case $check{delta} = delta$. Anyway thanks again!
    $endgroup$
    – Alakaei
    Jan 24 at 11:17













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084337%2ffourier-transform-of-e-ax2-in-mathbbr2%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

As you pointed out, the integral you wrote is not well defined. You have to think of $f(x,y) = e^{-ax^2}$ as a tempered distribution. Let $f_1(x) = e^{-ax^2}$ and $f_2(y) = 1$. The function $f$ can be written as $f_1 otimes f_2$. Therefore we have
$$mathcal F(f)(xi, eta) = mathcal F(f_1 otimes f_2)(xi, eta) = mathcal F(f_1)otimesmathcal F(f_2)(xi, eta).$$



The Fourier transform of $f_1(x) = e^{-ax^2}$ is $sqrt{frac{pi}{a}}e^{-frac{xi^2}{4a}}$ and the Fourier transform of $f_2(y) = 1$ is $2pi delta(eta)$.



So, we obtain the formula
$$mathcal F(f) =2pi sqrt{frac{pi}{a}}e^{-frac{xi^2}{4a}}delta(eta).$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Awesome thank you. Just a thing: I believe it's $check{delta}(eta) = delta( - eta)$. Is this correct?
    $endgroup$
    – Alakaei
    Jan 23 at 14:31












  • $begingroup$
    You are welcome. I'm not sure what do you mean by $check delta$. I just used that the Fouruer transform of $1$ is $2pi delta$, where $delta$ is the Dirac delta.
    $endgroup$
    – Hugocito
    Jan 23 at 19:59






  • 1




    $begingroup$
    Okay I said something correct, but trivial, and you are correct. I got confused because by the inversion formula is easy provable that if $u$ is a tempered distribution, then $mathcal{F}mathcal{F}(u) = 2pi check{u}$, where if $f$ is a function then $check{f}(x) = f(-x)$, if $u$ a distribution $check{u}(f) = u(check{f})$. But in Dirac's delta case $check{delta} = delta$. Anyway thanks again!
    $endgroup$
    – Alakaei
    Jan 24 at 11:17


















1












$begingroup$

As you pointed out, the integral you wrote is not well defined. You have to think of $f(x,y) = e^{-ax^2}$ as a tempered distribution. Let $f_1(x) = e^{-ax^2}$ and $f_2(y) = 1$. The function $f$ can be written as $f_1 otimes f_2$. Therefore we have
$$mathcal F(f)(xi, eta) = mathcal F(f_1 otimes f_2)(xi, eta) = mathcal F(f_1)otimesmathcal F(f_2)(xi, eta).$$



The Fourier transform of $f_1(x) = e^{-ax^2}$ is $sqrt{frac{pi}{a}}e^{-frac{xi^2}{4a}}$ and the Fourier transform of $f_2(y) = 1$ is $2pi delta(eta)$.



So, we obtain the formula
$$mathcal F(f) =2pi sqrt{frac{pi}{a}}e^{-frac{xi^2}{4a}}delta(eta).$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Awesome thank you. Just a thing: I believe it's $check{delta}(eta) = delta( - eta)$. Is this correct?
    $endgroup$
    – Alakaei
    Jan 23 at 14:31












  • $begingroup$
    You are welcome. I'm not sure what do you mean by $check delta$. I just used that the Fouruer transform of $1$ is $2pi delta$, where $delta$ is the Dirac delta.
    $endgroup$
    – Hugocito
    Jan 23 at 19:59






  • 1




    $begingroup$
    Okay I said something correct, but trivial, and you are correct. I got confused because by the inversion formula is easy provable that if $u$ is a tempered distribution, then $mathcal{F}mathcal{F}(u) = 2pi check{u}$, where if $f$ is a function then $check{f}(x) = f(-x)$, if $u$ a distribution $check{u}(f) = u(check{f})$. But in Dirac's delta case $check{delta} = delta$. Anyway thanks again!
    $endgroup$
    – Alakaei
    Jan 24 at 11:17
















1












1








1





$begingroup$

As you pointed out, the integral you wrote is not well defined. You have to think of $f(x,y) = e^{-ax^2}$ as a tempered distribution. Let $f_1(x) = e^{-ax^2}$ and $f_2(y) = 1$. The function $f$ can be written as $f_1 otimes f_2$. Therefore we have
$$mathcal F(f)(xi, eta) = mathcal F(f_1 otimes f_2)(xi, eta) = mathcal F(f_1)otimesmathcal F(f_2)(xi, eta).$$



The Fourier transform of $f_1(x) = e^{-ax^2}$ is $sqrt{frac{pi}{a}}e^{-frac{xi^2}{4a}}$ and the Fourier transform of $f_2(y) = 1$ is $2pi delta(eta)$.



So, we obtain the formula
$$mathcal F(f) =2pi sqrt{frac{pi}{a}}e^{-frac{xi^2}{4a}}delta(eta).$$






share|cite|improve this answer









$endgroup$



As you pointed out, the integral you wrote is not well defined. You have to think of $f(x,y) = e^{-ax^2}$ as a tempered distribution. Let $f_1(x) = e^{-ax^2}$ and $f_2(y) = 1$. The function $f$ can be written as $f_1 otimes f_2$. Therefore we have
$$mathcal F(f)(xi, eta) = mathcal F(f_1 otimes f_2)(xi, eta) = mathcal F(f_1)otimesmathcal F(f_2)(xi, eta).$$



The Fourier transform of $f_1(x) = e^{-ax^2}$ is $sqrt{frac{pi}{a}}e^{-frac{xi^2}{4a}}$ and the Fourier transform of $f_2(y) = 1$ is $2pi delta(eta)$.



So, we obtain the formula
$$mathcal F(f) =2pi sqrt{frac{pi}{a}}e^{-frac{xi^2}{4a}}delta(eta).$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 23 at 11:46









HugocitoHugocito

1,84411320




1,84411320








  • 1




    $begingroup$
    Awesome thank you. Just a thing: I believe it's $check{delta}(eta) = delta( - eta)$. Is this correct?
    $endgroup$
    – Alakaei
    Jan 23 at 14:31












  • $begingroup$
    You are welcome. I'm not sure what do you mean by $check delta$. I just used that the Fouruer transform of $1$ is $2pi delta$, where $delta$ is the Dirac delta.
    $endgroup$
    – Hugocito
    Jan 23 at 19:59






  • 1




    $begingroup$
    Okay I said something correct, but trivial, and you are correct. I got confused because by the inversion formula is easy provable that if $u$ is a tempered distribution, then $mathcal{F}mathcal{F}(u) = 2pi check{u}$, where if $f$ is a function then $check{f}(x) = f(-x)$, if $u$ a distribution $check{u}(f) = u(check{f})$. But in Dirac's delta case $check{delta} = delta$. Anyway thanks again!
    $endgroup$
    – Alakaei
    Jan 24 at 11:17
















  • 1




    $begingroup$
    Awesome thank you. Just a thing: I believe it's $check{delta}(eta) = delta( - eta)$. Is this correct?
    $endgroup$
    – Alakaei
    Jan 23 at 14:31












  • $begingroup$
    You are welcome. I'm not sure what do you mean by $check delta$. I just used that the Fouruer transform of $1$ is $2pi delta$, where $delta$ is the Dirac delta.
    $endgroup$
    – Hugocito
    Jan 23 at 19:59






  • 1




    $begingroup$
    Okay I said something correct, but trivial, and you are correct. I got confused because by the inversion formula is easy provable that if $u$ is a tempered distribution, then $mathcal{F}mathcal{F}(u) = 2pi check{u}$, where if $f$ is a function then $check{f}(x) = f(-x)$, if $u$ a distribution $check{u}(f) = u(check{f})$. But in Dirac's delta case $check{delta} = delta$. Anyway thanks again!
    $endgroup$
    – Alakaei
    Jan 24 at 11:17










1




1




$begingroup$
Awesome thank you. Just a thing: I believe it's $check{delta}(eta) = delta( - eta)$. Is this correct?
$endgroup$
– Alakaei
Jan 23 at 14:31






$begingroup$
Awesome thank you. Just a thing: I believe it's $check{delta}(eta) = delta( - eta)$. Is this correct?
$endgroup$
– Alakaei
Jan 23 at 14:31














$begingroup$
You are welcome. I'm not sure what do you mean by $check delta$. I just used that the Fouruer transform of $1$ is $2pi delta$, where $delta$ is the Dirac delta.
$endgroup$
– Hugocito
Jan 23 at 19:59




$begingroup$
You are welcome. I'm not sure what do you mean by $check delta$. I just used that the Fouruer transform of $1$ is $2pi delta$, where $delta$ is the Dirac delta.
$endgroup$
– Hugocito
Jan 23 at 19:59




1




1




$begingroup$
Okay I said something correct, but trivial, and you are correct. I got confused because by the inversion formula is easy provable that if $u$ is a tempered distribution, then $mathcal{F}mathcal{F}(u) = 2pi check{u}$, where if $f$ is a function then $check{f}(x) = f(-x)$, if $u$ a distribution $check{u}(f) = u(check{f})$. But in Dirac's delta case $check{delta} = delta$. Anyway thanks again!
$endgroup$
– Alakaei
Jan 24 at 11:17






$begingroup$
Okay I said something correct, but trivial, and you are correct. I got confused because by the inversion formula is easy provable that if $u$ is a tempered distribution, then $mathcal{F}mathcal{F}(u) = 2pi check{u}$, where if $f$ is a function then $check{f}(x) = f(-x)$, if $u$ a distribution $check{u}(f) = u(check{f})$. But in Dirac's delta case $check{delta} = delta$. Anyway thanks again!
$endgroup$
– Alakaei
Jan 24 at 11:17




















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084337%2ffourier-transform-of-e-ax2-in-mathbbr2%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
Read more

Partial Derivative Guidance.

Image
0 $begingroup$ I was given this tricky problem in which I'm confuse on how to solve it and I would like some help: Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$ . Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives. My attempt (which is wrong): Using chain rule: $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$ $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$ So, $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$ $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$ Thanks, calculus multivariable-calculus share | cite | improve this question edited Jan 19 at 19:38...
Read more

Understanding the size os this class of aleatory events

Image
0 1 $begingroup$ Hi I am having problems to understand this: I have: $Omega = [0,1]$ , $Lambda =$ all subsets with defined length Let $A_{0}$ = {A $subset$ [0,1]: A is finite union of intervals} 1º: My doubt is that $A_{0}$ is a algebra but its not a sigma algebra. Why? Also, A = $(0,1/2) cup (1/2,3/4) cup (3/4,7/8) cup ... cup(1-1/2^{n},1-1/2^{n+1})...$ 2º: So the class of aleatory events would be bigger than $A_{0}$ . Why? How the set A is greater than $A_{0}$ ? measure-theory share | cite | improve this question asked Jan 8 at 19:17 Diogo Bastos Diogo Bastos 116 6 ...
Read more