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--- magic_gaussian_integral [2014/03/21 11:13]
nikolaj
+++ magic_gaussian_integral [2015/04/15 13:43]
nikolaj
@@ Line 12: / Line 12: @@
 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.
+(Side 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 $\frac{1}{2}\pi^\frac{1}{2}{\mathrm{erf}}(d) = \sum_{k=0}^\infty\frac{1}{k!}(-1)^k\frac{d^{2k+1}}{2k+1}$ 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 $\mathbb R$.)
+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
 ^ $\int_{-\infty}^\infty\mathrm e^{-\tfrac{1}{2}a\,\phi^2+i\,\phi\,j}\mathrm d\phi =I_a\cdot\mathrm e^{-\tfrac{1}{2}j^2\,a^{-1}}$ ^