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 $\varphi$.

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 $1<p\leq 2$, 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.

See also

• Mellin transform
• Nachbin's theorem

References

• P. Flajolet, X. Gourdon, P. Dumas, Mellin transforms and asymptotics: Harmonic sums, Theoretical Computer Science, 144(1-2):3-58, June 1995
• McLachlan, N. W., Complex Variable Theory and Transform Calculus, Cambridge University Press, 1953.
• Polyanin, A. D. and Manzhirov, A. V., Handbook of Integral Equations, CRC Press, Boca Raton, 1998.
• Titchmarsh, E. C., Introduction to the Theory of Fourier Integrals, Oxford University Press, second edition, 1948.
• Yakubovich, S. B., Index Transforms, World Scientific, 1996.
• Zemanian, A. H., Generalized Integral Transforms, John Wiley & Sons, 1968.

External links

• Tables of Integral Transforms at EqWorld: The World of Mathematical Equations.
• Philippe Flajolet, Xavier Gourdon, Philippe Dumas, Mellin Transforms and Asymptotics: Harmonic sums.