I shall prove that the problems are in fact equivalent, in the sense that they both have the same optimal solution $x^*$. Let $P_1$ be the first optimization problem, and $P_2$ the second, and let $\alpha\succ 0$.
If $\tilde{x}$ is the optimal solution for $P_1$, then we know from the book that each $\tilde{x}_i$ satisfies the relation
$$
\tilde{x}_i = \max\{0,1/\tilde{\nu} - \alpha_i\},
$$
where $\tilde{\nu}$ is the optimal dual variable associated with the equality constraint.
Now, write $P_2$ equivalently as
$$\begin{split}
\text{minimize} &\quad t\\
P_2:\quad\text{subject to}&\quad t\geq -\log(\alpha_i + x_i),\quad i=1 ,\ldots,n\\
&\quad x\succeq0,\quad \mathbf{1}^\top x=1.
\end{split}.$$
Then, the Lagrangian is
$$L(t,x,\mu,\lambda,\nu) = t - \sum_{i=1}^n \mu_i(\log(\alpha_i + x_i) + t) - \lambda^\top x + \nu(\mathbf{1}^\top x - 1) =\\t(1 - \mathbf{1}^\top\mu) - \sum_{i=1}^n\mu_i\log(\alpha_i + x_i) - \lambda^\top x + \nu(\mathbf{1}^\top x - 1).$$
Using this, you can write the KKT conditions of $P_2$ for the primal and dual optimal variables $t^*,x^*,\mu^*,\lambda^*$ and $\nu^*$ as
$$x^*\succeq 0,\quad \mathbf{1}^\top x^* = 1,\quad \lambda^*\succeq0,~\mu^*\succeq 0,\quad \lambda_i^*x_i^*=0,\quad t^* + \log(\alpha_i + x_i^*) \geq 0,\quad \mu_i^*(t^* + \log(\alpha_i + x^*_i)) = 0,\quad i=1,\ldots,n,$$
and
$$\partial/\partial x_i \left[ L(t^*,x^*,\mu^*,\lambda^*,\nu^*)\right]= -\mu_i^*/(\alpha_i^* + x_i^*) - \lambda_i^* + \nu ^* = 0,~i=1,\ldots,n,\quad \mathbf{1}^\top\mu^* = 1.$$
The last equality is necessary, otherwise the dual becomes unbounded in $t^*$. Eliminating $\lambda^*$ yields:
$$x^*\succeq 0,\quad \mathbf{1}^\top x^* = 1,\quad x_i^*(\nu^* - \mu_i^*/(\alpha_i + x_i^*))=0,\quad \nu^*\geq \mu_i^*/(\alpha_i + x_i^*),\quad i=1\ldots,n,$$
and
$$\mu^*\succeq 0,\quad \mathbf{1}^\top\mu^* = 1,\quad t^* + \log(\alpha_i + x_i^*) \geq 0,\quad \mu_i^*(t^* + \log(\alpha_i + x^*_i)) = 0,\quad i=1,\ldots,n.$$
These are in fact very similar conditions to those of $P_1$. Now, let $M = \{i\in \{1,\ldots,n\} ~|~ \mu_i^* > 0\}$. This is a nonempty set, because $\mathbf{1}^\top \mu^* = 1$. For all $i\in M$, you can thus deduce (in a similar way to the book) that in fact
$$x_i^* = \max\{0, \mu_i^*/\nu^* - \alpha_i\}.$$
Furthermore, we have that $t^* = -\log(\alpha_i + x_i^*)$.
Now, let $M' = \{1,\ldots,n\}\setminus M$. In this case, we have that for all $i\in M'$:
$$
x_i^*\nu^* = 0,\quad \nu^* \geq 0,\quad t^* + \log(\alpha_i + x_i^*) \geq 0.
$$
We can't have $\nu^* = 0$, otherwise it would imply that $0 < \mu_j^*/(a_j + x_j^*) \leq 0$, for all $j\in M$, which is impossible. So, the only option is that $x_i^* = 0$ for all $i\in M'$.
Thus, the problem reduces to solving the equation:
$$
\sum_{i\in M} x_i^* = \sum_{i\in M}\max\{0,\mu_i^*/\nu^* - \alpha_i\} = 1.
$$
However, the choice of $\mu$ can be arbitrary, as long as it satisfies the KKT conditions. Let $K = \{i\in\{1,\ldots,n\}~|~ \tilde{x}_i > 0\}$, and suppose we set the multipliers to
$$
\mu_i^* = \begin{cases}
1/\lvert K \rvert & i \in K\\
0 &\text{otherwise,}
\end{cases}
$$
then $\mathbf{1}^\top \mu^* = 1, \mu^*\succeq 0$. Clearly, we also have $M = K$, which implies $x_i^* = \tilde{x}_i=0$ for all $i\in M'$. Now, setting $\nu^* = \tilde{\nu}/\lvert M \rvert > 0$, then this implies $x_i^* = \tilde{x}_i$ for all $i\in M$, and we get exactly the same solution as $P_1$. And since $\tilde{x}_i + \alpha_i = \tilde{x}_j + \alpha_j$ for all $i,j\in M, i\neq j$, then this implies
$$
\log(x_i^* + \alpha_i) = \log(x_j^* + \alpha_j),
$$
thus $t^* = -\log(x_i^* + \alpha_i)$ for all $i\in M$, so all of the KKT conditions are satisfied.
Therefore, solving $P_2$ is equivalent to solving $P_1$. The optimal value of $P_2$ is $-\log(\alpha_i + \tilde{x}_i)$, for any $\tilde{x}_i > 0$.
Best Answer
You're correct that the problem falls into the class treated by the calculus of variations. A nice modern approach to this subject is to treat the objective as a function on a topological vector space, which contains the space of functions over which you want to optimise the objective. The optimality conditions are then often quite analogous to the Karush-Kuhn-Tucker conditions for optimisation over a finite-dimensional vector space, although there can be some messy technical details which depend on the specific class of functions over which you want to optimise. David Luenberger's Optimization by Vector Space Methods contains a nice exposition of this approach.
For the specific problem you've given, the optimality conditions are similar to those of the finite-dimensional version. You introduce Lagrange multipliers $\ \lambda\in\mathbb{R}\ $ for the constraint $$ \int_{\mathcal{X}}f(x)=1\ , $$ and $\ \mu\in\mathcal{F}^*\ $ for the constraint $$ f(x)\ge0\ , $$ where $\ \mathcal{F}^*\ $ is the function-space dual of $\ \mathcal{F}\ $. This is where some of the messy technical details mentioned above come in. The duals of spaces of continuous functions are typically spaces of finitely additive measures, which can be difficult to deal with. A way out, in this case, is to relax the conditions on $\ f\ $ by allowing it to be any function that's square-integrable with respect to Lebesgue measure over $\ \mathcal{X}\ $. The space $\ \mathcal{H}\ $ of functions square-integrable over $\ \mathcal{X}\ $ is a real Hilbert space, which is self-dual. The Lagrange multiplier $\ \mu\in\mathcal{H}^*\ $ corresponding to the constraint $\ f(x)\ge0\ $ is now itself just a square-integrable function $\ \mu:\mathcal{X}\rightarrow\mathbb{R}\ $. The Lagrangian for the problem becomes $$ \int_{\mathcal{X}}-\log(\alpha(x)+f(x))dx+\lambda\left(\int_{\mathcal{X}}f(x)dx-1\right)-\int_{\mathcal{X}}\mu(x)f(x)dx\ , $$ and the optimality conditions are $$ \frac{-1}{\alpha(x)+f(x)}+\lambda-\mu(x)=0\\ \mu(x)\ge0,\ \ f(x)\ge0,\ \ \mu(x)f(x)=0,\ \ \int_{\mathcal{X}}f(x)=1\ , $$ which are very closely analogous to those of the finite dimensional case.
Now let $\ \mathcal{X}_1=\{ x\in\mathcal{X}\,|\,f(x)>0\,\}\ $. The complementary slackness conditions, $\ \mu(x)f(x)=0\ $, tell us that $\ \mu(x)=0\ $ for $\ x\in\mathcal{X}_1\ $, and the first optimality condition then gives \begin{align} \frac{1}{\alpha(x)+f(x)}&=\lambda\ \text{, or}\\ f(x)&=\frac{1}{\lambda}-\alpha(x)>0 \end{align} for $\ x\in\mathcal{X}_1\ $. For $\ x\in\mathcal{X}\setminus\mathcal{X}_1\ $, where $\ f(x)=0\ $, we must have \begin{align} \lambda-\frac{1}{\alpha(x)}&=\mu(x)\ge0\ \text{, or}\\ \frac{1}{\lambda}&\le\alpha(x)\ , \end{align} since $\ \lambda\ $ must be positive. It follows from this that $\ f(x)=$$\max\left(0,\frac{1}{\lambda}-\alpha(x)\right)\ $. The constraint $$ 1=\int_\mathcal{X}f(x)dx=\int_\mathcal{X}\max\left(0,\frac{1}{\lambda}-\alpha(x)\right)dx $$ now allows us to determine the value of $\ \lambda\ $. The function $\ \phi(\nu)=\int_\limits{\mathcal{X}}\max\left(0,\nu-\alpha(x)\right)dx\ $ is a strictly increasing continuous function of $\ \nu\ $, with $\ \phi(0)=0\ $ and $$\ \phi\left(\frac{1}{\int_\limits{\mathcal{X}}1dx}+\max_\limits{ x\in\mathcal{X}}\alpha(x)\right)\ge1\ ,$$ so there is a unique $\ \nu^*>0\ $ such that $\ \phi\big(\nu^*\big)=1\ $, and then $\ \lambda=\frac{1}{\nu^*}\ $.
Because the objective function is strictly concave in this case, the optimality conditions are both necessary and sufficient, so $\ f(x)=\max\left(0, \frac{1}{\nu^*}-\alpha(x)\right)\ $ achieves the minimum of the objective function over the class of square-integrable functions over $\ \mathcal{X}\ $. Since $\ f\ $ happens to be continuous, and all continuous functions over the compact set $\ \mathcal{X}\ $ are square-integrable, it must also achieve the minimum of the objective over the set of continuous functions on $\ \mathcal{X}\ $.