I am trying to derive the cumulant/moment generating function of
$$ f(x)=φ(x)(1+\frac{1}{2}\sin(2 π x)) $$
where φ(x) is the standard normal distribution . This distribution was first brought up by McCullagh 1994. But I could not figure out how its mgf/cgf was derived. Maybe the derivation has been posted somewhere already. I know the cumulant generating function is $$ cgf =t^2/2+\log[1+\frac{1}{2}e^{-2\pi^2} \sin(2\pi t)] $$ Just don't know how to get this results. Could anybody help? Maybe just a link. It is old question. --thanks a lot!
Since famously $\int_{\Bbb R}\varphi(x)\exp(ikx)dx=e^{-k^2/2}$, your distribution's MGF is$$\begin{align}Ee^{tX}&=e^{t^2/2}+\frac12\Im\int_{\Bbb R}\varphi(x)\exp(tx+2i\pi x)dx\\&\stackrel{y=x-t}{=}e^{t^2/2}+\frac12e^{t^2/2}\Im\left[e^{2i\pi t}\int_{\Bbb R}\varphi(y)\exp(2i\pi y)\right]\\&=e^{t^2/2}+\frac12e^{t^2/2}\Im\left[e^{2i\pi t-2\pi^2}\right]\\&=e^{t^2/2}\left(1+\frac12e^{-2\pi^2}\sin(2\pi t)\right).\end{align}$$Taking a logarithm gives the claimed CGF.