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?

Thanks in advance!

  • 1
    $\begingroup$ 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. $\endgroup$
    – Maxim
    Feb 17 '19 at 20:57
  • $\begingroup$ @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! $\endgroup$
    – zyy
    Feb 18 '19 at 4:34

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.


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