The operators ${\bf I}^p$ and ${\bf D}^{-p}$

It is obvious that the integration of R_q(t) gives R_q+1(t):

$${\bf I} R_q(t)=R_{q+1}(t)$$

and in general:

$${\bf I}^p R_q(t)=R_{q+p}(t) \mbox{\hspace{1.0cm}} (p\gt) .$$

Does it imply that: ${\bf D}^{-p}R_q(t)=R_{q+p}(t)$ (where ${\bf D}^{-p}$ is the inverse of ${\bf D}^p$)? If we apply the operator ${\bf D}^p$ to the sum R_p+q(t)+P_n(t), we get for $n\leq p-1$:

$${\bf D}^p[R_{q+p}(t)+P_n(t)]=R_q(t).$$

But $P_n(t) \stackrel{\infty}{\sim}0$, so:  

$${\bf D}^{-p}R_q(t) \stackrel{\infty}{\sim}R_{q+p}(t).$$ (14)