Differentiability of characteristic functions and moments of the corresponding measure Let $f$ be the characteristic function of a real-valued random variable $X$. It is known that if $f$ has a $k$-th order derivative (for some even $k$) then $\mu$ has a finite $k$-th order moment. Is there any reference or discussion the case when $k$ is odd? For example, if we only know that $f$ has its first order derivative, do we know that $\mu$ has finite first moment?
 A: A good reference regarding this type of results is the book by Eugène Lukacs "Characteristic Function". For example, chapter 2.3 "Characteristic functions and moments" provides results in this direction. Theorems 2.3.1, 2.3.2 and 2.3.3 might be what you are looking for.
A: Zygmund's example is the discrete random variable $X$ where
$\mathbb{P}(X=\pm n) = \frac{C}{n^2\log(n)}$ for integer $n \geq 2$. $C$ is a unique constant that makes this a probability distribution. You can calculate that $X$ does not have a finite first moment because
$$\mathbb{E}(|X|)= \sum_{n=2}^\infty \frac{2C}{n \log(n)} = +\infty$$. It diverges because $\int_2^\infty \frac{1}{x\log(x)}dx$ diverges.
The characteristic function is
$$\varphi(t) = C\sum_{n=2}^\infty \frac{\cos(nt)}{n^2 \log(n)}.$$
$\varphi(t)$ is differentiable at zero and the derivative is $\varphi'(0)=0$. Zygmund's argument does not rely on any fancy theory of trigonometric series. You can just check the differentiability directly.
Specifically, look at the quotients
$$\frac{\varphi(h) - \varphi(0)}{h} = C\sum_{n=2}^\infty \frac{\cos(nh) - 1}{hn^2\log(n)}$$.
Split this into two pieces for a $M \in \mathbb{N}$ to be chosen soon.
$$=C\sum_{n=2}^M \frac{\cos(nh) - 1}{hn^2\log(n)} + C\sum_{n=M+1}^\infty \frac{\cos(nh) - 1}{hn^2\log(n)}$$.
For the first sum, where $nh$ is small, we use the Remainder Theorem estimate
$$|\cos(nh) -1|\leq \frac{n^2h^2}{2}$$
and for the second sum, where $nh$ is big, we use the estimate
$$|\cos(nh) -1|\leq 2.$$
This all leads to the estimate, where $K$ is some constant independent of $h$ whose value changes line to line,
$$\left|\frac{\varphi(h) - \varphi(0)}{h}\right| \leq K \sum_{n=2}^M \frac{h}{\log(n)} + K\sum_{n=M+1}^\infty  \frac{1}{n^2\log(n) h}$$.
The first sum is trivially less than $KMh$. For the second sum,
$$\sum_{n=M+1}^\infty \frac{1}{n^2 \log(n) h} \leq \frac{1}{h\log(M)} \sum_{n=M+1}^\infty \frac{1}{n^2} \leq \frac{K}{hM\log(M)}.$$
These calculations prove that
$$\left|\frac{\varphi(h) - \varphi(0)}{h}\right| \leq KMh + \frac{1}{M \log(M) h}$$.
Given $h$, choose  integer $M=M(h)$ so that $\frac{1}{M \sqrt{\log(M)}}\approx h$. Then both terms go to $0$ as $h \to 0$, proving that $\varphi(t)$ is differentiable at $t=0$.
