Why oscillations in PI control?
For symplicity, consider the SISO linear system
\begin{align*}
\dot x &amp;= ax +bu\\
y &amp;= cx
\end{align*}
with $x$,$u$ and $y$ taking values in $\mathbb R$. Assume that you want to stabilize the system by making $y$ to converge to some fixed set point $\bar y$. Then necessarely there must be a point $(\bar x,\bar u)$ such that
\begin{align*}
0 &amp;= a\bar x+b\bar u,&amp; \bar y=c\bar x
\end{align*}
since $\bar x$ must be an equilibrium and such that $y=\bar y$. Trivially one has $$\bar u = -\dfrac{a}{bc}\bar x$$ One can define the error variables as $\tilde x=x-\bar x$, $\tilde y=y-\bar y$ and $\tilde u=u-\bar u$ that yield the error system
\begin{align*} \dot{\tilde x} &amp;= a\tilde x + b\tilde u,&amp; \tilde y&amp;=c \tilde x
\end{align*}
One can stabilize it by taking $\tilde u = k\tilde x$ such that $a+bk&lt;0$. That in fact ensures that asymptotically $\tilde y=0$ which implies $y=\bar y$. The overall control law is
$$
u = \tilde u + \bar u = k(x-\bar x) -\dfrac{a}{bc} \bar x = kx + \dfrac{1}{c}\left(-k-\dfrac{a}{b}\right)\bar y
$$
Thus as you can se, simple proportional is not enough, you also need a feed-forward control action given by
$$
u_{ss} = \dfrac{1}{c}\left(-k-\dfrac{a}{b}\right)\bar y
$$
If you define an integrator system like $\dot\sigma = y-\bar y = C(x-\bar x)$ then also $\sigma$ will have a set point $\bar \sigma$, and with $\tilde \sigma=\sigma-\bar\sigma$, the extended error system becomes
\begin{align*}
\begin{bmatrix}\dot {\tilde x}\\\dot{\tilde\sigma}\end{bmatrix} &amp;= \begin{bmatrix}a &amp; 0\\c &amp; 0\end{bmatrix}\begin{bmatrix}\tilde x\\\tilde{\sigma}\end{bmatrix} + \begin{bmatrix}b\\0\end{bmatrix}\tilde u, &amp; \tilde y&amp;=C\tilde x\end{align*}
Take $\tilde u=k_p \tilde x + k_i \tilde\sigma$ (proportional+integral), then the closed loop system reads
\begin{align*}
\begin{bmatrix}\dot {\tilde x}\\\dot{\tilde\sigma}\end{bmatrix} &amp;= \begin{bmatrix}a+bk_p &amp; bk_i\\c &amp; 0\end{bmatrix}\begin{bmatrix}\tilde x\\\tilde \sigma\end{bmatrix}, &amp; \tilde y&amp;=C\tilde x\end{align*}
You can chose $k_p$ and $k_i$ to make the closed loop matrix to be Hurwitz, and thus to stabilize the error system and you don't need any feedforward action, in fact the control law is
$$
u = \tilde u + \bar u = k_p(x - \bar x) + k_i (\sigma -\bar \sigma)
$$
Then with $$ \bar \sigma = \dfrac{k_p}{k_i}\bar x$$ the control law to becomes
$$
u = k_p x + k_i \sigma \qquad\qquad (1)
$$
and that's all. Moreover without integral control you need a feed-forward action which depends on the system parameters. If you have uncertainties on the parameters for sure you wil have an error on your set-point. with the integral action your final control law (1) does not depend on the parameters! and as long as the closed-loop system is stable you have null error at the steady state. Oscillations depend on how you chose $k_p$ and $k_i$. In fact if you chose them such that the closed loop matrix has complex eigenvalues then you can have socillations. For MIMO system the reasoning extends trivially.