# Mellin inversion theorem

In mathematics, the Mellin inversion formula (named after Hjalmar Mellin) tells us conditions under which the inverse Mellin transform, or equivalently the inverse two-sided Laplace transform, are defined and recover the transformed function.

## Method

If $\varphi(s)$ is analytic in the strip $a < \Re(s) < b$, and if it tends to zero uniformly as $\Im(s) \to \pm \infty$ for any real value c between a and b, with its integral along such a line converging absolutely, then if

$f(x)= \{ \mathcal{M}^{-1} \varphi \} = \frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} \varphi(s)\, ds$

we have that

$\varphi(s)= \{ \mathcal{M} f \} = \int_0^{\infty} x^s f(x)\,\frac{dx}{x}.$

Conversely, suppose f(x) is piecewise continuous on the positive real numbers, taking a value halfway between the limit values at any jump discontinuities, and suppose the integral

$\varphi(s)=\int_0^{\infty} x^s f(x)\,\frac{dx}{x}$

is absolutely convergent when $a < \Re(s) < b$. Then f is recoverable via the inverse Mellin transform from its Mellin transform $\varphi$[citation needed].

## Boundedness condition

We may strengthen the boundedness condition on $\varphi(s)$ if f(x) is continuous. If $\varphi(s)$ is analytic in the strip $a < \Re(s) < b$, and if $|\varphi(s)| < K |s|^{-2}$, where K is a positive constant, then f(x) as defined by the inversion integral exists and is continuous; moreover the Mellin transform of f is $\varphi$ for at least $a < \Re(s) < b$.

On the other hand, if we are willing to accept an original f which is a generalized function, we may relax the boundedness condition on $\varphi$ to simply make it of polynomial growth in any closed strip contained in the open strip $a < \Re(s) < b$.

We may also define a Banach space version of this theorem. If we call by $L_{\nu, p}(R^{+})$ the weighted Lp space of complex valued functions f on the positive reals such that

$\|f\| = \left(\int_0^\infty |x^\nu f(x)|^p\, \frac{dx}{x}\right)^{1/p} < \infty$

where ν and p are fixed real numbers with p>1, then if f(x) is in $L_{\nu, p}(R^{+})$ with $1 < p \le 2$, then $\varphi(s)$ belongs to $L_{\nu, q}(R^{+})$ with $q = p/(p-1)$ and

$f(x)=\frac{1}{2 \pi i} \int_{\nu-i \infty}^{\nu+i \infty} x^{-s} \varphi(s)\,ds.$

Here functions, identical everywhere except on a set of measure zero, are identified.

Since the two-sided Laplace transform can be defined as

$\left\{\mathcal{B} f\right\}(s) = \left\{\mathcal{M} f(- \ln x) \right\}(s)$

these theorems can be immediately applied to it also.