Given the pdf of a dirichlet distribution:
$$\frac{1}{\mathrm{B}(\boldsymbol\alpha)} \prod_{i=1}^K x_i^{\alpha_i - 1}$$ where $\boldsymbol\alpha = (\alpha_1, \ldots, \alpha_K)$, the standard way of solving for the mean of $X_i$ (at least from what I've seen online) is by having this integral:
$$\int_{0}^{1}\ldots\int_{0}^{1}x_i\frac{1}{\mathrm{B}(\boldsymbol\alpha)} \prod_{i=1}^K x_i^{\alpha_i - 1}dx_1\ldots dx_K$$ and then doing a trick to make this look like the a Dirichlet distribution with parameters $(\alpha_0, \ldots, \alpha_i + 1, \ldots, \alpha_k)$ and since that integrates to one you're left with $$\frac{\alpha_i}{\sum_{i=1}^K\alpha_i}$$
My Question
When you have this integral: $$\int_{0}^{1}\ldots\int_{0}^{1}x_i\frac{1}{\mathrm{B}(\boldsymbol\alpha)} \prod_{i=1}^K x_i^{\alpha_i - 1}dx_1\ldots dx_K$$ you can take out the $\frac{1}{\mathrm{B}(\boldsymbol\alpha)}$ since it is a constant and you're left with:
$$\frac{1}{\mathrm{B}(\boldsymbol\alpha)}\int_{0}^{1}\ldots\int_{0}^{1}x_i\prod_{i=1}^K x_i^{\alpha_i - 1}dx_1\ldots dx_K$$
which is equal to:
$$\frac{1}{\mathrm{B}(\boldsymbol\alpha)}\frac{1}{\left(\Pi_{j=1, j\not= i}^K\alpha_j\right)(\alpha_i + 1)}$$
and by definition of $\mathrm{B}$, this is the same as:
$$= \frac{\Gamma(\sum_{i=j}^K\alpha_j)}{\Pi_{i=j}^K\Gamma(\alpha_j)}\frac{1}{\left(\Pi_{j=1, j\not= i}^K\alpha_j\right)(\alpha_i + 1)}$$
and now we can use the fact that $z\Gamma(z) = \Gamma(z + 1)$ to get:
$$= \frac{\Gamma(\sum_{i=j}^K\alpha_j)}{\left(\Pi_{j=1, j\not= i}^K\Gamma(\alpha_j + 1)\right)(\Gamma(\alpha_i + 1) + \Gamma(\alpha_i))}$$
but I'm not sure how to go on from here. This should be equal to the mean since I'm doing the same operations, but I can't finish it. Is it just not possible to find the mean this way, or am I missing something or doing something wrong?