Meijer G-function transforms

Remarks

Example: $G^{1,0}_{0,1}(z \mid -; 0) = e^{-z}$, so with $m = q = 1$, $n = p = 0$ and $b_1 = 0$ the G-transform is the Laplace transform $g(y) = \int_0^\infty e^{-xy} f(x)\, dx$. Example: $G^{1,1}_{1,1}(z \mid 0; 0) = (1 + z)^{-1}$, so another choice gives the transform $\int_0^\infty f(x) (1 + xy)^{-1}\, dx$, which is a form of the Stieltjes transform. Example: $G^{1,0}_{0,2}(z \mid -; \nu/2, -\nu/2) = J_\nu(2\sqrt{z})$, a Bessel function kernel.

Non-example: for the Laplace case, $f(x) = e^{x^2}$ gives a divergent integral for every $y > 0$, so it has no G-transform with these parameters.

The Mellin transform turns the G-transform into a product: if $\mathcal{M}$ denotes the Mellin transform, then formally $\mathcal{M} g(s) = \mathcal{M} G(s)\, \mathcal{M} f(1 - s)$, and $$\mathcal{M} G(s) = \frac{\prod_{j=1}^{m} \Gamma(b_j + s) \prod_{j=1}^{n} \Gamma(1 - a_j - s)}{\prod_{j=m+1}^{q} \Gamma(1 - b_j - s) \prod_{j=n+1}^{p} \Gamma(a_j + s)}.$$