The Ramanujan master theorem states the following. If, for some analytic function $f$, there is a sufficiently nice function $\phi$ such that
$$f(x)=\sum\limits_{n=0}^{\infty}\frac{(-x)^n}{n!}\phi(n)$$
then
$$\int\limits_{0}^{\infty}x^{a-1}f(x)dx=\Gamma(a)\phi(-a)$$
A sufficient analytic function can be formulated as
$$f(x)=\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}f^{(n)}(0)=\sum\limits_{n=0}^{\infty}\frac{(-x)^n}{n!}(-1)^{n}f^{(n)}(0)$$
where the $\phi(n)$ in this case is $(-1)^{n}f^{(n)}(0)$
Let us say that
$$\int f(x)dx=F(x)+C$$
Assuming convergence, and writing $\lim\limits_{t\to\infty}g(t)=g(\infty)$ for compactness, we can get that
$$\int\limits_{0}^{\infty}f(x)dx=F(\infty)-F(0)$$
Then, by the Ramanujan master theorem, we can equivalently do
$$\int\limits_{0}^{\infty}f(x)dx=\int\limits_{0}^{\infty}x^{1-1}f(x)dx=\Gamma(1)\phi(-1)=(-1)^{-1}f^{(-1)}(0)=-F(0)$$
Given that $F(\infty)$ need not be $0$, such as with $\arctan(\infty)$ or $\tanh(\infty)$, am I misunderstanding what $f^{(-1)}$ comes to represent? Does it mean some "principal" antiderivative such that this equality holds? $F(\infty)-F(0)$ will be equal for all antiderivatives $F$, but $-F(0)$ won't be so.
I guess my issue is that we can get non-trivially different antiderivatives.
$$\frac{x^2}{2}+x+C_{1}=\int x+1 dx=\frac{1}{2}(x+1)^2+C_{2}$$
$$\frac{1}{2}\ln|x|+C_{1}=\int\frac{dx}{2x}=\frac{1}{2}\ln|2x|+C_{2}$$
Both examples would diverge if we tried to integrate them on $[0,\infty)$, but I'm using them to show that we can get naturally different expressions using valid means, and if we removed the constants of integration they would end up being not equal.