About $ F(u) = int_{-pi/2}^{+pi/2} ln(g(x) + u) dx $
$begingroup$
We know for $ u > 1 $
$$ int_{-pi/2}^{+pi/2} ln(sin(x) + u) dx
= pi left(lnleft(u + sqrt{u^2 -1}right) - ln(2)right) $$
Usually this is shown by using differentiation under the Integral sign or contour integration.
This made me wonder :
Consider the log-transform
For a given real continuous function $F(u)$ find a real continuous function $g(x) $ that only depends on $x$ ( not on $u$ ) such that
$$ F(u) = int_{-pi/2}^{+pi/2} ln(g(x) + u) dx $$
So
$$ mathcal{L}(F(u)) = g(u) $$
Where $mathcal{L}$ stands for “ log-transform “.
For instance
$$ mathcal{L}left[ pi left(lnleft(u + sqrt{u^2 -1}right) - ln(2)right) right] = sin(u). $$
Notice that in this case $sin(2u),sin(3u),sin(4u),... $ and $sin(-u),sin(-2u),sin(-3u),sin(-4u),... $ are also solutions !
Perhaps uniqueness comes from functions that are strictly increasing ? ( $mathcal{L} ... = exp(u) $? )
What is known about these (inverse) integral transforms ?
Is there an Integral representation for them ?
calculus representation-theory closed-form integral-transforms inverse-problems
edited Nov 28 '18 at 12:42
DavidG
2,1871720
asked Nov 27 '18 at 18:52
mickmick
5,12822064
$endgroup$
add a comment |
$begingroup$
We know for $ u > 1 $
$$ int_{-pi/2}^{+pi/2} ln(sin(x) + u) dx
= pi left(lnleft(u + sqrt{u^2 -1}right) - ln(2)right) $$
Usually this is shown by using differentiation under the Integral sign or contour integration.
This made me wonder :
Consider the log-transform
For a given real continuous function $F(u)$ find a real continuous function $g(x) $ that only depends on $x$ ( not on $u$ ) such that
$$ F(u) = int_{-pi/2}^{+pi/2} ln(g(x) + u) dx $$
So
$$ mathcal{L}(F(u)) = g(u) $$
Where $mathcal{L}$ stands for “ log-transform “.
For instance
$$ mathcal{L}left[ pi left(lnleft(u + sqrt{u^2 -1}right) - ln(2)right) right] = sin(u). $$
Notice that in this case $sin(2u),sin(3u),sin(4u),... $ and $sin(-u),sin(-2u),sin(-3u),sin(-4u),... $ are also solutions !
Perhaps uniqueness comes from functions that are strictly increasing ? ( $mathcal{L} ... = exp(u) $? )
What is known about these (inverse) integral transforms ?
Is there an Integral representation for them ?
calculus representation-theory closed-form integral-transforms inverse-problems
edited Nov 28 '18 at 12:42
DavidG
2,1871720
asked Nov 27 '18 at 18:52
mickmick
5,12822064
$endgroup$
$begingroup$
Pardon my ignorance, when you say 'log-transform' do you mean to take the log of the function? (Just want to be 100% clear before I give this a go). Btw - Great question.
$endgroup$
– DavidG
Nov 28 '18 at 23:25
$begingroup$
No log is taken DavidG. Just a name. Thanks
$endgroup$
– mick
Nov 29 '18 at 14:49
1
$begingroup$
Thanks for the clarification. Wanted to be sure before I proceeded.
$endgroup$
– DavidG
Nov 29 '18 at 23:12
add a comment |
$begingroup$
We know for $ u > 1 $
$$ int_{-pi/2}^{+pi/2} ln(sin(x) + u) dx
= pi left(lnleft(u + sqrt{u^2 -1}right) - ln(2)right) $$
Usually this is shown by using differentiation under the Integral sign or contour integration.
This made me wonder :
Consider the log-transform
For a given real continuous function $F(u)$ find a real continuous function $g(x) $ that only depends on $x$ ( not on $u$ ) such that
$$ F(u) = int_{-pi/2}^{+pi/2} ln(g(x) + u) dx $$
So
$$ mathcal{L}(F(u)) = g(u) $$
Where $mathcal{L}$ stands for “ log-transform “.
For instance
$$ mathcal{L}left[ pi left(lnleft(u + sqrt{u^2 -1}right) - ln(2)right) right] = sin(u). $$
Notice that in this case $sin(2u),sin(3u),sin(4u),... $ and $sin(-u),sin(-2u),sin(-3u),sin(-4u),... $ are also solutions !
Perhaps uniqueness comes from functions that are strictly increasing ? ( $mathcal{L} ... = exp(u) $? )
What is known about these (inverse) integral transforms ?
Is there an Integral representation for them ?
calculus representation-theory closed-form integral-transforms inverse-problems
edited Nov 28 '18 at 12:42
DavidG
2,1871720
asked Nov 27 '18 at 18:52
mickmick
5,12822064
$endgroup$
We know for $ u > 1 $
$$ int_{-pi/2}^{+pi/2} ln(sin(x) + u) dx
= pi left(lnleft(u + sqrt{u^2 -1}right) - ln(2)right) $$
Usually this is shown by using differentiation under the Integral sign or contour integration.
This made me wonder :
Consider the log-transform
For a given real continuous function $F(u)$ find a real continuous function $g(x) $ that only depends on $x$ ( not on $u$ ) such that
$$ F(u) = int_{-pi/2}^{+pi/2} ln(g(x) + u) dx $$
So
$$ mathcal{L}(F(u)) = g(u) $$
Where $mathcal{L}$ stands for “ log-transform “.
For instance
$$ mathcal{L}left[ pi left(lnleft(u + sqrt{u^2 -1}right) - ln(2)right) right] = sin(u). $$
Notice that in this case $sin(2u),sin(3u),sin(4u),... $ and $sin(-u),sin(-2u),sin(-3u),sin(-4u),... $ are also solutions !
Perhaps uniqueness comes from functions that are strictly increasing ? ( $mathcal{L} ... = exp(u) $? )
What is known about these (inverse) integral transforms ?
Is there an Integral representation for them ?
calculus representation-theory closed-form integral-transforms inverse-problems
calculus representation-theory closed-form integral-transforms inverse-problems
edited Nov 28 '18 at 12:42
DavidG
2,1871720
asked Nov 27 '18 at 18:52
mickmick
5,12822064
edited Nov 28 '18 at 12:42
DavidG
2,1871720
asked Nov 27 '18 at 18:52
mickmick
5,12822064
edited Nov 28 '18 at 12:42
DavidG
2,1871720
edited Nov 28 '18 at 12:42
DavidG
2,1871720
edited Nov 28 '18 at 12:42
DavidG
2,1871720
2,1871720
asked Nov 27 '18 at 18:52
mickmick
5,12822064
asked Nov 27 '18 at 18:52
mickmick
5,12822064
asked Nov 27 '18 at 18:52
mickmick
5,12822064
5,12822064
$begingroup$
Pardon my ignorance, when you say 'log-transform' do you mean to take the log of the function? (Just want to be 100% clear before I give this a go). Btw - Great question.
$endgroup$
– DavidG
Nov 28 '18 at 23:25
$begingroup$
No log is taken DavidG. Just a name. Thanks
$endgroup$
– mick
Nov 29 '18 at 14:49
1
$begingroup$
Thanks for the clarification. Wanted to be sure before I proceeded.
$endgroup$
– DavidG
Nov 29 '18 at 23:12
add a comment |
$begingroup$
Pardon my ignorance, when you say 'log-transform' do you mean to take the log of the function? (Just want to be 100% clear before I give this a go). Btw - Great question.
$endgroup$
– DavidG
Nov 28 '18 at 23:25
$begingroup$
No log is taken DavidG. Just a name. Thanks
$endgroup$
– mick
Nov 29 '18 at 14:49
1
$begingroup$
Thanks for the clarification. Wanted to be sure before I proceeded.
$endgroup$
– DavidG
Nov 29 '18 at 23:12
$begingroup$
Pardon my ignorance, when you say 'log-transform' do you mean to take the log of the function? (Just want to be 100% clear before I give this a go). Btw - Great question.
$endgroup$
– DavidG
Nov 28 '18 at 23:25
$begingroup$
Pardon my ignorance, when you say 'log-transform' do you mean to take the log of the function? (Just want to be 100% clear before I give this a go). Btw - Great question.
$endgroup$
– DavidG
Nov 28 '18 at 23:25
$begingroup$
No log is taken DavidG. Just a name. Thanks
$endgroup$
– mick
Nov 29 '18 at 14:49
$begingroup$
No log is taken DavidG. Just a name. Thanks
$endgroup$
– mick
Nov 29 '18 at 14:49
1
1
$begingroup$
Thanks for the clarification. Wanted to be sure before I proceeded.
$endgroup$
– DavidG
Nov 29 '18 at 23:12
$begingroup$
Thanks for the clarification. Wanted to be sure before I proceeded.
$endgroup$
– DavidG
Nov 29 '18 at 23:12
add a comment |
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$begingroup$
Pardon my ignorance, when you say 'log-transform' do you mean to take the log of the function? (Just want to be 100% clear before I give this a go). Btw - Great question.
$endgroup$
– DavidG
Nov 28 '18 at 23:25
$begingroup$
No log is taken DavidG. Just a name. Thanks
$endgroup$
– mick
Nov 29 '18 at 14:49
1
$begingroup$
Thanks for the clarification. Wanted to be sure before I proceeded.
$endgroup$
– DavidG
Nov 29 '18 at 23:12