|
|
(9 intermediate revisions by 2 users not shown) |
Line 1: |
Line 1: |
− | Instead of doing the substitution it should also be possible to write:
| + | 4 |
− | | + | |
− | $$ \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) g(y) e^{-i<k,x>} dy \: dx $$
| + | |
− | | + | |
− | as
| + | |
− | | + | |
− | $$ \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) g(y) e^{-i<k,x-y+y>} dy \: dx $$
| + | |
− | | + | |
− | $$ = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n } \int_{\mathbb{R}^n } f(x-y) e^{-i<k,x-y>} g(y) e^{-i<k,y>} dy \: dx $$
| + | |
− | | + | |
− | and then use Fubini
| + | |
− | | + | |
− | [[User:Un vieil homme|Un vieil homme]] ([[User talk:Un vieil homme|talk]]) 20:43, 31 December 2014 (CET)
| + | |
− | | + | |
− | In 1c)the derivation got exchanged with the integral without proofing that this is allowed:
| + | |
− | | + | |
− | $$ \frac{d}{dk} \widehat{F_c}(k) = \frac{d}{dk} (\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x)e^{-ikx} dx) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F_c(x) \frac{d}{dk} e^{-ikx} dx = \frac{-i}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-cx^{2}}xe^{-ikx} dx $$
| + | |
− | | + | |
− | [[User:Alex|Alex]] ([[User talk:Alex|talk]]) 14:55, 4 January 2015 (CET)
| + | |