LetThe integral is absolutely convergent. For a Schwartz function , Fubini's theorem and integration by parts in giveEquivalently, as a distributional derivative, and hence
Because is integrable on , the Dominated convergence theorem applied to the defining integral proves that is continuous on .
For , the change of variables formula givesand for the analogous formula is obtained with . On either open half-line the lower endpoint stays away from zero locally, so repeated differentiation under the integral sign proves smoothness. Thus
For , the preceding change of variables and one integration by parts giveSplit the last improper integral at one and add and subtract one on . Since ,
On either open half-line, part (i) says . Differentiating the formula in part (iii) makes all elementary terms cancel except the logarithm and the bracket. ThereforewhereandwhereBoth functions are smooth on their respective half-lines, proving the required assertion.
For , subtracting the two formulas from part (iv) yieldsUsing the stated Dirichlet integral and taking the one-sided limit gives
Remove the two reciprocal tails by settingThe assumed remainder at each end and local integrability on bounded intervals imply . Its Fourier transform is therefore continuous at zero. Since ,Parts (iv) and (v) show that both requested one-sided limits exist. If denotes the limit from positive frequencies and the limit from negative frequencies, then the common value cancels andThis is the universal jump caused by reciprocal tails.
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