|
2 | added 149 characters in body | ||
|
How about $$\frac{e^{-e^x}}{1+\epsilon x^2}$$ If you compute the Fourier transform you can shift the contour to height $\pm\pi/2$ to get an $e^{-|t|}$ times something decaying to 1, by Riemann-Lebesgue lemma Edit. Or you can look at a shift of your original example: $f(x+\log A)$ to get something on the lines of $$e^{-A e^x} e^{-B x^2} e^{C x}$$ |
||||
|
|
1 |
|
||
|
How about $$\frac{e^{-e^x}}{1+\epsilon x^2}$$ If you compute the Fourier transform you can shift the contour to height $\pm\pi/2$ to get an $e^{-|t|}$ times something decaying to 1, by Riemann-Lebesgue lemma |
||||
