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

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

we have that

<math>\varphi(s)= \{ \mathcal{M} f \} = \int_0^{\infty} x^{s-1} f(x)\,dx.</math>

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

<math>\varphi(s)=\int_0^{\infty} x^{s-1} f(x)\,dx</math>

is absolutely convergent when <math>a < \Re(s) < b</math>. Then <math>f</math> is recoverable via the inverse Mellin transform from its Mellin transform <math>\varphi</math>. 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.<ref>Template:Cite book</ref>

Boundedness condition

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

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

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

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

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

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

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

Since the two-sided Laplace transform can be defined as

<math> \left\{\mathcal{B} f\right\}(s) = \left\{\mathcal{M} f(- \ln x) \right\}(s)</math>

these theorems can be immediately applied to it also.

See also

• Mellin transform
• Nachbin's theorem

References

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External links

• Tables of Integral Transforms at EqWorld: The World of Mathematical Equations.