13

Ramunajan's Master Theorem states that if a complex-valued function $f(x)$ has an expansion of the form

$$\displaystyle f(x)=\sum _{k=0}^{\infty }{\frac {\,\varphi (k)\,}{k!}}(-x)^{k}$$

then the Mellin transform of $f(x)$ is given by

$$\displaystyle \int _{0}^{\infty }x^{s-1}\,f(x)\,\operatorname {d} x=\Gamma (s)\,\varphi (-s)$$

Here $\varphi(s)$ is some function (say analytic or integrable).

Now, what about the converse of this? Say that we know that the Mellin transform of $f(x)$ is equal to $\Gamma (s)\,\varphi (-s)$, is it then true that $f(x)$ has an infinite expansion in the form given above?

I couldn't find anything about this question on Wikipedia or somewhere else.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 27,005
Casimir Rönnlöf
  • 1,054

1 Answers1

5

For a partial converse to the Master Theorem, note that if $\mathcal M[f(x)]=\Gamma(s)\varphi(-s)$ then $$f(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}x^{-s}\Gamma(s)\varphi(-s)\,ds.$$ The poles of $\Gamma$ are simple and are the non-positive integers, so the residue at an integer $-t\le0$ is $$\lim_{s\to-t}(s+t)\Gamma(s)=\lim_{s\to-t}\frac{\Gamma(s+t+1)}{\prod\limits_{i=0}^{t+1}(s+i)}=\frac{(-1)^t}{t!}.$$ Thus if $\varphi$ has no singularities and does not have roots at the non-positive integers then the residue theorem gives $$f(x)=\sum_{t\ge0}\operatorname{Res}(x^{-s}\Gamma(s)\varphi(-s),-t)=\sum_{k=0}^\infty\frac{\varphi(k)(-x)^k}{k!}$$ which is the original statement.

In Berndt's Ramanujan's Quarterly Reports1, it is noted that

In the final section of the first report, Ramanujan derives certain expansions for four functions by assuming that a type of converse theorem to the Master Theorem holds. More specifically, he determines a power series for the integrand from the value of the integral. In fact, Ramanujan's converse to the Master Theorem follows from the inversion formula for Mellin transforms. Although Ramanujan proceeded formally, all of the results that he obtains are, indeed, correct.

(emphasis mine)

The four functions considered are

  1. $\left(2/(1+\sqrt{1+4x})\right)^n=p_*^{-n}$ where $p_*$ is the positive root of $p^2-p-x$, giving $\varphi(q)=n\Gamma(n+2q)/\Gamma(n+q+1)$;

  2. $\left(x+\sqrt{1+x^2}\right)^{-n}=e^{-n\operatorname{arcsinh}x}$, giving $\varphi(q)=n2^{q-1}\Gamma((n+q)/2)/\Gamma((n-q)/2+1)$;

  3. $\int_0^\infty a^{q-1}x^n\,da$ where $a\ge0$, $n>0$ and $x$ solves $\log x=ax$, giving $\varphi(q)=n(n+q)^{q-1}$;

  4. $\int_0^\infty a^{r-1}x^n\,da$ where $x$ solves $aqx^p+x^q=1$ with $a>0$, $0<q<p$ and $0<pr<n$, giving $\varphi(r)=nq^{r-1}\Gamma((n+pr)/q)/\Gamma((n+pr)/q-r+1)$.

Evidently in all these cases $\varphi$ is not analytic in the whole left-plane, but I suspect the cancelling of gamma terms with $\Gamma(-s)$ may be why the identity still holds.

Reference

[1] Berndt, B. C. (1984). Ramanujan's Quarterly Reports. Bulletin of the London Mathematical Society. 16(5):449-489.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 27,005