Categories: Complex analysis | Integral transforms | Theorems

Mellin inversion theorem

In mathematics, the Mellin inversion formula 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

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

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 $\varphi$ .

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 $L ν, p (R +)$ the weighted Lp space of complex valued functions f on the positive reals such that

$||f|| = (\int_0^\infty |x^\nu f(x)|^p \frac{dx}{x})^{1/p} < \infty$

where ν and p are fixed real numbers with p>1, then if f(x) is in $L ν, p (R +)$ with $1 < p \le 2$ , then $\varphi(s)$ belongs to $L ν, q (R +)$ 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 on a set of measure zero are identified.

Since the two-sided Laplace transform can be defined as

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

these theorems can be immediately applied to it also.

Mellin convolution theorem

If f and g are defined and integrable on the positive reals, and if $x k f (x)$ and $x k g (x)$ are absolutely integrable, we may define

$h(x) = (f \star g)(x) = \int_0^\infty f(t)g(\frac{x}{t})\frac{dt}{t}$

We then have that $x k h (x)$ is absolutely integrable on the positive reals, and

$\{ \mathcal{M} (f \star g ) \} = \{ \mathcal{M} f \} \{ \mathcal{M} g \}$

in a strip containing the line with abscissa k+1.

A converse can be defined for L¹ functions as well, but is more elegant for L². Suppose f is an element of $L k,2 (R +)$ and g is an element of $L m,2 (R +)$ , and suppose $R e (s) = k + m$ . Then

$\left\{ \mathcal{M} fg \right\}(s) = \frac{1}{2 \pi i}\int_{k - i \infty}^{k+i\infty} F(t)G(s-t)dt$

References

McLachlan, N. W., Complex Variable Theory and Transform Calculus, Cambridge University Press, 1953
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

Categories: Complex analysis | Integral transforms | Theorems