How to interpret the constant of integration/endpoints of the sum of derivatives linear operator Consider the linear operator expression
$$L[f] = f + f' + f'' + f''' + f'''' + ... $$
This doesn't always converge but for some expressions it does converge (such as polynomials or linear combinations of exponentials $e^{ax}$ where $|a| < 1$)
A natural question to ask is if this operator has a closed form. Using a technique almost identical to the proof of the geometric series combined with integration factors we can see that
$$ L[f] - f = \frac{d}{dx}[L[f]] $$
$$ \frac{d}{dx}[L[f]] - L[f] = -f $$
$$  e^{-x} \frac{d}{dx}[L[f]] - e^{-x} L[f] = -e^{-x} f $$
$$ e^{-x} L[f] = - \int e^{-x} f dx $$
$$ L[f] = -e^x \int e^{-x} f dx $$
So this "closed form" is not exactly well defined, theres an indefinite integral in there which means it varies by an arbitrary constant. Now if you take a nice polynomial (ex: $x^2$) And apply this formula without even understanding it you can end up with a correct answer:
$$ L[x^2] = x^2 + 2x + 2 $$
$$ -e^x \int e^{-x} x^2 dx  = -e^x \left( e^{-x} \left( x^2 + 2x + 2 \right) + C \right)$$
And now we just "set C to 0" and voila we have perfect agreement.
And so while it possible to "experimentally" check this formula works, setting C to 0 makes absolutely no sense in the general context of functions. How do you know what C is supposed to be 0? In this case setting $C$ to is 0 is equivalent to saying $f(0)=2$ and how on earth did you know that $f(0)=2$ and not $3$ or something else in general.  You need some initial conditions/ideas on where your bounds start/end.
Is there a way to make this formula into something more explicit / rigorous by either defining what the indefinite integral's bounds should be OR by describing a procedure (maybe its not guaranteed to halt in general) that can tell you what value of C is the right value?
A possible strategy:
If $f(x)$ has an asymptote as $x \rightarrow \pm \infty$ then perhaps the correct $C$ value is one that lets one of these asymptotic values be $0$. But this is such a frankenstein arbitrary kind of rule. It can be codified as, select $c$ such that
$$ \lim_{x \rightarrow \infty} \int_{c}^{x} e^{-x}f(x) dx  = 0$$
Then that is your integration bound OR if you prefer the language of constants then your additive constant $C = -e^{-c}f(c)$
 A: Per Ben Grossman's suggestion we can define $L_n = f + f' + ... f^{(n)}$ and then consider
$$ \frac{d}{dx} L_n[f] - L_n[f] = f^{(n+1)} - f  $$
We can try to find a closed form for $L_n$ in much the same way
$$  L_n[f] = e^x\int_{a}^{x} e^{-x} \left(f^{(n+1)} - f  \right) $$
Now we have to define this $\alpha$. We can break up the sum into:
$$ e^x\int_{a}^{x} e^{-x} f^{(n+1)} dx - e^x \int_{a}^{x} e^{-x}  f   dx$$
The left hand side expression $\int_{a}^{x} e^{-x} f^{(n+1)} dx$ can be evaluated using integration by parts as:
$$ -e^{-x} f^{(n+1)}|_a^{x}  + \int_{a}^{x} e^{-x} f^{(n+2)} dx $$
We can repeat this infinitely many times to yield
$$ -e^{-x} f^{(n+1)}|_a^{x} - e^{-x} f^{(n+2)}|_a^{x} -e^{-x} f^{(n+3)}|_a^{x} ... = \\ -e^{-x}f^{(n+1)}(x) + e^{-a}f^{(n+1)}(a)  - e^{-x}f^{(n+2)}(x)+e^{-a}f^{(n+2)}(a)+...$$
Now we can multiply by $e^x$ to make our lives a bit easier (And do a permutation of terms which we haven't rigorously shown to converge but c'est la vie)
$$e^x \int_{a}^{x} e^{-x} f^{(n+1)} dx  = -f^{(n+1)}(x)-f^{(n+2)}(x).... + e^x \left( e^{-a}f^{(n+1)}(a) + e^{-a}f^{(n+2)}(a) +... \right) $$
We can do the same now for the expression $-e^x \int_{a}^{x} e^{-x}  f   dx$
$$f^{(0)} + f^{(1)} + ... - e^x \left( e^{-a} f^{0}(a) + e^{-a} f^{1}(a) + ...  \right) $$
So now we look at the sum $ e^{x} \int_{a}^{x} e^{-x} f^{(n+1)} \  dx - e^{x} \int_{a}^{x} e^{-x} f \ dx $  which naturally has some massive cancellation yielding
$$ f^{(0)} + f^{(1)} + ... f^{(n)} - \left(e^{-a} f^{0}(a) + e^{-a} f^{1}(a) + ... e^{-a} f^{(n)}(a) \right) $$
So there many choices of $a$ to make $$\left(e^{-a} f^{0}(a) + e^{-a} f^{1}(a) + ... e^{-a} f^{(n)}(a) \right) = 0 $$
But if assume that $f$ is bounded in a certain vague sense (and that it's derivatives are bounded too) then letting $a = \infty$ should kill these terms thus we conclude that for sufficiently nice $f$
$$L_n[f] = e^x \int_{\infty}^{x} e^{-x} \left(f^{(n+1)} - f \right) \ dx $$
And letting $n \rightarrow \infty$ we then have
$$ L[f] =  -e^x \int_{\infty}^{x} e^{-x} \left( f \right) \ dx$$
