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  "body": "\n## Definition\n\nFor a function $g$ on $(0, \\infty)$ write $\\mathcal{M}g(s) = \\int_0^\\infty x^{s - 1} g(x) \\, dx$ for its [[Mellin transform|N5PVGRJE1T]].\nLet $k$ and $f$ be given functions on $(0, \\infty)$, let $\\lambda \\in \\C$, and consider the [[integral equation|R3SG1SN7JQ]]\n$$\\varphi(x) - \\lambda \\int_0^\\infty k\\!\\left(\\frac{x}{t}\\right) \\varphi(t) \\, \\frac{dt}{t} = f(x), \\quad x \u003e 0,$$\nor its first-kind form without the term $\\varphi(x)$.\nAssume that $\\mathcal{M}k$ and $\\mathcal{M}f$ converge absolutely in a common strip $a \u003c \\operatorname{Re} s \u003c b$.\nThe *Mellin transform method* consists of the following steps.\n\n1. Apply $\\mathcal{M}$; since the integral is the [[Mellin convolution|A2PMJG7J68]] of $k$ and $\\varphi$, the equation becomes $(1 - \\lambda \\mathcal{M}k(s)) \\, \\mathcal{M}\\varphi(s) = \\mathcal{M}f(s)$ in the strip.\n2. Solve for $\\mathcal{M}\\varphi(s) = \\mathcal{M}f(s) / (1 - \\lambda \\mathcal{M}k(s))$ (for the first kind, $\\mathcal{M}\\varphi = \\mathcal{M}f / \\mathcal{M}k$).\n3. Recover $\\varphi$ by the [[Mellin inversion formula|XJWC13NY9E]] $\\varphi(x) = \\frac{1}{2\\pi i} \\int_{c - i\\infty}^{c + i\\infty} x^{-s} \\, \\mathcal{M}\\varphi(s) \\, ds$ with $a \u003c c \u003c b$.\n\n## Remarks\n\nStep 1 rests on the identity $\\mathcal{M}\\left[\\int_0^\\infty k(x/t) \\varphi(t) \\, dt/t\\right](s) = \\mathcal{M}k(s) \\, \\mathcal{M}\\varphi(s)$, which follows from the substitution $x = t u$.\n\nExample: the first-kind equation $\\int_0^\\infty e^{-x/t} \\varphi(t) \\, \\frac{dt}{t} = f(x)$ with $f(x) = \\int_0^\\infty e^{-x/t - t} \\, \\frac{dt}{t}$.\nHere $k(u) = e^{-u}$ and $\\mathcal{M}k(s) = \\Gamma(s)$ for $\\operatorname{Re} s \u003e 0$.\nBy step 1 applied to $e^{-t}$, $\\mathcal{M}f(s) = \\Gamma(s)^2$, so step 2 gives $\\mathcal{M}\\varphi(s) = \\Gamma(s)$ and step 3 gives $\\varphi(t) = e^{-t}$.\n\nA kernel $h(x, t)$ that is homogeneous of degree $-1$, that is $h(cx, ct) = c^{-1} h(x, t)$ for all $c \u003e 0$, can be written $h(x, t) = \\frac{1}{t} h\\!\\left(\\frac{x}{t}, 1\\right)$, so the method applies to $\\int_0^\\infty h(x, t) \\varphi(t) \\, dt$.\nFor $h(x, t) = \\frac{1}{x + t}$ one gets $k(u) = \\frac{1}{1 + u}$ and $\\mathcal{M}k(s) = \\frac{\\pi}{\\sin \\pi s}$ for $0 \u003c \\operatorname{Re} s \u003c 1$.\n\nNon-example: the kernel $e^{-(x - t)^2}$ on $(0, \\infty)$ is not a function of $x / t$ times $\\frac{1}{t}$, so the Mellin transform does not turn the equation into a product.\n\nThe substitution $x = e^{u}$, $t = e^{v}$ turns the equation into a [[convolution integral equation|EY75YD988Z]] on $\\R$, and the method into the [[Fourier transform method|NRG03YP2PW]].\n",
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