= 1}^n \epsilon_k^{\delta_k} \right)^{1/n}. \\
\end{array}$$ Thus, $$\bar{c}(t) \exp \left( \int_t^T  \widehat{c}(s) ds \right) = \left(\overline{\epsilon^{\delta}}\right)^{-\frac{1}{1 +\widehat{\theta(\delta-1)}}} e^{- \frac{\widehat{\delta \rho}}{1 + \widehat{\theta(\delta -1)}  } (T - t)}.$$ Plugging this expression into , we get $$\label{derivci}
c_i(t) \exp \left( \int_t^T  c_i(s) ds \right) =  \newC_i e^{ \newomega_i (T - t)},$$ where we define $$\begin{aligned}
\newomega_i &:= \theta_i (\delta_i - 1)\frac{\widehat{\delta \rho}}{1 + \widehat{\theta(\delta -1)}  }- \delta_i \rho_i, \\
\newC_i &:= \epsilon_i^{- \delta_i} \left(\overline{\epsilon^{\delta}}\right)^{\frac{\theta_i(\delta_i -1 )}{1 +\widehat{\theta(\delta-1)}}} > 0.\end{aligned}$$ Integrate (\[derivci\]) from $t$ to $T$ and take the logarithm to get $$\int_t^T  c_i(s) ds = \begin{cases}
\log \left( 1 + \frac{\newC_i}{ \newomega_i} \left( e^{ \newomega_i (T - t)} - 1 \right) \right) &\text{if } \beta_i\neq 0 \\
 \log\left(\newC_i(T-t) + 1\right) &\text{if }  \beta_i = 0.
\end{cases} \label{intci}$$ This is indeed well defined because, when $\beta_i \neq 0$, the function $t \mapsto 1 + \frac{\newC_i}{ \newomega_i} \left( e^{ \newomega_i (T - t)} - 1 \right)$ is decreasing on $[0,T]$ and equal to $1$ at $t=T$. Differentiating both sides, we finally obtain $$c_i(t) = \begin{cases}
\left(\frac{1}{\newomega_i} + \left(\frac{1}{\newC_i} - \frac{1}{\newomega_i} \right)e^{- \newomega_i (T - t)} \right)^{-1} &\text{if } \beta_i\neq 0 \\
(T - t + \newC_i^{-1})^{-1} &\text{if } \beta_i = 0.
\end{cases}$$ In summary, 