$\DeclareMathOperator{Re}{Re}$ 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?

  • 3
    $\begingroup$ (+1) I appreciate this question hence your doubts made me thinking by myself and I am not able to solve this issue rigorously. $\endgroup$ – mrtaurho Feb 13 '19 at 19:07
  • $\begingroup$ It's actually more about analytically continuing the series expansion coefficients that are defined as a function of a natural number to real or complex numbers. In this case you need that the $c_k$ in $(-1)^kc_k$ which is equal to $1$ as a function of $k$ should be continued as function of real $s$ as the constant function equal to $1$. $\endgroup$ – Count Iblis Feb 13 '19 at 19:31

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 $$h(x) = \frac{1}{1+x}$$

  • $\begingroup$ I like the way you've established the identity, but I'm going to refrain from accepting this answer for now as it doesn't seem to address my question directly (I'm going to look over your answer more closely, so more might come of it than I see on just a cursory glance). $\endgroup$ – Clayton Feb 13 '19 at 20:45
  • $\begingroup$ @Clayton You can adapt the argument of my answer to obtain a proof of the RMT assuming $\phi(s)$ is analytic for $\Re(s) < 1$ and $\Gamma(s) \phi(-s)$ has exponential decay on vertical lines and $\Gamma(s) \phi(-s)-\sum_{k\ge 0} \frac{\phi(k)}{k!}\frac{1}{s+k}$ tends$ \to 0$ uniformly as $\Re(s) \to -\infty$ $\endgroup$ – reuns Feb 13 '19 at 22:04
  • $\begingroup$ I'm beginning to understand more, and I think your answer is helping with it. At least, based on looking at your answer, it looks like analytic continuation is vital here. Specifically, you're deducing that because $h(x)-\frac{1}{1+x}$ vanishes on $(0,1)$, they must be equal on $(0,\infty)$. I'm just trying to make the connection via the geometric series that mrtaurho uses in his answer to the question linked in the original post (that is, while what he does turns out to be correct, is it rigorous and if so, does its rigor follow from analytic continuation?) $\endgroup$ – Clayton Feb 13 '19 at 23:38

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.


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.

  • $\begingroup$ Is there any chance you have a more reputable source for the way RMT has been formulated here? Based on a few quick searches, it seems that the journal you've cited is unreliable (perhaps that paper is still of high quality, but I hadn't heard of the journal and it turns out that they've changed the name of the journal to Journal of Advances in Mathematics and Computer Science...this makes me feel a little more concerned about citing articles from such a journal). $\endgroup$ – Clayton Feb 13 '19 at 20:43
  • $\begingroup$ Except for the above concern, I think this answer the question rather well. Thank you. $\endgroup$ – Clayton Feb 13 '19 at 20:46
  • $\begingroup$ @Clayton Unfortunately not. WolframMathWorld does not go into that detail I would not call Wikipedia a more reliable source than the paper I linked. The problem, which I stumbled upon a while ago within a discussion with the user Did, is the uniqueness of the required expansion. Fortunately the linked paper redefines what expansion is needed. Anyway, I am glad that my answer helped you nevertheless. $\endgroup$ – mrtaurho Feb 13 '19 at 20:58
  • $\begingroup$ @Clayton I edited my answer after I have done some research on the issue of MacLaurin Expansions in the RMT or not. $\endgroup$ – mrtaurho Feb 13 '19 at 21:29
  • $\begingroup$ I'm not sure it was intended this way, but I think the first two links you provide lead to the same place (Ramanujan's Notebook). Either way, I'll look at these links closely, especially Berndt's book. $\endgroup$ – Clayton Feb 13 '19 at 21:52

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