My question today is how do you integrate this for an arbitrary real, continuous and differentiable function $g(x)$ (with $g(+ \infty) = 0$ and $g(0) = 0$, $g(x) \geq 0$) and a real number $a_0 > 0$:
$ I_0 = \int_0^{+ \infty} d x g(x) e^{- a_0 x \delta(x-x_0)} $
And then, how do you integrate this one (also real numbers):
$ I_0 = \int_0^{+ \infty} d x g(x) e^{- \sum_p a_p x \delta(x-x_p)} $
Where $\delta(x)$ is the Dirac delta function:
$\int \delta(x) f(x) dx = f(0)$
Thank you for reading
Example for g
For example we can have:
$$g(x) = x e^{-x} $$
What I tried
With absolutely no mathematical rigor (but Michael Penn is a genius for me + he knows how to backflip on flatground, I really like this guy), but this is how we sometimes discover something (this is an applied spectroscopy problem I had)! I tried a simplified integral (but this is almost the same):
$$ I = \int_0^{+ \infty} d x g(x) e^{- \delta(x-x_0)} $$
What do we have for $e^{- \delta(x-x_0)} $ ?
Simply:
$$ e^{- \delta(x-x_0)} = 0, \quad x = x_0 $$ $$ e^{- \delta(x-x_0)} = 1, \quad x \neq x_0 $$
This is like the "inverse" behavior of the Dirac delta. So let's try to integrate:
$$I = \int_0^{x_0^-} d x g(x) + \int_{x_0^-}^{x_0^+} 0 g(x) d x + \int_{x_0^+}^{+ \infty} d x g(x) $$
I propose 2 things:
We "pass to the limit" first proposition:
$$ I \stackrel{?}{=} \int_0^{+ \infty} d x g(x) $$
or we "supress the $g(x_0)$ because not include in the interval" second proposition:
$$ I \stackrel{?}{=} \int_0^{+ \infty} d x g(x) - g(x_0) $$
This seems really weird and certainly wrong but this is here to stimulate the motivation to answer to my question ^^.
Something that motivated me to use the second proposition is that when we do the Taylor expansion:
$$ I = \int_0^{+ \infty} d x g(x) \left( 1 - \delta(x-x_0) + \delta^2(x-x_0)/2 - ... \right) $$
and we can integrate the first two terms to find again the second proposition,
$$ I = \int_0^{+ \infty} d x g(x) - g(x_0) + \int_0^{+ \infty} d x g(x) \left( \delta^2(x-x_0)/2 - ... \right) $$
but I don't know how to do with the rest which seems to diverge...
I think you are on the right track.
$\int_0^\infty g(x)e^{-a_0x\delta(x-x_0)}dx =\int_0^{x_0-\epsilon}g(x)dx+\int_{x_0-\epsilon}^{x_0+\epsilon}g(x)e^{-a_0x\delta(x-x_0)}dx+\int_{x_0+\epsilon}^\infty g(x)dx$
$\int_{x_0-\epsilon}^{x_0+\epsilon}g(x)e^{-a_0x\delta(x-x_0)}dx\le \sup{(|g(x)|,|g(x_0)e^{-a_0x_0}}|) \cdot 2\epsilon$
So if $g(x)$ is bounded on the interval of integration for the middle integral, the limit of the integral as $\epsilon \to 0 $ is $0$.
Essentially this is integrating $g(x)$ with a jump discontinuity at $x_0$. Countably many jump discontinuities tend not to effect the value of the integral.