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












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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







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










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










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    @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






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edited Dec 8 '18 at 12:56







user9287212

















asked Dec 8 '18 at 9:21









user9287212user9287212

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  • $begingroup$
    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












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






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    $begingroup$

    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









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      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.

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