Let us have a function $f(x)$ such that
$$\int\limits_{\varepsilon}^{\varepsilon + a} f(x)dx = \int\limits_{2\varepsilon}^{2\varepsilon + a} f(x)dx = ... = \int\limits_0^a f(x)dx = const, \ \ \ \ \ \ \ \ \ \ (1)$$
where $\varepsilon$ is some small (compared to the studied span of $x$, say, $\varepsilon = \frac{x}{n}$) constant value of $x$, and $a$ is infinitesimally small. In other words, $f(x)$ probably can be visualized by a comb function and will have non-zero values only in the neighborhood of integer $x$. For all other $x$, function $f(x)$ will be zero.
Now, we need the value of the integral
$$\int\limits_0^n e^{-x} f(x) dx. \ \ \ \ \ \ \ \ \ \ (2)$$
The thing that I can't understand is this: if eq.(1) is true, then, why shouldn't we consider that the function $F(x) = e^{-x} f(x)$ has the same properties as $f(x)$, allowing us to write also
$$\int\limits_0^a F(x)dx = const? \ \ \ \ \ \ \ \ \ \ (3)$$
EDIT: An edit was made to simplify the question.
The conditions $$\int\limits_{\varepsilon}^{\varepsilon + a} f(x)dx = \int\limits_{2\varepsilon}^{2\varepsilon + a} f(x)dx = ... = \int\limits_0^a f(x)dx = const$$are satisfied by any constant function f(x)=c.
Thus $F(x) =f(x)e^x = ce^x$ and $$ \int\limits_0^a F(x)dx = c(e^a-1)$$