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Fourier Transform--Gaussian -- from Wolfram MathWorld

The Fourier transform of a Gaussian function f(x)=e^(-ax^2) is given by F_x[e^(-ax^2)](k) = int_(-infty)^inftye^(-ax^2)e^(-2piikx)dx (1) = int_(-infty)^inftye^(-ax^2)[cos(2pikx)-isin(2pikx)]dx (2) = int_(-infty)^inftye^(-ax^2)cos(2pikx)dx-iint_(-infty)^inftye^(-ax^2)sin(2pikx)dx. (3) The second integrand is odd, so integration over a symmetrical range gives 0. The value of the first integral is given by Abramowitz and Stegun (1972, p. 302, equation 7.4.6), so ...



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Fourier Transform--Gaussian -- from Wolfram MathWorld

https://mathworld.wolfram.com/FourierTransformGaussian.html

The Fourier transform of a Gaussian function f(x)=e^(-ax^2) is given by F_x[e^(-ax^2)](k) = int_(-infty)^inftye^(-ax^2)e^(-2piikx)dx (1) = int_(-infty)^inftye^(-ax^2)[cos(2pikx)-isin(2pikx)]dx (2) = int_(-infty)^inftye^(-ax^2)cos(2pikx)dx-iint_(-infty)^inftye^(-ax^2)sin(2pikx)dx. (3) The second integrand is odd, so integration over a symmetrical range gives 0. The value of the first integral is given by Abramowitz and Stegun (1972, p. 302, equation 7.4.6), so ...



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https://mathworld.wolfram.com/FourierTransformGaussian.html

Fourier Transform--Gaussian -- from Wolfram MathWorld

The Fourier transform of a Gaussian function f(x)=e^(-ax^2) is given by F_x[e^(-ax^2)](k) = int_(-infty)^inftye^(-ax^2)e^(-2piikx)dx (1) = int_(-infty)^inftye^(-ax^2)[cos(2pikx)-isin(2pikx)]dx (2) = int_(-infty)^inftye^(-ax^2)cos(2pikx)dx-iint_(-infty)^inftye^(-ax^2)sin(2pikx)dx. (3) The second integrand is odd, so integration over a symmetrical range gives 0. The value of the first integral is given by Abramowitz and Stegun (1972, p. 302, equation 7.4.6), so ...

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      Fourier Transform--Gaussian -- from Wolfram MathWorld
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      Fourier Transform--Gaussian
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      The Fourier transform of a Gaussian function f(x)=e^(-ax^2) is given by F_x[e^(-ax^2)](k) = int_(-infty)^inftye^(-ax^2)e^(-2piikx)dx (1) = int_(-infty)^inftye^(-ax^2)[cos(2pikx)-isin(2pikx)]dx (2) = int_(-infty)^inftye^(-ax^2)cos(2pikx)dx-iint_(-infty)^inftye^(-ax^2)sin(2pikx)dx. (3) The second integrand is odd, so integration over a symmetrical range gives 0. The value of the first integral is given by Abramowitz and Stegun (1972, p. 302, equation 7.4.6), so ...
    • description
      The Fourier transform of a Gaussian function f(x)=e^(-ax^2) is given by F_x[e^(-ax^2)](k) = int_(-infty)^inftye^(-ax^2)e^(-2piikx)dx (1) = int_(-infty)^inftye^(-ax^2)[cos(2pikx)-isin(2pikx)]dx (2) = int_(-infty)^inftye^(-ax^2)cos(2pikx)dx-iint_(-infty)^inftye^(-ax^2)sin(2pikx)dx. (3) The second integrand is odd, so integration over a symmetrical range gives 0. The value of the first integral is given by Abramowitz and Stegun (1972, p. 302, equation 7.4.6), so ...

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      Fourier Transform--Gaussian -- from Wolfram MathWorld
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      The Fourier transform of a Gaussian function f(x)=e^(-ax^2) is given by F_x[e^(-ax^2)](k) = int_(-infty)^inftye^(-ax^2)e^(-2piikx)dx (1) = int_(-infty)^inftye^(-ax^2)[cos(2pikx)-isin(2pikx)]dx (2) = int_(-infty)^inftye^(-ax^2)cos(2pikx)dx-iint_(-infty)^inftye^(-ax^2)sin(2pikx)dx. (3) The second integrand is odd, so integration over a symmetrical range gives 0. The value of the first integral is given by Abramowitz and Stegun (1972, p. 302, equation 7.4.6), so ...

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      Fourier Transform--Gaussian -- from Wolfram MathWorld
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      The Fourier transform of a Gaussian function f(x)=e^(-ax^2) is given by F_x[e^(-ax^2)](k) = int_(-infty)^inftye^(-ax^2)e^(-2piikx)dx (1) = int_(-infty)^inftye^(-ax^2)[cos(2pikx)-isin(2pikx)]dx (2) = int_(-infty)^inftye^(-ax^2)cos(2pikx)dx-iint_(-infty)^inftye^(-ax^2)sin(2pikx)dx. (3) The second integrand is odd, so integration over a symmetrical range gives 0. The value of the first integral is given by Abramowitz and Stegun (1972, p. 302, equation 7.4.6), so ...
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