Here, the Ramanujan Master Theorem is proven in the following way:
$$ \begin{split} \int_0^\infty x^{s-1}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!} \lambda(n)x^n & = \int_0^\infty x^{s-1}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!} E^n\lambda(0)x^n \\ & = \int_0^\infty x^{s-1}e^{-Ex}dx \;\lambda(0) = \frac{\Gamma(s)}{E^s}\lambda(0) = \Gamma(s)\lambda(-s). \end{split}$$ This is a very clever trick.
Questions
When can one assume that an operator to perform this trick exists? This obviously cannot be a number as $\lambda$ does not necessarily increase exponentially.
Why can't I write $$ E^{-s}\lambda(0) = \frac{E^{-s}\lambda(1)\lambda(0)}{\lambda(1)} = \frac{\lambda(1-s)\lambda(0)}{\lambda(0)} $$ instead? Or any other combination for that matter? Intuitively, a $\lambda$ cannot just be treated as a constant for the operator. But then again, $\lambda(0)$ is just a number which makes it confusing.
And for that matter, why can I pull out the Gamma function outside of the operator? There might be an argument $r$ for lambda that would make it $\lambda(r)=\Gamma(s)\lambda(0)$ and hence Gamma would have to be included in the argument of the operator, as $E^{-s}\Gamma(s)\lambda(0)=\lambda(r-s)$.
Generally, when can I say $E(a\lambda(arg)) = aE \lambda(arg) = a\lambda(arg+1)$?
Existence of Operator for the Trick: The operator $E$ here is not a number but a formal symbolic representation that acts on the function $\lambda(n)$. The assumption that such an operator can perform this trick hinges on the series and the integral converging and the operations being interchangeable under certain conditions. Specifically, the series inside the integral must converge uniformly for the swap between summation and integration to be valid. The operator exists in the formal sense that it manipulates the series in a way that adheres to the rules of calculus and analytical functions, often under the assumption that $\lambda(n)$ is sufficiently smooth or has specific properties that allow for such manipulation. Source : https://en.wikipedia.org/wiki/Shift_operator
Manipulation with $\lambda$ Function: Your question about writing $E^{-s}\lambda(0) = \frac{E^{-s}\lambda(1)\lambda(0)}{\lambda(1)} = \frac{\lambda(1-s)\lambda(0)}{\lambda(0)}$ touches on the properties of the $\lambda$ function and the operator $E$. The issue with this manipulation is that it assumes $\lambda$ behaves linearly or multiplicatively with respect to $E$, which is not necessarily true. The $\lambda$ function's behavior under the action of $E$ is not simply multiplication or division; it is defined by how $E$ acts on the indices of $\lambda$. The direct multiplication/division within the arguments of $\lambda$ outside of its established relationship with $E$ misinterprets the nature of both $\lambda$ and $E$.
Pulling Out the Gamma Function: The Gamma function $\Gamma(s)$ can be pulled out due to the specific form of the integral $\int_0^\infty x^{s-1}e^{-Ex}dx$ which directly corresponds to the definition of the Gamma function, $\Gamma(s) = \int_0^\infty x^{s-1}e^{-x}dx$, scaled by a factor of $E$. The operation is valid under the conditions that allow the interchange of integration and summation, and where the integral itself can be seen as defining a Gamma function. The argument about including $\Gamma(s)$ inside the operator or $\lambda$ misunderstands the nature of the proof where $\Gamma(s)$ emerges from the integration process itself, not from the action of $E$ on $\lambda$.
General Statement on Operator $E$: Saying $E(a\lambda(\text{arg})) = aE\lambda(\text{arg}) = a\lambda(\text{arg}+1)$ assumes a linear property of the operator $E$ and its action on $\lambda$. This is generally valid within the context of the operations defined by the proof, where $E$ acts to shift the index of $\lambda$. This linearity and shift are part of the formal definition of how $E$ interacts with $\lambda$ in this context but need to be carefully justified based on the properties of $\lambda$ and the convergence of the series and integrals involved.