I know that: $$\int_{-\infty}^{+\infty} \mathrm{d}t \, f(t) \delta(t) = f(0)$$
However, I cannot figure out the result of the integral below:
$$\int_{0}^{+\infty} \mathrm{d}t \, f(t) \delta(t) = ?$$
Is it $$\int_{0}^{+\infty} \mathrm{d}t \, f(t) \delta(t) = \frac{f(0)}{2}?$$
Please provide a source for the answer too.
The identity $$ \int_{-\infty}^{+\infty}dt\ f(t) \delta(t) = f(0)\tag{*} $$ is meaningless without context. Also this notation is a convenient abuse of notation, and not a standard (Riemann or Lebesgue) integral.
Let's say you are considering $\delta:\mathcal{S}(\mathbb{R})\to\mathbb{R}$ as a tempered distribution on the Schwartz class $\mathcal{S}(\mathbb{R})$. Then $(*)$ means nothing but the definition of $\delta$: $$ \delta(f)=f(0)\quad f\in\mathcal{S}(\mathbb{R}). $$ In this setting, $\int_{0}^{+\infty} \mathrm{d}t \, f(t) \delta(t) $ is not even a well-define notation.
Your question is a nice example demonstrating that it could be dangerous to think $\delta$ as a function of real variables.