InverseMellinTransform

The inverse Mellin transform — $f$ such that $\mathcal{M}\{f\}(s) = F(s)$ — as a rule table mirroring MellinTransform's nine pairs.

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InverseMellinTransform(F, s, x)

f(x)

such that

\int_0^\infty f(x)\,x^{s-1}\,dx = F(s)

Typed signatures

Type	Package	Applies
`(expression, expression, expression) -> expression`	`analytic`	always

Domain: Transforms

Details

Mirrors MellinTransform's nine pairs (each confirmed against wolframscript's own InverseMellinTransform), with one exception: $\pi\csc(\pi s)/s$ inverts to $\log(1+1/x)$ , NOT $\log(1+x)$ — Wolfram's default inversion contour for this particular $F(s)$ sits in the strip $0<\mathrm{Re}(s)<1$ , while $\log(1+x)$ 's own forward transform lives in $-1<\mathrm{Re}(s)<0$ ; the same $F(s)$ genuinely inverts to different $f(x)$ depending on which strip the (otherwise ambiguous) Mellin-Barnes contour sits in, and this matches Wolfram's actual default rather than assuming the round trip is symmetric.
The scaling theorem ( $F(s)/a^s \to f(ax)$ , any provably positive a) generalizes every pair; the power shift ( $F(s+c)\to x^c f(x)$ ) is supported only for the $\Gamma(s)\leftrightarrow e^{-x}$ pair (matching the one shifted example the forward table lists, $\Gamma(a+s)\to x^a e^{-x}$ ) — combining a shift and a scale together is declined (unverified).
Beyond the nine: $1/(s+b)\to x^b\,\theta(1-x)$ , its scaling $a^s/(s+b)\to(x/a)^b\,\theta(1-x/a)$ , and $\Gamma(cs)^2\to(2/c)K_0(2x^{1/2c})$ ( $\Gamma(s)^2\to 2K_0(2\sqrt x)$ ), the last two for a provably positive a or c.
Also: $\Gamma((s+c+1)/2)/(s+c)\to\sqrt\pi\,x^c\operatorname{erfc}(x)$ ( $c$ a half-integer or integer, any multiple of the denominator), and $\Gamma(\kappa s)\Gamma(1-2\kappa s)/\Gamma(1-\kappa s)\to m(1+4x^m)^{-1/2}$ for $\kappa=1/m$ , the pair of $(1+x)^{-1/2}$ after Gamma's duplication formula and a scaling.
Declined: anything outside these shapes.