Integral of Dirac delta function from zero to infinity 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.
 A: 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. 
A: See the question Delta function integrated from zero and Andrew's answer to the question.
According to both Bracewell and Andrew, your guess of $f(0)/2$ is right.  See also my own question "Is distribution theory necessary for most users of the Dirac delta function?" and my answer to the question "Proof of Dirac Delta's sifting property".  
