I was asked in an interview to compute $E[\cos X]$ where $X$ is lognormal. I tried using lognormal's characteristic function (Taylor series representation, which is divergent) and $\cos X=\frac{e^{iX}+e^{-iX}}{2}$, but that only leads to an infinite series. I suppose the interviewer is asking for some kind of analytical solution rather than an infinite series.
Thanks a lot!
The interviewer may have only wanted that you identify the characteristic function of the log-normal distribution, which is known to have no closed-form expression. In practice, various divergent series representations and approximation formulae are used to evaluate the characteristic function numerically. In the standard case $\mu = 0$, $\sigma = 1$, the approximation $$ \varphi(t) \approx \frac{\exp\!\big({-\frac12}(W^2(-it) + 2\, W(-it))\big)}{\sqrt{1 + W(-it)}} $$ of the characteristic function leads to the approximation $$ E[\cos X] = \tfrac12\big( \varphi(1) + \varphi(-1)\big) \approx 0.339598\, . $$ Otherwise, we may have used the fact that $Y = \ln X$ has a normal distribution. Thus, we need to compute $E[\cos e^Y]$ where $Y$ is normally distributed. In the standard case $\mu = 0$, $\sigma = 1$, we get $E[\cos e^Y] \approx 0.419794$.