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?