First, define a probability distribution with density:
\begin{align*} \newcommand\bm\boldsymbol \newcommand{\mbl}{\mathbf{l}} f(\mbl \mid \bm \alpha) = \frac{1}{B(\bm \alpha + \bm 1)}\exp\left(\mbl \cdot \bm\alpha\right) \end{align*} where $\mathbf l \in (-\infty,0]^k, \sum_i\exp(l_i) = 1$
I'm having trouble finishing this integral (to calculate the convolution):
\begin{align*} \newcommand{\mba}{\mathbf{a}} \newcommand{\mbb}{\mathbf{b}} \newcommand{\mbl}{\mathbf{l}} \newcommand{\mbp}{\mathbf{p}} \newcommand{\mbv}{\mathbf{v}} \newcommand{\mbx}{\mathbf{x}} f(\mbl \mid \mba, \mbb) &= \int_{l_1}^0 \dotsi \int_{l_k}^0 f(\mbx \mid \mba)f(\mbl-\mbx \mid \mbb)d\mbx \newline &\propto \int_{l_1}^0 \dotsi \int_{l_k}^0 \exp(\mbx \cdot \mba + (\mbl-\mbx) \cdot \mbb)d\mbx \newline &= \int_{l_1}^0 \dotsi \int_{l_k}^0 \exp(\mbx \cdot (\mba - \mbb) + \mbl \cdot \mbb)d\mbx \newline &= \prod_{i=1}^k\frac{\exp(a_il_i) - \exp(b_il_i)}{a_i-b_i} ??? \end{align*}
Edit:
After looking at mpiktas's answer, the final line should have been \begin{align*} &= \prod_{i=1}^k\frac{\exp(b_il_i) - \exp(a_il_i)}{a_i-b_i} \end{align*} which ensures that the terms are all positive.