oh, the integral is definitely scale invariant, sorry ...but then isn't scale invariance exactly what your differentiation under the integral sign proves? /me is confused
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Notice that exp(a exp(i x) ) = 1 + a exp(i x) + a^2 exp(2 i x)/2 + ... exp(a exp(-i x) ) = 1 + a exp(-i x) + a^2 exp(-2 i x)/2 + ... Then in_0^{2pi} [exp(a exp(i x) ) + exp( a exp(- i x )) ]/2 dx= 2(2pi)/2 = 2pi because int_0^{2pi} exp(-i k x) dx = 0 correct me if I am wrong
I’ll have to get back to you, I’m about to head out.. but indeed looks promising
