Since $X_t:=\int_0^ts \, dB_s$ is a process with independent increments, its distribution is infinitely divisible and its variance is $c_t=\frac{1}{3}t^3$.
I think, its characteristic function $E[\exp(iuX_t)]=\exp(\Psi_t(u))$, where$$\Psi_t(u)=-\frac{1}{2}u^2c_t=-\frac{1}{2}u^2\cdot\frac{1}{3}t^3.$$
I am not sure about this. Anyone could help me to confirm the result? Thanks for your comments and suggestions!
Yeah, this is correct.
It is well-known that
$$X_t := \int_0^t s \, dB_s$$
is a Gaussian process. It follows from the fact that $(X_t)_{t \geq 0}$ is a martingale that
$$\mathbb{E}(X_t) = \mathbb{E}(X_0) = 0.$$
On the other hand, Itô's isometry yields that
$$\mathbb{E} \left( X_t^2 \right) = \int_0^t s^2 \, ds = \frac{t^3}{3}.$$
Consequently, $X_t \sim N(0,t^3/3)$. This, in turn, gives us the characteristic function of $X_t$:
$$\mathbb{E}e^{i \xi X_t} = \exp \left( -\frac{1}{2} \frac{t^3}{3} |\xi|^2 \right).$$