The invmellin function computes the inverse Mellin transform (F(s)) of expr (f(t)), a linear transformation from $C\left(C\right)\to C\left[0,\infty \right)$ defined by the contour integral:

In this integral, c is assumed to be real. Also note that Maple currently does not handle general contour integrals. The above contour integral definition is only used to provide the information below on properties of the inverse Mellin transform.

The function $F\left(s\right)$ returned is defined only on the positive real axis.

There are multiple transforms $F\left(s\right)$ for a given $f\left(t\right)$, corresponding to the cases where $\mathrm{c1}<\mathrm{\Re }\left(t\right)$ $<\mathrm{c2}$ for various boundaries $\mathrm{c1}$ and $\mathrm{c2}$.  The range is specified by the parameter ran.  This parameter is optional. If the range parameter is not given, it is assumed to be $-\mathrm{\infty }..\mathrm{\infty }$.

All constants are assumed to be complex unless otherwise specified.

The invmellin function attempts to simplify an expression according to a set of heuristics, and then to match the result against internal lookup tables of patterns.  These tables are of expressions containing algebraic, Bessel, exponential, GAMMA, trigonometric, as well as other functions. The user can add their own functions to invmellin's lookup tables with the function addtable.

Other functions that can be transformed are linear combinations of products of integer powers of t; rational polynomials; terms of the form $a^{-t}$ where $0<a$; some definite integrals of functions whose transforms are known; derivatives of functions whose transforms are known; convolutions of two functions $f\left(t\right)$ and $g\left(t\right)$ whose transforms are known; and functions of the form $f\left(at+b\right)$ and $f\left(a{t}^n\right)$ with $0<a,b$ complex, and $n$ a positive integer where the transforms of $f\left(t\right)$ and $f\left(t^n\right)$ are known.

If the option opt is set to 'NO_INT', then the program will not resort to integration of the original problem if all other methods fail. This will increase the speed at which the transform will run.

invmellin recognizes the Dirac-delta (or unit-impulse) function as Dirac(t) and Heaviside's unit step function as Heaviside(t).

The command with(inttrans,invmellin) allows the use of the abbreviated form of this command.

$\mathrm{with}\left(\mathrm{inttrans}\right)\&colon;$

$\mathrm{assume}\left(0<a\right)$

$\mathrm{assume}\left(b\&comma;\mathrm{complex}\right)\&colon;$

$\mathrm{assume}\left(c\&comma;\mathrm{complex}\right)\&colon;$

$\mathrm{assume}\left(n\&comma;\mathrm{posint}\right)\&colon;$

$G≔\mathrm{mellin}\left(g\left(x\right)\&comma;x\&comma;y\right)$

$G≔\mathrm{`?`}$

$\mathrm{invmellin}\left(G\&comma;y\&comma;z\right)$

$g\left(z\right)$

$\mathrm{with}\left(\mathrm{inttrans}\right)\&colon;$

$\mathrm{addtable}\left(\mathrm{invmellin}\&comma;F\left(t\right)\&comma;f\left(s\right)\&comma;t\&comma;s\&comma;\mathrm{invmellin}=-\mathrm{\infty }..\mathrm{\infty }\right)\&colon;$

$\mathrm{addtable}\left(\mathrm{invmellin}\&comma;\mathrm{F1}\left(t\right)\&comma;\mathrm{f1}\left(s\right)\&comma;t\&comma;s\&comma;\mathrm{invmellin}=-\mathrm{\infty }..\mathrm{\infty }\right)\&colon;$

$\mathrm{addtable}\left(\mathrm{invmellin}\&comma;\mathrm{F2}\left(t\right)\&comma;\mathrm{f2}\left(s\right)\&comma;t\&comma;s\&comma;\mathrm{invmellin}=-\mathrm{\infty }..\mathrm{\infty }\right)\&colon;$

$\mathrm{invmellin}\left(F\left(x\right)\&comma;x\&comma;y\right)$

$f\left(y\right)$

$\mathrm{invmellin}\left(\mathrm{F1}\left(x\right)\&comma;x\&comma;y\right)$

$\mathrm{f1}\left(y\right)$

$\mathrm{invmellin}\left(\mathrm{F2}\left(x\right)\&comma;x\&comma;y\right)$

$\mathrm{f2}\left(y\right)$

$\mathrm{invmellin}\left(b\mathrm{F1}\left(z\right)+c\mathrm{F2}\left(z\right)\&comma;z\&comma;x\right)$

$\mathrm{b~}\mathrm{f1}\left(x\right)+\mathrm{c~}\mathrm{f2}\left(x\right)$

$\mathrm{invmellin}\left(F\left(az+b\right)\&comma;z\&comma;x\right)$

$\frac{x^{\frac{\mathrm{b~}}{\mathrm{a~}}}f\left(x^{\frac{1}{\mathrm{a~}}}\right)}{\mathrm{a~}}$

$\mathrm{invmellin}\left(F\left(a{z}^n\right)\&comma;z\&comma;x\right)$

$\mathrm{Typesetting}:-\mathrm{mfrac}\left({ℳ}^{-1}\left(F\left(z^{\mathrm{n~}}\right)\&comma;z\&comma;x^{\frac{1}{{\mathrm{a~}}^{\frac{1}{\mathrm{n~}}}}}\right)\&comma;{\mathrm{a~}}^{\frac{1}{\mathrm{n~}}}\right)$

$\mathrm{invmellin}\left(\mathrm{D}\left(F\right)\left(z\right)\&comma;z\&comma;x\right)$

$\ln \left(x\right)f\left(x\right)$

$\mathrm{invmellin}\left(zF\left(z\right)\&comma;z\&comma;x\right)$

$-x\left(\frac{\&DifferentialD;}{\&DifferentialD;x}f\left(x\right)\right)$

$\mathrm{invmellin}\left(a^{-z}F\left(z\right)\&comma;z\&comma;x\right)$

${ℳ}^{-1}\left({\&ExponentialE;}^{-z\ln \left(\mathrm{a~}\right)}F\left(z\right)\&comma;z\&comma;x\right)$

$\mathrm{invmellin}\left(\frac{\mathrm{\Gamma }\left(2-z\right)}{\mathrm{\Gamma }\left(1-z\right)}F\left(z-1\right)\&comma;z\&comma;x\right)$

$\frac{\&DifferentialD;}{\&DifferentialD;x}f\left(x\right)$

$\mathrm{invmellin}\left(\mathrm{F1}\left(z\right)\mathrm{F2}\left(1-z\right)\&comma;z\&comma;x\right)$

${\int }_0^{\mathrm{\infty }}\mathrm{f1}\left(x\mathrm{\_U}\right)\mathrm{f2}\left(\mathrm{\_U}\right)\&DifferentialD;\mathrm{\_U}$

$\mathrm{invmellin}\left(1\&comma;z\&comma;x\right)$

$\mathrm{Dirac}\left(x-1\right)$

$\mathrm{invmellin}\left(z\&comma;z\&comma;x\right)$

$-x\mathrm{Dirac}\left(1\&comma;x-1\right)$

$\mathrm{invmellin}\left(\exp \left(a{z}^2\right)\&comma;z\&comma;x\right)$

$\frac{{\&ExponentialE;}^{-\frac{{\ln \left(\frac{1}{x}\right)}^2}{4\mathrm{a~}}}}{2\sqrt{\mathrm{\pi }\mathrm{a~}}}$

$\mathrm{invmellin}\left(\mathrm{\Gamma }\left(z\right)\&comma;z\&comma;x\right)$

Warning, invmellin can transform GAMMA(t) if Re(t)<>0, Re(t) > -1

${ℳ}^{-1}\left(\mathrm{\&Gamma;}\left(z\right)\&comma;z\&comma;x\right)$

$\mathrm{invmellin}\left(\mathrm{\Gamma }\left(z\right)\&comma;z\&comma;x\&comma;0..\mathrm{\infty }\right)$

${\&ExponentialE;}^{-x}$

$\mathrm{invmellin}\left(\mathrm{\Gamma }\left(z\right)\&comma;z\&comma;x\&comma;-1..0\right)$

${\&ExponentialE;}^{-x}-1$