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.

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

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

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 $\varphi$ .

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 $L_{\nu, p}(R^{%2B})$ 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 $L_{\nu, p}(R^{%2B})$ with $1 < p \le 2$ , then $\varphi(s)$ belongs to $L_{\nu, q}(R^{%2B})$ with $q = p/(p-1)$ and

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

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.

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.

Mellin inversion theorem

See also

References

External links