I am reading a physics paper in which they claim that:
$$\int_{t_0}^t \delta(t-t') c(t')=\frac{1}{2} c(t).$$
They just say about this equation:
"The result will always hold when $\int_{-\infty}^{+\infty} d\omega e^{-i \omega(t-t')} =2 \pi \delta(t-t')$ is achieved as the limit of an integral over a function going smoothly to zero at $\pm \infty$ which is essentially what we are doing here."
I know the Dirac is not a function but a distribution, so we can easily make mistakes if we forget about this fact.
However, is there a real reason why if the delta is evaluated on the "edge" of the integral, then I will have $1/2$ the function evaluated on the given point?
Is it purely a kind of choice of convention we can make?
How can I give a mathematical sense to it?
(I know some very basics about distribution theory so I would appreciate a very easy to understand answer if it is based on it).
[edit] :
The answer from reuns helped me but I still have a problem I don't understand. Consider :
$$ \int_{t_i}^{t_f} f(t') \delta(t'-t_f) dt' = \int_{-\infty}^{+ \infty} f(t') \Theta(t'-t_i) \Theta(t_f-t') \delta(t'-t_f) dt'$$
From his explanation, we now that this integral is equal to : $\frac{g(+t_f)+g(-t_f)}{2}$ where $g(t)=f(t) \Theta(t-t_i) \Theta(t_f-t)$. So in the end it is equal to $f(t_f)/2$. Fine.
But now, we also can write:
$$ \int_{t_i}^{t_f} f(t') \delta(t'-t_f) dt' = -\int_{t_f}^{t_i} f(t') \delta(t'-t_f) dt'$$ $$= -\int_{+\infty}^{-\infty} f(t') \Theta(t'-t_i) \Theta(t_f-t') \delta(t'-t_f) dt'$$
And using the same result, I would end up with $-f(t_f)/2$ : extra minus in this result.
However, what I did is just to use the relation : $\int_a^b = -\int_b^a$. Am I not allowed to use it with distributions ? Where is the mistake ?
You can define the $\delta$ distribution as $$\int_{-\infty}^\infty \delta(t) f(t)dt = \lim_{n \to \infty} \int_{-\infty}^\infty \frac{1_{|t| < 1/n}}{2n} f(t)dt $$ whenever the limit converges.
For $f$ piecewise continuous it does converge and the result is $\lim_{t \to 0} \frac{f(t)+f(-t)}{2}$.
Then the goal is to show those things are compatible with the Fourier transform in the sense that if $f$ is piecewise continuous and $\in L^1$ then $\lim_{t \to 0} \frac{f(t)+f(-t)}{2} = \lim_{A \to \infty}\frac{1}{2\pi} \int_{-A}^A \hat{f}(\omega)d\omega$ so that in the dual space of integrables piecewise continuous functions $\delta(t) = \lim_{A \to \infty} \frac1{2\pi} \int_{-A}^A e^{i \omega t}d\omega$.
What they are saying in your paper is that we also have $\delta(t) = \lim_{A \to \infty}\frac1{2\pi}\int_{-\infty}^\infty \varphi_A(\omega) e^{i \omega t}d\omega$ for any sequence $\varphi_A$ converging to $1$ and satisfying a few properties.