In general, the problem $$ \min_u \int_0^\infty (x^T Q x + u^TRu) \, dt \quad\text{such that}\quad \begin{cases}\dot x(t) = Ax(t)+Bu(t) \\ x(0) = x_0\end{cases} $$ has an associated (stationary) Riccati equation given by $$ 0 = Q+A^TS+SA-SBR^{-1}BS, \quad S\text{ constant} $$ In particular, consider the $1$-dimensional case where $Q=B=1$, $R=\gamma>0$ and $A=\alpha>0$, that is $$ \min_u \int_0^\infty (x^2 + \gamma u^2) \, dt \quad\text{such that}\quad \begin{cases}\dot x(t) = \alpha x(t)+u(t) \\ x(0) = 0\end{cases} $$
The Riccati equation for this problem is $$ S^2-2\alpha\gamma S-\gamma = 0 \quad\Rightarrow\quad S = \alpha\gamma\pm\sqrt{\alpha^2\gamma^2+\gamma} $$ Since $Q$ and $R$ are positive, the optimal solution of the problem is given by $$ u(t) = -R^{-1}B^TSx(t) = -\frac{S}{\gamma}x(t) = \Big(-\alpha\mp\underbrace{\sqrt{\alpha^2+1/\gamma}}_{:= k}\Big)x(t) $$ So $$ \dot x(t) = \mp kx(t),\quad \int_0^t \frac{dx(s)}{x(s)} = \mp k \int_0^t ds,\quad x(t) = x(0)\exp(\mp kt) $$ and plugging the initial condition $x(0)=0$ we get $x(t)=u(t)\boxed{=0},\forall t$.
However, on my notes, I read that the solution to $$ \min_u \int_0^T(x^2 + \gamma u^2) \, dt \quad\text{such that}\quad \begin{cases}\dot x(t) = A x(t)+Bu(t) \\ x(0) = x_0\end{cases} $$ is $$ x(t)=x_0\exp\bigg(\int_0^t A\,dS\bigg)+\int_0^t \exp\bigg(\int_0^{t-s}Adr\bigg)Bu(s)\,ds $$ and applying it in our case we would have $x(t) = 0+e^{\alpha t}\int_0^t e^{-\alpha s}u(s)\,ds\boxed{\ne0}$, which contradicts the previous result.
Which one is correct?