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  "body": "\n## Definition\n\nFix integers $0 \\le m \\le q$ and $0 \\le n \\le p$, and complex parameters $a_1, \\dots, a_p$ and $b_1, \\dots, b_q$.\nLet $G(z) = G^{m,n}_{p,q}(z \\mid a_1, \\dots, a_p; b_1, \\dots, b_q)$ be the [[Meijer G-function|19JNNN5KV7]] with these parameters.\nLet $f \\colon (0, \\infty) \\to \\C$ be such that the integral below converges for $y$ in a set $Y \\subseteq (0, \\infty)$.\nThe *G-transform* of $f$ with these parameters is the function\n$$g(y) = \\int_0^\\infty G(xy)\\, f(x)\\, dx, \\qquad y \\in Y.$$\n\n## Remarks\n\nExample: $G^{1,0}_{0,1}(z \\mid -; 0) = e^{-z}$, so with $m = q = 1$, $n = p = 0$ and $b_1 = 0$ the G-transform is the [[Laplace transform|0V49JEZW06]] $g(y) = \\int_0^\\infty e^{-xy} f(x)\\, dx$.\nExample: $G^{1,1}_{1,1}(z \\mid 0; 0) = (1 + z)^{-1}$, so another choice gives the transform $\\int_0^\\infty f(x) (1 + xy)^{-1}\\, dx$, which is a form of the [[Stieltjes transform|RG1HFAYD98]].\nExample: $G^{1,0}_{0,2}(z \\mid -; \\nu/2, -\\nu/2) = J_\\nu(2\\sqrt{z})$, a Bessel function kernel.\n\nNon-example: for the Laplace case, $f(x) = e^{x^2}$ gives a divergent integral for every $y \u003e 0$, so it has no G-transform with these parameters.\n\nThe [[Mellin transform|N5PVGRJE1T]] 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\n$$\\mathcal{M} G(s) = \\frac{\\prod_{j=1}^{m} \\Gamma(b_j + s) \\prod_{j=1}^{n} \\Gamma(1 - a_j - s)}{\\prod_{j=m+1}^{q} \\Gamma(1 - b_j - s) \\prod_{j=n+1}^{p} \\Gamma(a_j + s)}.$$\n",
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