Fourier transformation of $e^{-ax^2}$












-1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – José Carlos Santos
    Dec 8 '18 at 9:25










  • $begingroup$
    Your answer is not correct. If you show us your work we can show where the mistake is.
    $endgroup$
    – Kavi Rama Murthy
    Dec 8 '18 at 12:26










  • $begingroup$
    @Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
    $endgroup$
    – user9287212
    Dec 8 '18 at 12:58


















-1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – José Carlos Santos
    Dec 8 '18 at 9:25










  • $begingroup$
    Your answer is not correct. If you show us your work we can show where the mistake is.
    $endgroup$
    – Kavi Rama Murthy
    Dec 8 '18 at 12:26










  • $begingroup$
    @Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
    $endgroup$
    – user9287212
    Dec 8 '18 at 12:58
















-1












-1








-1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 12:56







user9287212

















asked Dec 8 '18 at 9:21









user9287212user9287212

11




11












  • $begingroup$
    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.
    $endgroup$
    – José Carlos Santos
    Dec 8 '18 at 9:25










  • $begingroup$
    Your answer is not correct. If you show us your work we can show where the mistake is.
    $endgroup$
    – Kavi Rama Murthy
    Dec 8 '18 at 12:26










  • $begingroup$
    @Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
    $endgroup$
    – user9287212
    Dec 8 '18 at 12:58




















  • $begingroup$
    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.
    $endgroup$
    – José Carlos Santos
    Dec 8 '18 at 9:25










  • $begingroup$
    Your answer is not correct. If you show us your work we can show where the mistake is.
    $endgroup$
    – Kavi Rama Murthy
    Dec 8 '18 at 12:26










  • $begingroup$
    @Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
    $endgroup$
    – user9287212
    Dec 8 '18 at 12:58


















$begingroup$
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.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 9:25




$begingroup$
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.
$endgroup$
– José Carlos Santos
Dec 8 '18 at 9:25












$begingroup$
Your answer is not correct. If you show us your work we can show where the mistake is.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 12:26




$begingroup$
Your answer is not correct. If you show us your work we can show where the mistake is.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 12:26












$begingroup$
@Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
$endgroup$
– user9287212
Dec 8 '18 at 12:58






$begingroup$
@Kavi Rama Murthy I made a typo, my calculations should start from $ frac{1}{2pi}$instead of $frac{1}{sqrt{2pi}}$
$endgroup$
– user9287212
Dec 8 '18 at 12:58












1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $0$: $$frac{dphi}{dt}+tphi(t)=int_{Bbb R}frac{1}{sqrt{2pi}}(ix+t)expbigg(-frac{x^2}{2}+itxbigg)dx=Bigg[-frac{i}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)Bigg]_{-infty}^infty.$$






share|cite|improve this answer









$endgroup$













    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%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









    0












    $begingroup$

    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 $0$: $$frac{dphi}{dt}+tphi(t)=int_{Bbb R}frac{1}{sqrt{2pi}}(ix+t)expbigg(-frac{x^2}{2}+itxbigg)dx=Bigg[-frac{i}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)Bigg]_{-infty}^infty.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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 $0$: $$frac{dphi}{dt}+tphi(t)=int_{Bbb R}frac{1}{sqrt{2pi}}(ix+t)expbigg(-frac{x^2}{2}+itxbigg)dx=Bigg[-frac{i}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)Bigg]_{-infty}^infty.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $0$: $$frac{dphi}{dt}+tphi(t)=int_{Bbb R}frac{1}{sqrt{2pi}}(ix+t)expbigg(-frac{x^2}{2}+itxbigg)dx=Bigg[-frac{i}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)Bigg]_{-infty}^infty.$$






        share|cite|improve this answer









        $endgroup$



        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 $0$: $$frac{dphi}{dt}+tphi(t)=int_{Bbb R}frac{1}{sqrt{2pi}}(ix+t)expbigg(-frac{x^2}{2}+itxbigg)dx=Bigg[-frac{i}{sqrt{2pi}}expbigg(-frac{x^2}{2}+itxbigg)Bigg]_{-infty}^infty.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 9:28









        J.G.J.G.

        26.3k22541




        26.3k22541






























            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%2f3030868%2ffourier-transformation-of-e-ax2%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more