〈 Berger | Dillon 〉

This derivation needs LOTS of justification. It diverges absolutely and in L_1. I but exists as an indefinite integral. Do the \int_{-R}^S. A substitution u = ix means that u runs from -iR to +iS. Now rotate the contour, proving that the curves from e.g. -iR to -R -> 0.

I agree with you. A lot of things I tweet need further justification, but I try to fit everything on a single slide and also make it accessible to a wider audience.

For the substitution in integral a), maybe I'm missing something, but u^2 = i*x^2, so why do you have -u^2 after the substitution and not u^2?

I find this pretty believable! The integral of cos(nx) over a finite interval goes to zero as n increases - nonrigorously, the widths of the spikes go to zero much more quickly than the number of spikes increases. Away from zero cos(x^2) looks like cos(nx) for very large n

Actually now that I think about it I’m pretty sure cos(nx) on an interval doesn’t go to zero in the L1 norm as n -> inf. So it’s not the “width of the spikes.” It is true that the integral of just cos(nx) goes to zero