Continuous probability distribution with no first moment but the characteristic function is differentiable
Below, there is a discrete example because I misread the word continuous.
However, if $X$ has the density $f$ given by $$f(x)=\begin{cases}0&\mbox{ if }|x|\leqslant 2,\\ \frac C{x^2\log |x|}&\mbox{ if }|x|>2,\end{cases}$$ then $X$ is not integrable and the characteristic function is given by $$\varphi(t)=2C\int_2^{+\infty}\frac{\cos(tx)}{x^2\log x}\mathrm{d}x.$$
- Let us show that $\varphi$ is differentiable at $0$ and $\varphi'(0)=0$. We start from
$$\frac{\varphi(t)-\varphi(0)}{2tC}=\int_2^{1/t}\frac{\cos(tx)-1}{x^2\log x}\frac{dx}t+\int_{1/t}^{+\infty}\frac{\cos(tx)-1}{x^2\log x}\frac{dx}t=:A(t)+B(t).$$
Then using $|\cos u-1|\leqslant u^2$
$$|A(t)|\leqslant \int_2^{1/t}\frac{t^2x^2}{x^2\log x}\frac{dx}t=t\int_2^{1/t}\frac{dx}{\log x}.$$
Let $y:=\log x$, then $x=e^y$ and $dx=e^ydy$:
$$|A(t)|\leqslant t\int_{\log 2}^{-\log t}\frac{e^y}y\mathrm dy,$$
and splitting the interval at $R$, we get
$$|A(t)|\leqslant te^R/\log 2+R^{-1},$$
hence $\lim_{to\to 0}A(t)=0$.
Since
$$|B(t)|\leqslant \frac 1t\int_{1/t}^\infty\frac{2}{x^2\log(x)}dx,$$
we obtain after the substitution $s=1/x$ that
$$|B(t)|\leqslant -\frac 2t\int_0^t\frac{ds}{\log s}.$$
An other transformation $tu=s$ yields
$$|B(t)|\leqslant -\int_0^1\frac{du}{\log u+\log t},$$ which converges to $0$ as $t$ goes to $0$. - A similar argument can be used to show the differentiability at each point.
This example is taken from Stoyanov's book Counter-examples in probability.
Take $X$ such that $$\mathbb P(X=(-1)^kk)=\frac C{k^2\log k},\quad k\geqslant 2.$$
Then $\mathbb E|X|$ is infinite, while the characteristic function
$$\varphi_X(t)=\sum_{k\geqslant 2}\frac C{k^2\log k}e^{it(-1)^kk}$$
is differentiable everywhere: compute $\varphi_X(t+h)-\varphi_X(t)$ approximating $e^{ih(-1)^kk}-1$ by $h(-1)^kk+o(h)$.