I was searching for an elementary proof of the Ramanujan Master Theorem and I found a page from Ramanujan's Notebook on wikipedia which contained the proof. I think that it has some gaps, so can anyone please explain the proof. enter image description here

Any help will be appreciated. Thanks.

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    $\begingroup$ Berndt discusses Ramanujan's proof in Ramanujan's Notebooks, Part 1, page 298. $\endgroup$ – nospoon Jun 8 '16 at 12:26
  • $\begingroup$ @nospoon : can you tell us if it assumes $\phi(z)$ is entire, or meromorphic ? $\endgroup$ – reuns Jun 8 '16 at 13:10
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    $\begingroup$ @nospoon Thanks! Can you add that as an answer so that it no longer showed unanswered? $\endgroup$ – Henry Jun 8 '16 at 13:13

As noted in the comments, Berndt discusses Ramanujan's Master theorem and some of its applications in Ramanujan's Notebooks, Part I (1985)(page 297).

He presents and explains Ramanujan's own proof (which is what you seek), and its difficulties/legitimacy.

His discussion is complete with a rigorous proof by Hardy of a version of this theorem, including the specific conditions under which it applies.

EDIT. Hardy's version: Hardy's version.

  • $\begingroup$ ok but what are those conditions ? $\endgroup$ – reuns Jun 8 '16 at 13:43
  • $\begingroup$ @user1952009 see edit. $\endgroup$ – nospoon Jun 8 '16 at 14:18
  • $\begingroup$ he uses the residue theorem, and then we'll need the reflection formula $\frac{\pi}{\sin(\pi s)} = \Gamma(s) \Gamma(1-s)$ for proving OP's formula. so I think there might be a simpler proof. and he doesn't consider after $\Psi(x) = \sum_k \frac{\psi(k)}{k!} (-x)^k$ instead of $\Psi(x) = \sum_k \psi(k) (-x)^k$ ? $\endgroup$ – reuns Jun 8 '16 at 14:39
  • $\begingroup$ Thanks! I saw the proof in the book. $\endgroup$ – Henry Jun 11 '16 at 14:53

There is only one gap which is actually a rather trivial step if you have seen it before. The first line is just the integral representation of $n!$, the second line is the theorem we want to prove. The third and fourth line is a shorthand proof of the integral representation of $n!$, it says that minus the primitive function evaluated at zero yields $n!$. So, until this point nothing interesting has happened apart from the statement of the theorem. Then what he does amounts to noting that:

$$\int_{0}^{\infty}x^{n-1}\exp\left(-r^kx\right)\;\mathrm{d}x = \frac{(n-1)!}{r^{nk}}$$


$$\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}\int_{0}^{\infty}x^{n-1}\exp\left(-r^kx\right)\;\mathrm{d}x = (n-1)!\sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}\frac{1}{r^{nk}}$$

which is what you get if you sum the general term of the expressions given at lines 5 till 7. The summation on the right hand side yields (assuming the function is analytic):

$$(n-1)! f\left(\frac{1}{r^n}\right)$$

This is the right hand side of line 10 after the statement "Adding up all the results we have". The gap is how he obtained the left hand side. But this is just a matter of expanding the exponential in the integral in powers and interchanging the two summations. This yields:

$$\int_{0}^{\infty}\sum_{m=0}^{\infty}\sum_{k=0}^{\infty}(-1)^m r^{km} x^m\frac{f^{(k)}(0)}{k!m!}x^{n-1}\;\mathrm{d}x $$

The summation over $k$ yields:

$$\int_{0}^{\infty}\sum_{m=0}^{\infty}(-1)^m x^m\frac{f\left( r^{m}\right)}{m!}x^{n-1}\;\mathrm{d}x $$

which is the left hand side of line ten. We can write line ten as:

$$\int_{0}^{\infty}x^{n-1}\sum_{m=0}^{\infty}(-1)^m x^m\frac{f\left( r^{m}\right)}{m!}\;\mathrm{d}x = (n-1)! f\left(\frac{1}{r^n}\right)$$

Then, as the next line says, it's a matter of putting $ f\left(r^n\right) = \phi(n)$ to obtain the theorem:

$$\int_{0}^{\infty}x^{n-1}\sum_{m=0}^{\infty}(-1)^m \frac{\phi\left(m\right)}{m!}x^m \;\mathrm{d}x = (n-1)!\phi\left(-n\right)$$

Formal derivations like this that involve only manipulations of power series can be handled more efficiently using umbral calculus methods, a special case of Ramanujan's master theorem was actually obtained by Glaisher using such methods (the last statement cr. 4 for $n = 0$). Umbral calculus involves deliberately confusing subscripts of coefficients as exponents performing the computations and then at the end putting the indices back where they belong. So, in the integral:

$$\int_{0}^{\infty}x^{n-1}f(x) \;\mathrm{d}x$$


$$f(x) = \sum_{k=0}^{\infty}(-1)^k\frac{c_k}{k!}x^k$$

we would write $f(x)$ as:

$$f(x) = \sum_{k=0}^{\infty}(-1)^k\frac{c^k}{k!}x^k = \exp(-c x)$$

This yields:

$$\int_{0}^{\infty}x^{n-1}f(x) \;\mathrm{d}x = \int_{0}^{\infty}x^{n-1}\exp(-c x) \;\mathrm{d}x = \frac{(n-1)!}{c^n}$$

And then we change the exponent back to an index, which yields the theorem:

$$\int_{0}^{\infty}x^{n-1}f(x) \;\mathrm{d}x = (n-1)! c_{-n}$$

  • $\begingroup$ did Ramajuan even prove it rigorously, or just made a formal derivation and then applied to particular cases where the derivation could be made rigorous ? $\endgroup$ – reuns Jun 13 '16 at 0:59
  • $\begingroup$ anyway +1 for having decoded what he wrote $\endgroup$ – reuns Jun 13 '16 at 1:00
  • $\begingroup$ @user1952009 (I just upvoted your question, not sure why it was downvoted.. )I don't think Ramanujan had a rigorous version, otherwise Hardy would have had something to say about that... $\endgroup$ – Count Iblis Jun 13 '16 at 1:02

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