Mixing
Is there a preference if one of the functions in convolution of Mellin transform is divergent?
$begingroup$
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!
integration convolution mellin-transform
edited yesterday
Robert Howard
2,2012932
asked Jan 23 at 0:14
zyyzyy
1136
$endgroup$
add a comment |
$begingroup$
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!
integration convolution mellin-transform
edited yesterday
Robert Howard
2,2012932
asked Jan 23 at 0:14
zyyzyy
1136
$endgroup$
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 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 at 4:34
add a comment |
$begingroup$
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!
integration convolution mellin-transform
edited yesterday
Robert Howard
2,2012932
asked Jan 23 at 0:14
zyyzyy
1136
$endgroup$
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!
integration convolution mellin-transform
integration convolution mellin-transform
edited yesterday
Robert Howard
2,2012932
asked Jan 23 at 0:14
zyyzyy
1136
edited yesterday
Robert Howard
2,2012932
asked Jan 23 at 0:14
zyyzyy
1136
edited yesterday
Robert Howard
2,2012932
edited yesterday
Robert Howard
2,2012932
edited yesterday
Robert Howard
2,2012932
2,2012932
asked Jan 23 at 0:14
zyyzyy
1136
asked Jan 23 at 0:14
zyyzyy
1136
asked Jan 23 at 0:14
zyyzyy
1136
1136
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 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 at 4:34
add a comment |
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 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 at 4:34
1
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 at 20:57
$begingroup$
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 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 at 4:34
$begingroup$
@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 at 4:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Feb 18 at 15:28
MaximMaxim
5,7631220
$endgroup$
add a comment |
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$begingroup$
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.
answered Feb 18 at 15:28
MaximMaxim
5,7631220
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 18 at 15:28
MaximMaxim
5,7631220
$endgroup$
add a comment |
$begingroup$
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.
answered Feb 18 at 15:28
MaximMaxim
5,7631220
$endgroup$
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.
answered Feb 18 at 15:28
MaximMaxim
5,7631220
answered Feb 18 at 15:28
MaximMaxim
5,7631220
answered Feb 18 at 15:28
MaximMaxim
5,7631220
answered Feb 18 at 15:28
MaximMaxim
5,7631220
5,7631220
add a comment |
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$begingroup$
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 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 at 4:34