Assume the prior is given as $$ \operatorname{Dir}(\boldsymbol{\mu} \mid \boldsymbol{\alpha})=\frac{\Gamma\left(\sum_{k=1}^K \alpha_k\right)}{\Gamma\left(\alpha_1\right) \cdots \Gamma\left(\alpha_K\right)} \prod_{k=1}^K \mu_k^{\alpha_k-1} $$ and the likelihood is given as $$ \operatorname{Mult}(\mathbf{y} \mid \boldsymbol{\mu}, N)= \frac{N!}{y_1!y_2!\ldots y_K!} \prod_{k=1}^K \mu_k^{y_k} $$ The marginal likelihood is given as $$ p\left(\mathbf{y} \mid \boldsymbol{\alpha}\right)=\int p\left(\mathbf{y} \mid \boldsymbol{\mu}\right) p\left(\boldsymbol{\mu} \mid \boldsymbol{\alpha}\right) d \boldsymbol{\mu} $$ Using this marginal likelihood function we can perform MLE to get an estimation of $\boldsymbol{\alpha}$. I am wondering whether the likelihood function is a concave one. Below is my try to prove the concavity.
Try 1: I derived the formula for the integral appearing in the target function. $$ \int_{\sum^N_{i=1} \mu_i = 1} \prod^N_{i=1} \mu_i^{z_i -1} d\mu = \frac{\prod^N_{i=1}\Gamma(z_i)}{\Gamma(\sum^N_{i=1}z_i)} \frac{z_1+z_2}{\sum^N_{i=1}z_i} $$ However, I found it too complex when I substituted this into the integral and tried to prove the log-concavity.
Try 2: It suffices to prove that $$ \forall \boldsymbol{\mu}, p(\boldsymbol{\mu}|\boldsymbol{\alpha}) \text{ is concave}. $$ then the linearity of the integral will propagate the concavity. However, this also seems too difficult since we needs to prove the concavity of Dirichlet density function w.r.t. its parameter space.
Any ideas?