Mellin inversion theorem

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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

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

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

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

where ν and p are fixed real numbers with p>1, then if f(x) is in $L ν, p (R +)$ 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

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

these theorems can be immediately applied to it also.

1 Example calculation: the inverse transform of phi(s) = pi/sin(pi s)
2 Mellin convolution theorem
3 External links
4 References

[edit] Example calculation: the inverse transform of phi(s) = pi/sin(pi s)

The contours used in the inverse Mellin transform of pi/sin(pi s)

As an example, consider the transform pair

$f(x) = \frac{1}{1+x}$

and

$\varphi(s) = \frac{\pi}{\sin \pi s}.$

The fundamental strip was computed on the page for the Mellin transform and seen to be $\langle 0, 1\rangle.$ Hence the inverse of $\varphi(s)$ is given by the integral

$\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{\pi}{x^s \, \sin\pi s} ds,$

where $c$ is a real number such that $0 < c < 1.\,$ We will compute this integral in two different ways, using the two contours shown on the right. The contour on the left yields an expansion of $f (x)$ about zero, and the one on the right, about infinity. This is a general feature of Mellin inversion which is frequently used in the evaluation of harmonic sums and applications of the Mellin-Perron summation formula.

[edit] The two contours

Let $s=\sigma + i t.\,$ The contour for the expansion at zero consists of a vertical line segment along the line $\sigma=c\,$ , from $c-iT\,$ to $c+iT\,$ , where $T$ is a parameter that goes to infinity, and the limit along this segment approaches the Mellin inversion integral. The parameter $T$ is parameterized in turn by $j$ , so that it avoids the poles of $\varphi(s)$ , which are at the integers. We take $T_j=(2j+1)/2,\,$ with $j$ a positive integer at least two. The contour continues with a horizontal line segment from $c+iT\,$ to $-T+iT\,$ , a vertical line segment from $-T+iT\,$ to $-T-iT\,$ , and a final horizontal line segment from $-T-iT\,$ to $c-iT.\,$

The contour for the expansion at infinity consists of the vertical line segment from $c+iT\,$ to $c-iT\,$ , the horizontal line segment from $c-iT\,$ to $T-iT\,$ , the vertical line segmemt from $T-iT\,$ to $T+iT\,$ , and finally, the horizontal line segment from $T+iT\,$ to $c+iT\,$ . If we can show that all contributions except the one from the central vertical segment go to zero as $j$ goes to infinity, then we will have evaluated the Mellin inversion integral in two different ways, and it will be equal to the sum of the residues of the poles inside the respective contour, by the Cauchy residue theorem. We start by computing the residues at the poles.

[edit] Residues at the poles

The poles of $\varphi(s)$ are at the integers $n;$ they are simple, and the residues are $(-1)^n.\,$ To see this, note that

$\frac{\pi}{\sin \pi s} = \frac{2\pi i}{e^{i\pi s} - e^{-i\pi s}} = \frac{2\pi i e^{i\pi s}}{e^{2 i\pi s} - 1}$

and

$\lim_{s\rightarrow n} (s-n) \frac{2\pi i e^{i\pi s}}{e^{2 i\pi s} - 1} = 2\pi i (-1)^n \lim_{s\rightarrow n} \frac{1}{2 i\pi e^{2 i\pi s}} = (-1)^n.$

The next step is to examine the behavior of $\varphi(s)$ along vertical and horizontal line segments. This will help establish bounds on the integrals along line segments not equal to the central segment at $c$ .

[edit] Behavior along horizontal line segments

We compute two bounds, one for the upper and one for the lower half-plane. We have

$\left| \frac{\pi}{\sin \pi s} \right| = \left| \frac{2\pi i e^{i\pi s}}{e^{2 i\pi s} - 1} \right| = 2\pi \frac{e^{-\pi t}}{\left| e^{2 i\pi s} - 1 \right|}$

and

$\left| \frac{\pi}{\sin \pi s} \right| = \left| \frac{2\pi i e^{-i\pi s}}{1 - e^{-2 i\pi s}} \right| = 2\pi \frac{e^{\pi t}}{\left| 1 - e^{-2 i\pi s} \right|}.$

We will use the first of these for the upper half-plane, and the second one, for the lower half-plane.

Now observe that when

$t > \frac{\log 2}{2 \pi} \quad \mbox{then} \quad \frac{3}{2} > \left| e^{2 i\pi \sigma - 2 \pi t} - 1 \right| > \frac{1}{2},$

because along a horizontal line segment, the term $e^{2 i\pi \sigma}\,$ merely rotates around the unit circle. Similarly, when

$t < -\frac{\log 2}{2 \pi} \quad \mbox{then} \quad \frac{3}{2} > \left| 1 - e^{-2 i\pi \sigma + 2 \pi t} \right| > \frac{1}{2},$

because the term $e^{-2 i\pi \sigma}\,$ also rotates around the unit circle.

The conclusion is that along a horizontal segment in the upper half plane, we have

$\left| \frac{\pi}{\sin \pi s} \right| < 4\pi e^{-\pi t}$

and in the lower half plane,

$\left| \frac{\pi}{\sin \pi s} \right| < 4\pi e^{\pi t}$

[edit] Behavior along vertical line segments

Here we show that the maximum modulus of $\varphi(s)$ along a vertical line segment is attained on the real axis.

We have

$\left| \frac{\pi}{\sin \pi s} \right| = \left| \frac{2\pi i}{e^{i \pi s} - e^{-i \pi s}} \right|.$

Now

$e^{i \pi s} - e^{-i \pi s} = e^{i \pi\sigma} e^{-\pi t} - e^{-i \pi\sigma} e^{\pi t}\,$

which is

$\cos(\pi\sigma) \left( e^{-\pi t} - e^{\pi t} \right) + i \sin(\pi\sigma) \left( e^{-\pi t} + e^{\pi t} \right)$

so that

$\left| e^{i \pi s} - e^{-i \pi s} \right| = \cos(\pi\sigma)^2 \left( e^{- 2 \pi t} -2 + e^{2 \pi t} \right) + \sin(\pi\sigma)^2 \left( e^{- 2 \pi t} +2 + e^{2 \pi t} \right)$

which is

$e^{- 2 \pi t} + e^{2 \pi t} + 2 \left( \sin(\pi\sigma)^2 - \cos(\pi\sigma)^2 \right).$

With $σ$ constant, this clearly has a global minimum at $t=0\,$ , so that finally,

$\left| \frac{\pi}{\sin \pi s} \right| \le \left| \frac{\pi}{\sin \pi \sigma} \right|.$

We are now in a position to evaluate the integrals, making repeated use of the ML inequality, which states that

$\left| \int_C f(z) dz \right| \le ML$

where $M$ is the maximum modulus of $f (z)$ on the curve $C$ and $L$ is the length of $C .$

[edit] Expansion at zero

This corresponds to the left contour and is valid for $|x|<1.\,$ We apply the ML inequality to the integrals along the three extra segments.

Upper horizontal segment:

$(c + T) |x|^c 4\pi e^{-\pi T},\,$ vanishes when

T

goes to infinity because the exponential term dominates and $|x|<1.\,$

Left vertical segment:

$2 T |x|^T \left| \frac{\pi}{\sin (-\pi T)} \right| = 2 \pi T |x|^T ,$ vanishes because $|x|<1.\,$

Lower horizontal segment:

$(c + T) |x|^c 4\pi e^{-\pi T},\,$ vanishes as above.

It follows that the expansion of the original function at zero is

$\sum_{n\le 0} (-1)^n x^{-n} = 1 - x + x^2 - x^3 + \cdots = \frac{1}{1+x} = f(x).$

[edit] Expansion at infinity

This corresponds to the right contour and is valid for $|x|>1.\,$

Upper horizontal segment:

$(T - c) |x|^c 4\pi e^{-\pi T},\,$ vanishes as above.

Right vertical segment:

$2 T |x|^{-T} \left| \frac{\pi}{\sin (\pi T)} \right| = 2 \pi T |x|^{-T} ,$ vanishes because $|x|>1.\,$

Lower horizontal segment:

$(T - c) |x|^c 4\pi e^{-\pi T},\,$ vanishes as above.

It follows that the expansion of the original function at infinity is

$- \sum_{n\ge 1} (-1)^n x^{-n} = \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x^3} - \frac{1}{x^4} + \cdots = \frac{1}{x} \frac{1}{1+1/x} = \frac{1}{1+x} = f(x).$

(The initial minus sign occurs because the inversion integral is traversed from top to bottom.)

[edit] 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

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

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

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

[edit] External links

Tables of Integral Transforms at EqWorld: The World of Mathematical Equations.
Marko Riedel, Applications of the Mellin-Perron Formula in Number Theory.
Philippe Flajolet, Xavier Gourdon, Philippe Dumas, Mellin Transforms and Asymptotics: Harmonic sums.

[edit] 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.