Mellin transform under a power substitution

Remarks

For $a > 0$, substitute $u = x^{a}$, so $x = u^{1/a}$ and $dx = \frac{1}{a} u^{1/a - 1} \, du$: $$\int_0^\infty x^{s-1} f(x^{a}) \, dx = \frac{1}{a} \int_0^\infty u^{(s-1)/a} \, u^{1/a - 1} f(u) \, du = \frac{1}{a} \int_0^\infty u^{s/a - 1} f(u) \, du.$$ For $a < 0$ the substitution reverses the orientation of $(0, \infty)$, which turns the factor $1/a$ into $1/\abs{a}$. The case $a = -1$ is the rule for the inversion $x \mapsto 1/x$. Example: $f(x) = e^{-x}$ and $a = 2$ give $\int_0^\infty x^{s-1} e^{-x^2} \, dx = \frac{1}{2} \Gamma(s/2)$ for $\operatorname{Re} s > 0$; at $s = 1$ this is $\int_0^\infty e^{-x^2} \, dx = \frac{1}{2}\Gamma(1/2) = \frac{\sqrt{\pi}}{2}$. The hypothesis $a \ne 0$ is needed: for $a = 0$, $g$ is the constant $f(1)$, which is Mellin convergent at no $s$ unless $f(1) = 0$.

Going deeper

For a student

This rule computes many transforms from a few, for instance Gaussian integrals from the gamma function. It is used in the Riemann proof of the functional equation, where the theta function $\sum_n e^{-\pi n^2 x}$ meets $\Gamma(s/2)$ rather than $\Gamma(s)$. Combined with the rules for dilations and for multiplication by powers, it describes the Mellin transform of $x^{b} f(c x^{a})$ for all $a \ne 0$, $b \in \C$ and $c > 0$.