Ramanujan's Master Theorem relation to Analytic Continuation

Question

$\DeclareMathOperator{Re}{Re}$ To provide some background, this is a question based on establishing the identity $$\int_0^\infty \frac{v^{s-1}}{1+v}\,dv=\frac{\pi}{\sin \pi s},\qquad 0<\Re s<1$$using Ramanujan's Master Theorem (RMT). See the question asked here for the full details related to it.

In this answer by user @mrtaurho, we use the geometric series expansion $$\frac{1}{1+v}=\sum_{k=0}^\infty (-v)^k.$$However, the geometric series clearly isn't valid for $|v|\geq 1$, and it was explained that the radius of convergence didn't play a role in the proof of RMT and so doesn't matter too much here, only that the underlying structure is revealed when considering the geometric series representation.

This made me think of analytic continuation since we only need an identity to be valid in some region to be able to extend a function outside of that region; it seems like the geometric series is only valid for $|v|<1$ but this allows us to extend the connection beyond just $(0,1)$.

So my question is this:

Is there some connection between RMT and analytic continuation? Or perhaps the connection is finer than RMT and is only a small part of some detail in its proof?

Answer 1

You can show $$\int_0^\infty \frac{x^{s-1}}{1+x}dx=\frac{\pi}{\sin \pi s},\qquad \Re (s) \in (0,1)$$ in term of the analytic-ness of the inverse Fourier/Laplace/Mellin transform of meromorphic functions with exponential decay, the argument shouldn't very different to the proof of Ramanujan master theorem.

1. $\frac{\pi}{\sin(\pi s)}$ is Schwartz on vertical lines without poles thus for $\Re(s) \in (0,1)$, $\frac{\pi}{\sin(\pi s)} = \int_0^\infty x^{s-1}h(x) dx$ with $h(e^u)e^{\sigma u}$ Schwartz for $\sigma \in (0,1)$. Moreover $\frac{\pi}{\sin(\pi s)}$ has an exponential decay on those vertical lines, thus $h$ is analytic on $(0,\infty)$.

2. Let $$F(s) = \int_0^\infty x^{s-1}\frac{1_{x < 1}}{1+x}dx = \sum_{k=0}^\infty (-1)^k \int_0^1 x^{s-1+k}dx = \sum_{k=0}^\infty \frac{(-1)^k}{s+k} $$

   $\frac{\pi}{\sin(\pi s)}-F(s)$ is analytic for $\Re(s) < 1$ and it converges uniformly to $0$ as $\Re(s) \to - \infty$ and it is $L^2$ on vertical lines $\Re(s) \in (0,1)$. Thus $\frac{\pi}{\sin(\pi s)}-F(s) = \int_0^\infty x^{s-1}g(x)1_{x > 1}dx$ for some function $g $.

3. Since $\frac{\pi}{\sin(\pi s)}-F(s)=\int_0^\infty x^{s-1} (h(x)-\frac{1_{x < 1}}{1+x})dx$ then $$g(x)1_{x > 1} = h(x)-\frac{1_{x < 1}}{1+x}$$ so $h(x)- \frac{1}{1+x}$ vanishes on $(0,1)$ and since $h$ is analytic it implies $$h(x) = \frac{1}{1+x}$$

Answer 2

Lets take a closer in the formulation of Ramanujan's Master Theorem as it is written here.

If $F(x)$ is expanded in the form of Maclaurin's series $$F(x)=\sum_{n=0}^\infty\left\{(-1)^n\frac{\mathrm d^nF(x)}{\mathrm dx^n}\right\}_{x=0}\frac{(-x)^n}{n!}$$ then Ramanujan asserts that the value of $I=\int_0^\infty x^{s-1}F(x)\mathrm dx$ can be found from the coefficient of $\frac{(-x)^n}{n!}$ in the expansion of $F(x)$. Conversely Ramanujan claims that if the value of $I$ is known, then the Maclaurin's coefficient of $F(x)$ can be found.

Rephrasing the above slightly we are left with the claim that the Mellin Transform of a function which posses a MacLaurin Expansion of the aforementioned form can directly be deduced from the corresponding coefficient. The crucial point here is $-$ and I have to admit that I was not aware of this either for to long $-$ has to be in fact a MacLaurin Expansion; nothing else will work nor is allowed to be used here.

In the case of our well-known geometric series we are somewhat tricked by the assumption that we are dealing with a geometric series and not with the MacLaurin Expansion of $f(x)=\frac1{1+x}$. However, this is precisely what we are doing. It is not hard to be shown that the $n$th derivative of $f(x)$ are given explicitly by

$$f^{(n)}(x)=\frac{\mathrm d^n}{\mathrm dx^n}\frac1{1+x}=(-1)^n\frac{n!}{(1+x)^{n+1}}$$

Now, by plugging this general formula in the series from above, we get

$$F(x)=\sum_{n=0}^\infty\left\{(-1)^n\left[(-1)^n\frac{n!}{(1+x)^{n+1}}\right]\right\}_{x=0}\frac{(-x)^n}{n!}=\sum_{n=0}^\infty (-x)^n$$

So $f(x)$ fulfills the conditions to be tackeld with Ramanujan's Master Theorem hence we can actually obtain a suitable Maclaurin's series. Of course, we could also observe that

$$1-x+x^2-x^3+\cdots=\sum_{n=0}^\infty (-x)^n=\frac1{1+x}~~~|x|<1$$

Which would be our well-known geometric series. However, it is more of a coincidence rather than a general fact that two, so differently obtained series are in fact the same. I have to admit that we can deduce the radius of convergence of our MacLaurin Series and would come to the same result that $|x|<1$ but that is not of relevance since we are more interested in the structure which on the other hand is precisely prescripted.

EDIT

For myself I cannot judge the reliability of a mathematical source so I will leave this part to you. After some research I found this book Theory of Differential Equations in Engineering and Mechanics aswell as this book Ramanujan's Notebook and this article An Analogue of Ramanujan’s Master Theorem all refering to a MacLaurin Expansion instead of just a series expansion of the form[...]. Other sources only rely on the indefinite series expansion as aforementioned. However, I will hope this can perhaps clear your concerns.