# Is there a preference if one of the functions in convolution of Mellin transform is divergent?

The convolution of Mellin transform is

$$\sigma \left( x \right) = \int _x^1 f \left( \epsilon \right) h \left( \frac{x}{\epsilon} \right) \frac{1}{\epsilon} \mathrm{d} \epsilon ,$$

if both $$f \left( \epsilon \right)$$ and $$h \left( \epsilon \right)$$ vanish out of range $$\epsilon \in \left( 0, 1 \right)$$.

My question here is, if function $$h \left( \zeta \right)$$ is divergent at $$\zeta = 1$$, would there be a difference if we instead write

$$\sigma \left( x \right) = \int _x^1 h \left( \epsilon \right) f \left( \frac{x}{\epsilon} \right) \frac{1}{\epsilon} \mathrm{d} \epsilon ?$$

Though the convolution is symmetric for functions $$f \left( \epsilon \right)$$ and $$h \left( \epsilon \right)$$, but is this still true if one of the functions has a divergence within the interval?

• It doesn't make a difference for $x \in (0, 1)$. We're computing the improper integrals $$f*h = \lim_{y \downarrow x} \int_y^1 f(t) h {\left( \frac x t \right)} \frac {dt} t = \lim_{y \downarrow x} G(y), \\ h*f = \lim_{y \uparrow 1} \int_x^y h(t) f {\left( \frac x t \right)} \frac {dt} t = \lim_{y \uparrow 1} \int_{x/y}^1 f(t) h {\left( \frac x t \right)} \frac {dt} t = \lim_{y \uparrow 1} G {\left( \frac x y \right)}.$$ Either both limits do not exist or they are the same. Feb 17 '19 at 20:57
• @Maxim You are absolutely correct, you did substitution of $\tau \equiv \frac{x}{t}$ in $h * f$. If you make this an answer, I can accept it!
Fix an $$x \in (0, 1)$$. If $$h$$ has a singularity at $$1$$ and we define the convolution as an improper integral, we obtain $$f*h = \lim_{y \downarrow x} \int_y^1 f(t) h {\left( \frac x t \right)} \frac {dt} t = \lim_{y \downarrow x} G(y), \\ h*f = \lim_{y \uparrow 1} \int_x^y h(t) f {\left( \frac x t \right)} \frac {dt} t = \lim_{y \uparrow 1} \int_{x/y}^1 f(t) h {\left( \frac x t \right)} \frac {dt} t = \lim_{y \uparrow 1} G {\left( \frac x y \right)}.$$ When $$y$$ tends to $$1$$ from below, $$x/y$$ tends to $$x$$ from above and $$x/y \neq x$$ (since $$x \neq 0$$). Under these conditions, the limit composition rule holds: either $$\lim_{y \uparrow 1} G {\left( \frac x y \right)} = \lim_{y \downarrow x} G(y)$$ or both limits do not exist.