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

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

we have that

${\displaystyle \varphi (s)=\{{\mathcal {M}}f\}=\int _{0}^{\infty }x^{s-1}f(x)\,dx.}$

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

${\displaystyle \varphi (s)=\int _{0}^{\infty }x^{s-1}f(x)\,dx}$

is absolutely convergent when ${\displaystyle a<\Re (s)<b}$. Then ${\displaystyle f}$ is recoverable via the inverse Mellin transform from its Mellin transform ${\displaystyle \varphi }$. These results can be obtained by relating the Mellin transform to the Fourier transform by a change of variables and then applying an appropriate version of the Fourier inversion theorem.^[1]

Boundedness condition

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

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

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

${\displaystyle \|f\|=\left(\int _{0}^{\infty }|x^{\nu }f(x)|^{p}\,{\frac {dx}{x}}\right)^{1/p}<\infty }$

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

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

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

Since the two-sided Laplace transform can be defined as

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

these theorems can be immediately applied to it also.

See also

• Mellin transform
• Nachbin's theorem