Does Wikipedia misstate Glasser's master theorem












8












$begingroup$


Citing from Wikipedia and one of its references, MathWorld, following hold:



Glasser's master Theorem For $f$ integrable, $Phi(x) = |a|x - sum_{i=1}^N frac{|alpha_i|}{x-beta_i}$ and $a$, $alpha_i$, $beta_i$ arbitrary real constants the identity
begin{equation}
mathrm{PV}int_{-infty}^infty f(Phi(x)) dx =
mathrm{PV} int_{-infty}^infty f(x) dx
label{Glasser}
tag{1}
end{equation}
holds.



Now consider
begin{align*}
Phi_1(x) &= |a|x - sum_{i=1}^N frac{|alpha_i|}{x-beta_i} \
Phi_2(x) &= x - sum_{i=1}^N frac{|aalpha_i|}{x-|a|beta_i}
end{align*}
Then, by Glasser's theorem ref{Glasser}
$$mathrm{PV}int_{-infty}^infty f(Phi_1(x)) dx =
mathrm{PV} int_{-infty}^infty f(x) dx =
mathrm{PV}int_{-infty}^infty f(Phi_2(x)).$$



However, under the change of variables $y = |a| x$
begin{equation}
mathrm{PV}int_{-infty}^infty f(Phi_1(x)) dx =
frac{1}{|a|}mathrm{PV} int_{-infty}^infty f(Phi_2(y)) dy.
label{my Idea}
tag{2}
end{equation}



Thus, I assume Glasser's theorem only holds for $|a| = 1$; a quick numerical check seems to support Eq. ref{my Idea}. Is Wikipedia and MathWorld wrong about this?






integration proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    Good question! Maybe that’s a flaw, but I’ve seen this version of master theorem too many times that I cannot believe it is wrong.
    $endgroup$
    – Szeto
    Jul 26 '18 at 15:11










  • $begingroup$
    sos440.blogspot.com/2017/01/glassers-master-theorem.html?m=1
    $endgroup$
    – Bob
    Jul 26 '18 at 15:31










  • $begingroup$
    The paper of Glasser: jstor.org/stable/2007531?seq=1#page_scan_tab_contents
    $endgroup$
    – Calvin Khor
    Jul 27 '18 at 14:06
















8












$begingroup$


Citing from Wikipedia and one of its references, MathWorld, following hold:



Glasser's master Theorem For $f$ integrable, $Phi(x) = |a|x - sum_{i=1}^N frac{|alpha_i|}{x-beta_i}$ and $a$, $alpha_i$, $beta_i$ arbitrary real constants the identity
begin{equation}
mathrm{PV}int_{-infty}^infty f(Phi(x)) dx =
mathrm{PV} int_{-infty}^infty f(x) dx
label{Glasser}
tag{1}
end{equation}
holds.



Now consider
begin{align*}
Phi_1(x) &= |a|x - sum_{i=1}^N frac{|alpha_i|}{x-beta_i} \
Phi_2(x) &= x - sum_{i=1}^N frac{|aalpha_i|}{x-|a|beta_i}
end{align*}
Then, by Glasser's theorem ref{Glasser}
$$mathrm{PV}int_{-infty}^infty f(Phi_1(x)) dx =
mathrm{PV} int_{-infty}^infty f(x) dx =
mathrm{PV}int_{-infty}^infty f(Phi_2(x)).$$



However, under the change of variables $y = |a| x$
begin{equation}
mathrm{PV}int_{-infty}^infty f(Phi_1(x)) dx =
frac{1}{|a|}mathrm{PV} int_{-infty}^infty f(Phi_2(y)) dy.
label{my Idea}
tag{2}
end{equation}



Thus, I assume Glasser's theorem only holds for $|a| = 1$; a quick numerical check seems to support Eq. ref{my Idea}. Is Wikipedia and MathWorld wrong about this?






integration proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    Good question! Maybe that’s a flaw, but I’ve seen this version of master theorem too many times that I cannot believe it is wrong.
    $endgroup$
    – Szeto
    Jul 26 '18 at 15:11










  • $begingroup$
    sos440.blogspot.com/2017/01/glassers-master-theorem.html?m=1
    $endgroup$
    – Bob
    Jul 26 '18 at 15:31










  • $begingroup$
    The paper of Glasser: jstor.org/stable/2007531?seq=1#page_scan_tab_contents
    $endgroup$
    – Calvin Khor
    Jul 27 '18 at 14:06














8












8








8


1



$begingroup$


Citing from Wikipedia and one of its references, MathWorld, following hold:



Glasser's master Theorem For $f$ integrable, $Phi(x) = |a|x - sum_{i=1}^N frac{|alpha_i|}{x-beta_i}$ and $a$, $alpha_i$, $beta_i$ arbitrary real constants the identity
begin{equation}
mathrm{PV}int_{-infty}^infty f(Phi(x)) dx =
mathrm{PV} int_{-infty}^infty f(x) dx
label{Glasser}
tag{1}
end{equation}
holds.



Now consider
begin{align*}
Phi_1(x) &= |a|x - sum_{i=1}^N frac{|alpha_i|}{x-beta_i} \
Phi_2(x) &= x - sum_{i=1}^N frac{|aalpha_i|}{x-|a|beta_i}
end{align*}
Then, by Glasser's theorem ref{Glasser}
$$mathrm{PV}int_{-infty}^infty f(Phi_1(x)) dx =
mathrm{PV} int_{-infty}^infty f(x) dx =
mathrm{PV}int_{-infty}^infty f(Phi_2(x)).$$



However, under the change of variables $y = |a| x$
begin{equation}
mathrm{PV}int_{-infty}^infty f(Phi_1(x)) dx =
frac{1}{|a|}mathrm{PV} int_{-infty}^infty f(Phi_2(y)) dy.
label{my Idea}
tag{2}
end{equation}



Thus, I assume Glasser's theorem only holds for $|a| = 1$; a quick numerical check seems to support Eq. ref{my Idea}. Is Wikipedia and MathWorld wrong about this?






integration proof-verification







share|cite|improve this question











$endgroup$




Citing from Wikipedia and one of its references, MathWorld, following hold:



Glasser's master Theorem For $f$ integrable, $Phi(x) = |a|x - sum_{i=1}^N frac{|alpha_i|}{x-beta_i}$ and $a$, $alpha_i$, $beta_i$ arbitrary real constants the identity
begin{equation}
mathrm{PV}int_{-infty}^infty f(Phi(x)) dx =
mathrm{PV} int_{-infty}^infty f(x) dx
label{Glasser}
tag{1}
end{equation}
holds.



Now consider
begin{align*}
Phi_1(x) &= |a|x - sum_{i=1}^N frac{|alpha_i|}{x-beta_i} \
Phi_2(x) &= x - sum_{i=1}^N frac{|aalpha_i|}{x-|a|beta_i}
end{align*}
Then, by Glasser's theorem ref{Glasser}
$$mathrm{PV}int_{-infty}^infty f(Phi_1(x)) dx =
mathrm{PV} int_{-infty}^infty f(x) dx =
mathrm{PV}int_{-infty}^infty f(Phi_2(x)).$$



However, under the change of variables $y = |a| x$
begin{equation}
mathrm{PV}int_{-infty}^infty f(Phi_1(x)) dx =
frac{1}{|a|}mathrm{PV} int_{-infty}^infty f(Phi_2(y)) dy.
label{my Idea}
tag{2}
end{equation}



Thus, I assume Glasser's theorem only holds for $|a| = 1$; a quick numerical check seems to support Eq. ref{my Idea}. Is Wikipedia and MathWorld wrong about this?






integration proof-verification




integration proof-verification






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edited Jul 26 '18 at 15:06







manthano

















asked Jul 26 '18 at 14:16









manthanomanthano

21518




21518












  • $begingroup$
    Good question! Maybe that’s a flaw, but I’ve seen this version of master theorem too many times that I cannot believe it is wrong.
    $endgroup$
    – Szeto
    Jul 26 '18 at 15:11










  • $begingroup$
    sos440.blogspot.com/2017/01/glassers-master-theorem.html?m=1
    $endgroup$
    – Bob
    Jul 26 '18 at 15:31










  • $begingroup$
    The paper of Glasser: jstor.org/stable/2007531?seq=1#page_scan_tab_contents
    $endgroup$
    – Calvin Khor
    Jul 27 '18 at 14:06


















  • $begingroup$
    Good question! Maybe that’s a flaw, but I’ve seen this version of master theorem too many times that I cannot believe it is wrong.
    $endgroup$
    – Szeto
    Jul 26 '18 at 15:11










  • $begingroup$
    sos440.blogspot.com/2017/01/glassers-master-theorem.html?m=1
    $endgroup$
    – Bob
    Jul 26 '18 at 15:31










  • $begingroup$
    The paper of Glasser: jstor.org/stable/2007531?seq=1#page_scan_tab_contents
    $endgroup$
    – Calvin Khor
    Jul 27 '18 at 14:06
















$begingroup$
Good question! Maybe that’s a flaw, but I’ve seen this version of master theorem too many times that I cannot believe it is wrong.
$endgroup$
– Szeto
Jul 26 '18 at 15:11




$begingroup$
Good question! Maybe that’s a flaw, but I’ve seen this version of master theorem too many times that I cannot believe it is wrong.
$endgroup$
– Szeto
Jul 26 '18 at 15:11












$begingroup$
sos440.blogspot.com/2017/01/glassers-master-theorem.html?m=1
$endgroup$
– Bob
Jul 26 '18 at 15:31




$begingroup$
sos440.blogspot.com/2017/01/glassers-master-theorem.html?m=1
$endgroup$
– Bob
Jul 26 '18 at 15:31












$begingroup$
The paper of Glasser: jstor.org/stable/2007531?seq=1#page_scan_tab_contents
$endgroup$
– Calvin Khor
Jul 27 '18 at 14:06




$begingroup$
The paper of Glasser: jstor.org/stable/2007531?seq=1#page_scan_tab_contents
$endgroup$
– Calvin Khor
Jul 27 '18 at 14:06










1 Answer
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$begingroup$

Thank you for noticing this. Your argument is correct. (An alternative way to see the error is simply to set every $α_i$ to 0.) I fixed the Wikipedia page (it now uses $x-a$ in place of $|a|x$) and submitted a correction to MathWorld.






share|cite|improve this answer









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    $begingroup$

    Thank you for noticing this. Your argument is correct. (An alternative way to see the error is simply to set every $α_i$ to 0.) I fixed the Wikipedia page (it now uses $x-a$ in place of $|a|x$) and submitted a correction to MathWorld.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Thank you for noticing this. Your argument is correct. (An alternative way to see the error is simply to set every $α_i$ to 0.) I fixed the Wikipedia page (it now uses $x-a$ in place of $|a|x$) and submitted a correction to MathWorld.






      share|cite|improve this answer









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        3





        $begingroup$

        Thank you for noticing this. Your argument is correct. (An alternative way to see the error is simply to set every $α_i$ to 0.) I fixed the Wikipedia page (it now uses $x-a$ in place of $|a|x$) and submitted a correction to MathWorld.






        share|cite|improve this answer









        $endgroup$



        Thank you for noticing this. Your argument is correct. (An alternative way to see the error is simply to set every $α_i$ to 0.) I fixed the Wikipedia page (it now uses $x-a$ in place of $|a|x$) and submitted a correction to MathWorld.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 9 '18 at 14:48









        Dmytro TaranovskyDmytro Taranovsky

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