Mellin transform
= Mellin transform
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{title2=$\mathcal M$}
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= Mellin transformation
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The <Mellin transform> is an <integral transform> suited to changes of scale: replacing $f(x)$ by $f(ax)$ multiplies its transform by $a^{-s}$ for $a>0$. For a <locally integrable function> $f:(0,\infty)\to\mathbb C$, define
$$
\mathcal Mf(s)=\int_0^\infty x^{s-1}f(x)\,dx,\qquad s=\sigma+i\tau,
$$
where the <improper integral> converges. The <complex number> $x^{s-1}=\exp((s-1)\log x)$ uses the real <natural logarithm> because $x>0$. Its <absolute value> is $x^{\sigma-1}$, so <absolute convergence> depends on the <real part> $\sigma$ rather than the <imaginary part> $\tau$. The <Mellin fundamental strip> records the open range of such <real parts>.<NIST Digital Library of Mathematical Functions>[Chapter 2, §2.5(i), equation 2.5.1 and the following strip-of-analyticity paragraph]{ref}
A familiar example is the <exponential function> $f(x)=e^{-x}$:
$$
\mathcal M\{e^{-x}\}(s)=\Gamma(s),\qquad \operatorname{Re}s>0.
$$
This is exactly the <Gamma integral>. Near zero $e^{-x}$ tends to one, so <absolute convergence> requires $\sigma>0$; at infinity $x^r e^{-x}\to0$ for every real $r$, so there is no upper boundary. The <Gamma function> has a <meromorphic continuation> with <simple poles> at $0,-1,-2,\ldots$, but a finite continued value elsewhere in $\sigma\leq0$ does not make the defining <improper integral> converge.<NIST Digital Library of Mathematical Functions>[Chapter 5, §5.2(i), equation 5.2.1, its Re z > 0 condition, and the following analytic-continuation paragraph]{ref}
The <Mellin transform of an interval indicator> gives an elementary example with <absolute convergence> for every <complex number> $s$. The <Mellin transform in logarithmic coordinates> explains why changes of scale become <translations>, connecting the <scaling property of the Mellin transform> to the <Fourier transform>.