 Mellin Inversion Theorem
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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) , 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) . Then f is recoverable via the inverse Mellin transform from its Mellin transform $\varphi$ [].

## 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) , and if $|\varphi (s)| , 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) .

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) .

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 ? and p are fixed real numbers with p>1, then if f(x) is in $L_{\nu ,p}(R^{+})$ with $1 , 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.