Q: Let $X$ be a random variable with characteristic function $\phi_X$, suppose $\phi_X\in L^1$ can we say that $X\in L^1$?
We know that, $X\in L^1$ implies that $\phi_X$ has a bounded continuous derivative of first order. Now, if there exists a characteristic function $\phi$ which is integrable but has no bounded continuous derivative, then the answer to the question is No. And I guess this is the case, but I cannot find such a characteristic function. And maybe the answer is yes. Any help would be appreciated.
The answer is NO. If $X$ has Cauchy distribution then $\phi_X(t)=e^{-|t|}$ which is integrable. But $X$ does not have finite mean. The density function is $f_X(x)=\frac 1 {\pi} \frac 1 {1+x^{2}}$ and it is easy to check that $\int |x|f_X(x)dx=\infty$.