Fourier transformation of $e^{-ax^2}$
Find the Fourier transformation of $e^{-ax^2}$, for any constant $a>0$. In general, $$int frac{1}{sqrt{(2pi sigma^2)}}e^{-frac{(x-mu)^2}{2sigma^2}} e^{-iλx }dx = e^{-imuλ-frac{1}{2}sigma^2t^2 }$$ holds.
Apparently, the result can be used to solve the heat equation. I have tried to do it as below $$ frac {1}{2pi} int_{-infty}^{infty} e^{-ax^2}e^{iλx} dx $$ After some calculations, I got $frac {1}{sqrt{4pi a}}e^{-frac{λ^2}{4a}} $. But I looked up on the internet and there are some notes show that it should be $frac {1}{sqrt{2 a}}e^{-frac{λ^2}{4a}} $.
Edit: I found that the notes I saw on the internet use $ frac {1}{sqrt{2pi}} int_{-infty}^{infty} e^{-ax^2}e^{iλx} dx $ instead of $ frac {1}{2pi} $. Maybe that's the reason. But why the notes provided by my lecturer used $ frac {1}{2pi} $? Are there no standard form?
fourier-transform heat-equation
asked Dec 8 at 9:21
user9287212
11
add a comment |
Find the Fourier transformation of $e^{-ax^2}$, for any constant $a>0$. In general, $$int frac{1}{sqrt{(2pi sigma^2)}}e^{-frac{(x-mu)^2}{2sigma^2}} e^{-iλx }dx = e^{-imuλ-frac{1}{2}sigma^2t^2 }$$ holds.
Apparently, the result can be used to solve the heat equation. I have tried to do it as below $$ frac {1}{2pi} int_{-infty}^{infty} e^{-ax^2}e^{iλx} dx $$ After some calculations, I got $frac {1}{sqrt{4pi a}}e^{-frac{λ^2}{4a}} $. But I looked up on the internet and there are some notes show that it should be $frac {1}{sqrt{2 a}}e^{-frac{λ^2}{4a}} $.
Edit: I found that the notes I saw on the internet use $ frac {1}{sqrt{2pi}} int_{-infty}^{infty} e^{-ax^2}e^{iλx} dx $ instead of $ frac {1}{2pi} $. Maybe that's the reason. But why the notes provided by my lecturer used $ frac {1}{2pi} $? Are there no standard form?
fourier-transform heat-equation
asked Dec 8 at 9:21
user9287212
11
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 8 at 9:25
Your answer is not correct. If you show us your work we can show where the mistake is.
– Kavi Rama Murthy
Dec 8 at 12:26
@Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
– user9287212
Dec 8 at 12:58
add a comment |
Find the Fourier transformation of $e^{-ax^2}$, for any constant $a>0$. In general, $$int frac{1}{sqrt{(2pi sigma^2)}}e^{-frac{(x-mu)^2}{2sigma^2}} e^{-iλx }dx = e^{-imuλ-frac{1}{2}sigma^2t^2 }$$ holds.
Apparently, the result can be used to solve the heat equation. I have tried to do it as below $$ frac {1}{2pi} int_{-infty}^{infty} e^{-ax^2}e^{iλx} dx $$ After some calculations, I got $frac {1}{sqrt{4pi a}}e^{-frac{λ^2}{4a}} $. But I looked up on the internet and there are some notes show that it should be $frac {1}{sqrt{2 a}}e^{-frac{λ^2}{4a}} $.
Edit: I found that the notes I saw on the internet use $ frac {1}{sqrt{2pi}} int_{-infty}^{infty} e^{-ax^2}e^{iλx} dx $ instead of $ frac {1}{2pi} $. Maybe that's the reason. But why the notes provided by my lecturer used $ frac {1}{2pi} $? Are there no standard form?
fourier-transform heat-equation
asked Dec 8 at 9:21
user9287212
11
Find the Fourier transformation of $e^{-ax^2}$, for any constant $a>0$. In general, $$int frac{1}{sqrt{(2pi sigma^2)}}e^{-frac{(x-mu)^2}{2sigma^2}} e^{-iλx }dx = e^{-imuλ-frac{1}{2}sigma^2t^2 }$$ holds.
Apparently, the result can be used to solve the heat equation. I have tried to do it as below $$ frac {1}{2pi} int_{-infty}^{infty} e^{-ax^2}e^{iλx} dx $$ After some calculations, I got $frac {1}{sqrt{4pi a}}e^{-frac{λ^2}{4a}} $. But I looked up on the internet and there are some notes show that it should be $frac {1}{sqrt{2 a}}e^{-frac{λ^2}{4a}} $.
Edit: I found that the notes I saw on the internet use $ frac {1}{sqrt{2pi}} int_{-infty}^{infty} e^{-ax^2}e^{iλx} dx $ instead of $ frac {1}{2pi} $. Maybe that's the reason. But why the notes provided by my lecturer used $ frac {1}{2pi} $? Are there no standard form?
fourier-transform heat-equation
fourier-transform heat-equation
asked Dec 8 at 9:21
user9287212
11
asked Dec 8 at 9:21
user9287212
11
edited Dec 8 at 12:56
asked Dec 8 at 9:21
user9287212
11
asked Dec 8 at 9:21
user9287212
11
asked Dec 8 at 9:21
user9287212
11
11
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 8 at 9:25
Your answer is not correct. If you show us your work we can show where the mistake is.
– Kavi Rama Murthy
Dec 8 at 12:26
@Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
– user9287212
Dec 8 at 12:58
add a comment |
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 8 at 9:25
Your answer is not correct. If you show us your work we can show where the mistake is.
– Kavi Rama Murthy
Dec 8 at 12:26
@Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
– user9287212
Dec 8 at 12:58
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 8 at 9:25
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 8 at 9:25
Your answer is not correct. If you show us your work we can show where the mistake is.
– Kavi Rama Murthy
Dec 8 at 12:26
Your answer is not correct. If you show us your work we can show where the mistake is.
– Kavi Rama Murthy
Dec 8 at 12:26
@Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
– user9287212
Dec 8 at 12:58
@Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
– user9287212
Dec 8 at 12:58
add a comment |
1 Answer
1
active
oldest
votes
By linear transformations it's sufficient to check the case $mu=0,,sigma=1$, so we want to prove $$phi(t):=int_{Bbb R}frac{1}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)dx=exp-frac{t^2}{2}.$$By inspection $phi(0)=1$, so it suffices to prove the following is $0$: $$frac{dphi}{dt}+tphi(t)=int_{Bbb R}frac{1}{sqrt{2pi}}(ix+t)expbigg(-frac{x^2}{2}+itxbigg)dx=Bigg[-frac{i}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)Bigg]_{-infty}^infty.$$
answered Dec 8 at 9:28
J.G.
22.3k22034
add a comment |
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
});
}
});
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%2f3030868%2ffourier-transformation-of-e-ax2%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
By linear transformations it's sufficient to check the case $mu=0,,sigma=1$, so we want to prove $$phi(t):=int_{Bbb R}frac{1}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)dx=exp-frac{t^2}{2}.$$By inspection $phi(0)=1$, so it suffices to prove the following is $0$: $$frac{dphi}{dt}+tphi(t)=int_{Bbb R}frac{1}{sqrt{2pi}}(ix+t)expbigg(-frac{x^2}{2}+itxbigg)dx=Bigg[-frac{i}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)Bigg]_{-infty}^infty.$$
answered Dec 8 at 9:28
J.G.
22.3k22034
add a comment |
By linear transformations it's sufficient to check the case $mu=0,,sigma=1$, so we want to prove $$phi(t):=int_{Bbb R}frac{1}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)dx=exp-frac{t^2}{2}.$$By inspection $phi(0)=1$, so it suffices to prove the following is $0$: $$frac{dphi}{dt}+tphi(t)=int_{Bbb R}frac{1}{sqrt{2pi}}(ix+t)expbigg(-frac{x^2}{2}+itxbigg)dx=Bigg[-frac{i}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)Bigg]_{-infty}^infty.$$
answered Dec 8 at 9:28
J.G.
22.3k22034
add a comment |
By linear transformations it's sufficient to check the case $mu=0,,sigma=1$, so we want to prove $$phi(t):=int_{Bbb R}frac{1}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)dx=exp-frac{t^2}{2}.$$By inspection $phi(0)=1$, so it suffices to prove the following is $0$: $$frac{dphi}{dt}+tphi(t)=int_{Bbb R}frac{1}{sqrt{2pi}}(ix+t)expbigg(-frac{x^2}{2}+itxbigg)dx=Bigg[-frac{i}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)Bigg]_{-infty}^infty.$$
answered Dec 8 at 9:28
J.G.
22.3k22034
By linear transformations it's sufficient to check the case $mu=0,,sigma=1$, so we want to prove $$phi(t):=int_{Bbb R}frac{1}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)dx=exp-frac{t^2}{2}.$$By inspection $phi(0)=1$, so it suffices to prove the following is $0$: $$frac{dphi}{dt}+tphi(t)=int_{Bbb R}frac{1}{sqrt{2pi}}(ix+t)expbigg(-frac{x^2}{2}+itxbigg)dx=Bigg[-frac{i}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)Bigg]_{-infty}^infty.$$
answered Dec 8 at 9:28
J.G.
22.3k22034
answered Dec 8 at 9:28
J.G.
22.3k22034
answered Dec 8 at 9:28
J.G.
22.3k22034
answered Dec 8 at 9:28
J.G.
22.3k22034
22.3k22034
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%2f3030868%2ffourier-transformation-of-e-ax2%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
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Dec 8 at 9:25
Your answer is not correct. If you show us your work we can show where the mistake is.
– Kavi Rama Murthy
Dec 8 at 12:26
@Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
– user9287212
Dec 8 at 12:58