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magic_gaussian_integral [2014/03/21 11:13] nikolaj |
magic_gaussian_integral [2015/04/15 13:42] nikolaj |
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| That whole integral is just a showoff generalization of the one dimensional integral | That whole integral is just a showoff generalization of the one dimensional integral | ||
| - | ^ $I_a:=\int_{-\infty}^\infty\mathrm e^{-\tfrac{1}{2}a\,\phi^2}\mathrm d\phi =(2\pi)^{1/2}a^{-1/2}$ ^ | + | ^ $I_a:=\int_{-\infty}^\infty{\mathrm e}^{-\tfrac{1}{2}a\,\phi^2}\mathrm d\phi =(2\pi)^{1/2}a^{-1/2}$ ^ |
| - | which is obtained with the basic trick of switching to polar coordinates. Adding a imaginary linear term in the exponent results in the statement that the Fourier transform of a Gaussian curve is again a Gaussian curve | + | which is obtained, for example, with the basic trick of switching to polar coordinates. |
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| + | Remark: Having found the normalization of ${\mathrm e}^{-\tfrac{1}{2}a\,\phi^2}$ lets us define the function ${\mathrm{erf}}$ with $\lim_{d\to\infty}{\mathrm{erf}}(d)=1$. Then we can show that $\sum_{k=0}^\infty\frac{1}{k!}(-1)^k\frac{d^{2k+1}}{2k+1}=\frac{1}{2}\pi^\frac{1}{2}{\mathrm{erf}}(d)$ and with $\int_0^d(-\phi^2)^k{\mathrm d}\phi = (-1)^k\frac{d^{2k+1}}{2k+1}$ we can evaluate the initial integral on $[-d,d]$, not just all of $\mathbb R$. | ||
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| + | Adding a imaginary linear term in the exponent results in the statement that the Fourier transform of a Gaussian curve is again a Gaussian curve | ||
| ^ $\int_{-\infty}^\infty\mathrm e^{-\tfrac{1}{2}a\,\phi^2+i\,\phi\,j}\mathrm d\phi =I_a\cdot\mathrm e^{-\tfrac{1}{2}j^2\,a^{-1}}$ ^ | ^ $\int_{-\infty}^\infty\mathrm e^{-\tfrac{1}{2}a\,\phi^2+i\,\phi\,j}\mathrm d\phi =I_a\cdot\mathrm e^{-\tfrac{1}{2}j^2\,a^{-1}}$ ^ | ||
