How Does Ramanujan’s Master Theorem Work? I am trying to understand Ramanujan' Master theorem, but I really can't seem to wrap my head around it. So the theorem states that if you have a function $F$ such that you can write it as

$$F(x)=\sum_{k=0}^\infty \frac {w(k)(-x)^k}{k!}$$ 

for some analytic or integrable function $w$, then the Mellin transform of $F$ is

$$I=\int_0^\infty x^{n-1}F(x)dx= \Gamma(n)w(-n).$$

However, there are some related parts that confuse me:


*

*The definition of $w$ by the series in the first expression just requires $w$ to take certain values for $k=0,1,\ldots,\infty$, but not for any values in $\mathbb{R}\setminus\mathbb{N}_0$. Is this really enough to uniquely define $w$? If so why?


A typical example would be $w(k)=1$ for all $k$. In that case, we get $F(x)=e^{-x}$ and $I=\Gamma(n)$. However, if I instead use $w(k)=cos(2\pi k)$, then I still get $F(x)=e^{-x}$. However, for this $w$ I get $I(n)=\Gamma(n)cos(-2\pi n)$, which agrees on $n\in \mathbb{Z}$ but not anywhere else. Obviously, these can't both be correct. Both $w$ are integrable, so why is one of the choices of $w$ incorrect? 


*(this is likely related to the first) Why do we require the minus in the expression? If we define some function $\hat{w}$ such that $\hat{w}(k)=(-1)^k w(k)$ for $k\in\mathbb{N}_0$, then writing the series in terms of $\hat{w}$ would give the more familiar Maclauren series.


As an example take $w(k)=cos(2\pi k)$. No if I choose $\hat{w}(k)=(cos(\pi k))^2$ (note the factor $2$ is missing), then

$$F(x)=\sum_{k=0}^\infty \frac {\hat{w}(k)(x)^k}{k!}$$ 

is a Maclauren series for $e^{-x}$. Why can't we get a similar result for the Maclauren series?
 A: 
You should think to the residue theorem and Mellin inversion. The decay and analycity conditions implies that knowing $\varphi(-k)$ is enough to know $\varphi(s)$.

Let $ \varphi(s)$ be analytic on $\Re(s) \le 0$, such that $e^{-|t|\pi / 2 }\varphi(it)$ has exponential decay, $\varphi(0) = 0$, and such that for some $x_0 > 0$ $$\lim_{T \to \pm \infty} \int_{-\infty}^0 |\Gamma(\sigma+iT) \varphi(\sigma+iT) x_0^{-\sigma}| d\sigma = 0,\qquad \lim_{\sigma \to -\infty} x_0^{-\sigma}\int_{-\infty}^\infty |\Gamma(\sigma+it) \varphi(\sigma+it)| dt = 0$$
then


*

*for all $x > 0$ $$f(x)=\frac{1}{2i\pi}\int_{-i\infty}^{i\infty} \Gamma(s)\varphi(s)x^{-s}ds$$ $f$ is analytic and bounded on $x > 0$

*for $x \in (0,x_0]$ $$\frac{1}{2i\pi}\int_{-i\infty}^{i\infty} \Gamma(s)\varphi(s)x^{-s}ds = \sum Res(\Gamma(s)\varphi(s)x^{-s}) = \sum_{k=1}^\infty \varphi(-k) \frac{(-1)^k}{k!}x^k$$
Thus $f(x), x >0$ is the analytic continuation of $\sum_{k=1}^\infty \frac{\varphi(-k) (-1)^k}{k!}x^k , |x| < |x_0|$ and by inverse Fourier/Mellin transform, for $\Re(s) \in (-1,0)$ $$\Gamma(s)\varphi(s) = 
 \int_0^\infty f(x) x^{s-1}dx$$
