The probability mass function of the Binomial distribution is given by
$$\begin{equation}
f(x)=\binom{n}{x} p^x (1-p)^{n-x},
\end{equation}$$
where $p \in [0,1]$ and $x=\{0,1,\dots,n\}$ (finite support).
The Mellin transform of $f$ is therefore
$$\begin{align}
\mathcal{M}\{f\}(s)&=\int_0^{\infty}x^{s-1} \binom{n}{x} p^x (1-p)^{n-x}\mathrm{d}x\\
&=\sum_{k=0}^{n}k^{s-1} \binom{n}{k} p^k (1-p)^{n-k}.
\end{align}$$
Since $f$ is a probability mass function, the integral becomes a sum. Now I'd like to apply the inverse Mellin transform to get back $f$:
$$\begin{align}
\mathcal{M}^{-1}\{\mathcal{M}\{f\}\}(x) &= \frac{1}{2 \pi i} \int_{c-i\infty}^{c+i\infty} x^{-s} \left( \sum_{k=0}^{n}k^{s-1} \binom{n}{k} p^k (1-p)^{n-k} \right) \mathrm{d}s \\
&=\frac{1}{2 \pi i} \sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} \int_{c-i\infty}^{c+i\infty} x^{-s} k^{s-1} \mathrm{d}s.
\end{align}$$
The problem is that the integral $\int_{c-i\infty}^{c+i\infty} x^{-s} k^{s-1} \mathrm{d}s$ doesn't seem to converge. Essentially it's equal to
$$
\begin{align}
\int_{c-i \infty}^{c+i\infty} x^{-s}k^{s-1} \mathrm{d} s &= \frac{1}{x} \int_{c-i \infty}^{c+i\infty} \left(\frac{k}{x}\right)^{s-1} \mathrm{d}s\\
&=\frac{i}{x} \int_{-\infty}^{\infty} e^{it \ln \left({\frac{k}{x}} \right)} \mathrm{d}t.
\end{align}
$$
So the integrand is forever oscillatory, and doesn't seem to have a value. I did read that it "equals" the dirac Delta distribution, but I'm not sure how it'll fit in with the other terms since that's not really a function. But at the same time, since $f$ has support on a finite set, it kind of makes sense for it to show up?
I'm pretty lost, how can I recover $f$?
Best Answer
Its not the Mellin transform of
$$f(x)=\binom{n}{x}\, p^x\, (1-p)^{n-x}\tag{1},$$
but rather the Mellin transform of
$$g(x)=\sum\limits_{k=0}^n f(x)\, \delta (x-k)\tag{2}$$
which, assuming the convention $\int\limits_0^\infty \delta(x)\, dx=1$ (versus $\int\limits_0^\infty \delta(x)\, dx=\frac{1}{2}$ for example), is
$$\mathcal{M}_x[g(x)](s)=\sum\limits_{k=0}^n \left(\int_0^{\infty} f(x)\, \delta (x-k)\, x^{s-1} \, dx\right)\\=\sum\limits_{k=0}^n f(k)\, k^{s-1}=\sum\limits_{k=0}^n \binom{n}{k}\, p^k\, (1-p)^{n-k}\, k^{s-1}\tag{3}$$
and this explains the presence of the Dirac delta terms in the inverse Mellin transform.
But you still need to normalize it such that $$\sum\limits_{k=0}^n \left(\int\limits_{-\infty}^{\infty} f(x)\, \delta (x-k) \, dx\right)=\sum\limits_{k=0}^n f(k)=\sum\limits_{k=0}^n \binom{n}{k}\, p^k\, (1-p)^{n-k}=1\tag{4}.$$
The remainder of this answer illustrates an alternate perspective on the problem.
Assuming the definition
$$f(k)=\left\{\begin{array}{cc} \binom{n}{k}\, p^k\, (1-p)^{n-k} & 0\leq k\leq n \\ 0 & k>n \\ \end{array}\right.\tag{5}$$
consider the summatory step function
$$h(x)=\sum\limits_{k=0}^x f(k)\tag{6}$$
which can be viewed as sort of a Cumulative Distribution Function (CDF) with first-order derivative
$$h'(x)=\sum\limits_{k=0}^x f(k)\,\delta(x-k)\tag{7}$$
which can be viewed as sort of a Probability Density Function (PDF).
Now consider the Mellin transform
$$H(s)=s\, \mathcal{M}_x[h(x)](-s)=s\, \int_0^{\infty} h(x)\, x^{-s-1} \, dx\\=2 \pi\, f(0)\, s\, \delta(-i s)+\sum\limits_{k=1}^n \frac{f(k)}{k^s}\tag{8}$$
which is equivalent to the Mellin transform
$$H(s)=\mathcal{M}_x[h'(x)](1-s)=\int_0^{\infty} h'(x)\, x^{-s} \, dx\\=2 \pi\, f(0)\, s\, \delta(-i s)+\sum\limits_{k=1}^n \frac{f(k)}{k^s}\tag{9}$$
The function $h(x)$ is recovered from the inverse Mellin transform
$$h(x)=\mathcal{M}^{-1}_s\left[\frac{H(s)}{s}\right]\left(\frac{1}{x}\right)=\frac{1}{2 \pi i} \int_{\alpha-i \infty}^{\alpha+i \infty} \frac{H(s)}{s}\, x^s \, ds=\sum\limits_{k=0}^x f(k)\tag{10}$$
where $\alpha=0$ must be used for the $2 \pi\, f(0)\, s\, \delta(-i s)$ term of $H(s)$ and $\alpha>0$ must be used for the remaining terms of $H(s)$.
Formulas (6) to (10) above are more consistent with the way a summatory step function
$$r(x)=\sum\limits_{n=1}^x a(k)\tag{11}$$
is related to its corresponding Dirichlet series
$$R(s)=\sum\limits_{n=1}^\infty \frac{a(k)}{k^s}\tag{12}.$$