Mellin transform | Academic Cesspit

A Mellin transform of a function $f:(0,\infty)\to\mathbb{R}$ is of the form

$\displaystyle (Mf)(a)=\int_0^\infty x^{a-1}f(x)\,dx$

for $a>0.$ In order to implement residue calculus to evaluate this, one needs to make some assumptions on $f.$ The following proposition will tell us what these assumptions should be.

Proposition. Let $f$ have finitely many poles in $\mathbb{C}$ all lying in $\tilde{\mathbb{C}}=\mathbb{C}-\{z:\mbox{Re}\,z\geq 0\},$ $a\in\mathbb{C}-\mathbb{Z},$ and both $\lim_{z\to 0}f(z)z^a=0$ and $\lim_{z\to\infty}f(z)z^a=0.$ Then

$\displaystyle\int_0^\infty x^{a-1}f(x)\,dx=\frac{2\pi i}{1-e^{2\pi ia}}\sum_{p\in\tilde{\mathbb{C}}}\mbox{res}_p(z^{a-1}f(z)).$

Proof. We first choose a nonprincipal branch of the logarithm compatible with this situation. Let $z=|z|e^{i\theta}$ for $\theta\in[0,2\pi)$ and then set $\ln z=\ln |z| +i\theta$ and $z^a=\exp(a\ln z).$ Then these functions are holomorphic on $\tilde{\mathbb{C}}.$ One other lemma we need is the following: if $I$ is a compact interval on the positive real axis and $\varepsilon>0,$ then $\lim_{\varepsilon\to 0}(x+i\varepsilon)^a=x^a$ and $\lim_{\varepsilon\to 0}(x-i\varepsilon)^a=x^ae^{2\pi ia},$ and the convergence is uniform.

Note that we have $\lim_{\varepsilon\to 0}\ln(x+i\varepsilon)=\ln z$ and $\lim_{\varepsilon\to 0}\ln(x-i\varepsilon)=\lim_{\varepsilon\to 0}\ln|x-i\varepsilon|+i\theta=\ln (x)+2\pi i,$ which gives the result of the lemma.

Now to integrate the function, we consider the complex function $z^{a-1}f(z)$ and the contour given by

with $\gamma=\gamma_1+\gamma_2+\gamma_3+\gamma_4.$ Then by the residue theorem

$\displaystyle\lim_{\varepsilon\to 0}\lim_{r\to 0}\lim_{s\to\infty}\int_\gamma z^{a-1}f(z)\,dz=2\pi i\sum_{p\in\tilde{\mathbb{C}}}\mbox{res}_p(z^{a-1}f(z)),$

For $\gamma_2$ we have

$\displaystyle\left|\int_{\gamma_2}z^{a-1}f(z)\,dz\right|\leq 2\pi s|z^{a-1}f(z)|=2\pi |z^af(z)|$

since $|z|=s$ on $\gamma_2.$ We have a similar bound for $\gamma_4.$ Now due to the assumption of the proposition, it follows that $|z^af(z)|\to 0$ and hence $\int_{\gamma_2}z^{a-1}f$ and $\int_{\gamma_4}z^{a-1}f$ go to $0$ as $\varepsilon\to 0$ and $r,s\to\infty.$