Multidimensional angular integrals with complex exponentials












1












$begingroup$


I am doing some multidimensional integrations, the angular parts of which are one of the following types:
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi', frac{e^{-im'phi'} e^{imphi}}{a + b cos(phi - phi')}
end{align}
or
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi', frac{e^{-im'phi'} e^{imphi}}{[a + b cos(phi - phi')]^2}.
end{align}



For $m, m' = 0$ the first integral simplifies to
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi' frac{1}{a + bcos (phi - phi')}
end{align}
which I calculate to be
begin{align}
frac{4pi^2}{sqrt{a^2-b^2}}
end{align}
However I am not sure how to do the integrals when $m, m' ne 0$. I suspect that they will be zero but I am unable to prove that.
I was thinking that maybe I can substitute $phi - phi'$ with $z$, and then integrate over $phi$ and $z$. For example if I do so in the first integral then that becomes
begin{align}
int_0^{2pi} dphi ,e^{-i(m - m')phi} int_phi^{phi-2pi} dz frac{e^{-im'z}}{a + b cos z}.
end{align}
I am unable to progress from there.






integration multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are $m, m'$ integers or arbitrary reals? For $m, m' in mathbb Z$ and $0 < |b| < a$, the integral is $$frac {4 pi^2} {sqrt {a^2 - b^2}} left( frac {sqrt {a^2 - b^2} - a} b right)^{|m|} delta_{m, m'}.$$
    $endgroup$
    – Maxim
    Dec 5 '18 at 16:05












  • $begingroup$
    @Maxim How do you get that? When I use Mathematica to do the integral I get zero for non-zero integers $m, m'$.
    $endgroup$
    – monstergroup42
    Dec 5 '18 at 23:05
















1












$begingroup$


I am doing some multidimensional integrations, the angular parts of which are one of the following types:
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi', frac{e^{-im'phi'} e^{imphi}}{a + b cos(phi - phi')}
end{align}
or
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi', frac{e^{-im'phi'} e^{imphi}}{[a + b cos(phi - phi')]^2}.
end{align}



For $m, m' = 0$ the first integral simplifies to
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi' frac{1}{a + bcos (phi - phi')}
end{align}
which I calculate to be
begin{align}
frac{4pi^2}{sqrt{a^2-b^2}}
end{align}
However I am not sure how to do the integrals when $m, m' ne 0$. I suspect that they will be zero but I am unable to prove that.
I was thinking that maybe I can substitute $phi - phi'$ with $z$, and then integrate over $phi$ and $z$. For example if I do so in the first integral then that becomes
begin{align}
int_0^{2pi} dphi ,e^{-i(m - m')phi} int_phi^{phi-2pi} dz frac{e^{-im'z}}{a + b cos z}.
end{align}
I am unable to progress from there.






integration multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are $m, m'$ integers or arbitrary reals? For $m, m' in mathbb Z$ and $0 < |b| < a$, the integral is $$frac {4 pi^2} {sqrt {a^2 - b^2}} left( frac {sqrt {a^2 - b^2} - a} b right)^{|m|} delta_{m, m'}.$$
    $endgroup$
    – Maxim
    Dec 5 '18 at 16:05












  • $begingroup$
    @Maxim How do you get that? When I use Mathematica to do the integral I get zero for non-zero integers $m, m'$.
    $endgroup$
    – monstergroup42
    Dec 5 '18 at 23:05














1












1








1





$begingroup$


I am doing some multidimensional integrations, the angular parts of which are one of the following types:
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi', frac{e^{-im'phi'} e^{imphi}}{a + b cos(phi - phi')}
end{align}
or
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi', frac{e^{-im'phi'} e^{imphi}}{[a + b cos(phi - phi')]^2}.
end{align}



For $m, m' = 0$ the first integral simplifies to
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi' frac{1}{a + bcos (phi - phi')}
end{align}
which I calculate to be
begin{align}
frac{4pi^2}{sqrt{a^2-b^2}}
end{align}
However I am not sure how to do the integrals when $m, m' ne 0$. I suspect that they will be zero but I am unable to prove that.
I was thinking that maybe I can substitute $phi - phi'$ with $z$, and then integrate over $phi$ and $z$. For example if I do so in the first integral then that becomes
begin{align}
int_0^{2pi} dphi ,e^{-i(m - m')phi} int_phi^{phi-2pi} dz frac{e^{-im'z}}{a + b cos z}.
end{align}
I am unable to progress from there.






integration multivariable-calculus definite-integrals







share|cite|improve this question











$endgroup$




I am doing some multidimensional integrations, the angular parts of which are one of the following types:
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi', frac{e^{-im'phi'} e^{imphi}}{a + b cos(phi - phi')}
end{align}
or
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi', frac{e^{-im'phi'} e^{imphi}}{[a + b cos(phi - phi')]^2}.
end{align}



For $m, m' = 0$ the first integral simplifies to
begin{align}
int_0^{2pi} dphi int_0^{2pi} dphi' frac{1}{a + bcos (phi - phi')}
end{align}
which I calculate to be
begin{align}
frac{4pi^2}{sqrt{a^2-b^2}}
end{align}
However I am not sure how to do the integrals when $m, m' ne 0$. I suspect that they will be zero but I am unable to prove that.
I was thinking that maybe I can substitute $phi - phi'$ with $z$, and then integrate over $phi$ and $z$. For example if I do so in the first integral then that becomes
begin{align}
int_0^{2pi} dphi ,e^{-i(m - m')phi} int_phi^{phi-2pi} dz frac{e^{-im'z}}{a + b cos z}.
end{align}
I am unable to progress from there.






integration multivariable-calculus definite-integrals




integration multivariable-calculus definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 5:40







monstergroup42

















asked Dec 5 '18 at 5:13









monstergroup42monstergroup42

155




155












  • $begingroup$
    Are $m, m'$ integers or arbitrary reals? For $m, m' in mathbb Z$ and $0 < |b| < a$, the integral is $$frac {4 pi^2} {sqrt {a^2 - b^2}} left( frac {sqrt {a^2 - b^2} - a} b right)^{|m|} delta_{m, m'}.$$
    $endgroup$
    – Maxim
    Dec 5 '18 at 16:05












  • $begingroup$
    @Maxim How do you get that? When I use Mathematica to do the integral I get zero for non-zero integers $m, m'$.
    $endgroup$
    – monstergroup42
    Dec 5 '18 at 23:05


















  • $begingroup$
    Are $m, m'$ integers or arbitrary reals? For $m, m' in mathbb Z$ and $0 < |b| < a$, the integral is $$frac {4 pi^2} {sqrt {a^2 - b^2}} left( frac {sqrt {a^2 - b^2} - a} b right)^{|m|} delta_{m, m'}.$$
    $endgroup$
    – Maxim
    Dec 5 '18 at 16:05












  • $begingroup$
    @Maxim How do you get that? When I use Mathematica to do the integral I get zero for non-zero integers $m, m'$.
    $endgroup$
    – monstergroup42
    Dec 5 '18 at 23:05
















$begingroup$
Are $m, m'$ integers or arbitrary reals? For $m, m' in mathbb Z$ and $0 < |b| < a$, the integral is $$frac {4 pi^2} {sqrt {a^2 - b^2}} left( frac {sqrt {a^2 - b^2} - a} b right)^{|m|} delta_{m, m'}.$$
$endgroup$
– Maxim
Dec 5 '18 at 16:05






$begingroup$
Are $m, m'$ integers or arbitrary reals? For $m, m' in mathbb Z$ and $0 < |b| < a$, the integral is $$frac {4 pi^2} {sqrt {a^2 - b^2}} left( frac {sqrt {a^2 - b^2} - a} b right)^{|m|} delta_{m, m'}.$$
$endgroup$
– Maxim
Dec 5 '18 at 16:05














$begingroup$
@Maxim How do you get that? When I use Mathematica to do the integral I get zero for non-zero integers $m, m'$.
$endgroup$
– monstergroup42
Dec 5 '18 at 23:05




$begingroup$
@Maxim How do you get that? When I use Mathematica to do the integral I get zero for non-zero integers $m, m'$.
$endgroup$
– monstergroup42
Dec 5 '18 at 23:05










1 Answer
1






active

oldest

votes


















0












$begingroup$

I'm going to assume that $m, m' in mathbb Z$ and $0 < |b| < a$. Due to periodicity, the inner integral becomes
$$int_0^{2 pi} frac {e^{-i m' phi'}} {a + b cos (phi - phi')} dphi' =
e^{-i m' phi} int_0^{2 pi} frac 1 {sqrt {a^2 - b^2}}
frac {1 - r^2} {1 - 2 r cos phi' + r^2} e^{-i m' phi'} dphi',$$
where $r = (sqrt {a^2 - b^2} - a)/b$. This, up to a constant factor, is the $n$th Fourier series coefficient of a Poisson kernel, which is known and equals $r^{|n|}$. Then
$$int_0^{2 pi} int_0^{2 pi}
frac {e^{i (m phi - m' phi')}} {a + b cos (phi - phi')} dphi' dphi =
frac {2 pi r^{|m'|}} {sqrt {a^2 - b^2}}
int_0^{2 pi} e^{i (m - m') phi} dphi =
frac {4 pi^2 r^{|m|}} {sqrt {a^2 - b^2}} delta_{m, m'}.$$
The $n$th Fourier coefficient for the second integrand is the convolution of $r^{|n|}$ with itself:
$$sum_{k in mathbb Z} r^{|k|} r^{|n - k|} =
left( |n| + frac {1 + r^2} {1 - r^2} right) r^{|n|}, \
int_0^{2 pi} int_0^{2 pi} frac
{e^{i (m phi - m' phi')}}
{(a + b cos (phi - phi'))^2} dphi' dphi =
frac {4 pi^2 r^{|m|}} {a^2 - b^2}
left( |m| + frac {1 + r^2} {1 - r^2} right) delta_{m, m'}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $m, m'$ are indeed integers. As far as I understand, for the Poisson kernel, $r in [0, 1)$. I do not think that is guaranteed for the r that you have presented here. Also do you have any suggestions for the second integral?
    $endgroup$
    – monstergroup42
    Dec 6 '18 at 7:32












  • $begingroup$
    Please see the update. It is sufficient to require $|r| < 1$, which holds for $0 < |b| < a$.
    $endgroup$
    – Maxim
    Dec 6 '18 at 8:26











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









0












$begingroup$

I'm going to assume that $m, m' in mathbb Z$ and $0 < |b| < a$. Due to periodicity, the inner integral becomes
$$int_0^{2 pi} frac {e^{-i m' phi'}} {a + b cos (phi - phi')} dphi' =
e^{-i m' phi} int_0^{2 pi} frac 1 {sqrt {a^2 - b^2}}
frac {1 - r^2} {1 - 2 r cos phi' + r^2} e^{-i m' phi'} dphi',$$
where $r = (sqrt {a^2 - b^2} - a)/b$. This, up to a constant factor, is the $n$th Fourier series coefficient of a Poisson kernel, which is known and equals $r^{|n|}$. Then
$$int_0^{2 pi} int_0^{2 pi}
frac {e^{i (m phi - m' phi')}} {a + b cos (phi - phi')} dphi' dphi =
frac {2 pi r^{|m'|}} {sqrt {a^2 - b^2}}
int_0^{2 pi} e^{i (m - m') phi} dphi =
frac {4 pi^2 r^{|m|}} {sqrt {a^2 - b^2}} delta_{m, m'}.$$
The $n$th Fourier coefficient for the second integrand is the convolution of $r^{|n|}$ with itself:
$$sum_{k in mathbb Z} r^{|k|} r^{|n - k|} =
left( |n| + frac {1 + r^2} {1 - r^2} right) r^{|n|}, \
int_0^{2 pi} int_0^{2 pi} frac
{e^{i (m phi - m' phi')}}
{(a + b cos (phi - phi'))^2} dphi' dphi =
frac {4 pi^2 r^{|m|}} {a^2 - b^2}
left( |m| + frac {1 + r^2} {1 - r^2} right) delta_{m, m'}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $m, m'$ are indeed integers. As far as I understand, for the Poisson kernel, $r in [0, 1)$. I do not think that is guaranteed for the r that you have presented here. Also do you have any suggestions for the second integral?
    $endgroup$
    – monstergroup42
    Dec 6 '18 at 7:32












  • $begingroup$
    Please see the update. It is sufficient to require $|r| < 1$, which holds for $0 < |b| < a$.
    $endgroup$
    – Maxim
    Dec 6 '18 at 8:26
















0












$begingroup$

I'm going to assume that $m, m' in mathbb Z$ and $0 < |b| < a$. Due to periodicity, the inner integral becomes
$$int_0^{2 pi} frac {e^{-i m' phi'}} {a + b cos (phi - phi')} dphi' =
e^{-i m' phi} int_0^{2 pi} frac 1 {sqrt {a^2 - b^2}}
frac {1 - r^2} {1 - 2 r cos phi' + r^2} e^{-i m' phi'} dphi',$$
where $r = (sqrt {a^2 - b^2} - a)/b$. This, up to a constant factor, is the $n$th Fourier series coefficient of a Poisson kernel, which is known and equals $r^{|n|}$. Then
$$int_0^{2 pi} int_0^{2 pi}
frac {e^{i (m phi - m' phi')}} {a + b cos (phi - phi')} dphi' dphi =
frac {2 pi r^{|m'|}} {sqrt {a^2 - b^2}}
int_0^{2 pi} e^{i (m - m') phi} dphi =
frac {4 pi^2 r^{|m|}} {sqrt {a^2 - b^2}} delta_{m, m'}.$$
The $n$th Fourier coefficient for the second integrand is the convolution of $r^{|n|}$ with itself:
$$sum_{k in mathbb Z} r^{|k|} r^{|n - k|} =
left( |n| + frac {1 + r^2} {1 - r^2} right) r^{|n|}, \
int_0^{2 pi} int_0^{2 pi} frac
{e^{i (m phi - m' phi')}}
{(a + b cos (phi - phi'))^2} dphi' dphi =
frac {4 pi^2 r^{|m|}} {a^2 - b^2}
left( |m| + frac {1 + r^2} {1 - r^2} right) delta_{m, m'}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $m, m'$ are indeed integers. As far as I understand, for the Poisson kernel, $r in [0, 1)$. I do not think that is guaranteed for the r that you have presented here. Also do you have any suggestions for the second integral?
    $endgroup$
    – monstergroup42
    Dec 6 '18 at 7:32












  • $begingroup$
    Please see the update. It is sufficient to require $|r| < 1$, which holds for $0 < |b| < a$.
    $endgroup$
    – Maxim
    Dec 6 '18 at 8:26














0












0








0





$begingroup$

I'm going to assume that $m, m' in mathbb Z$ and $0 < |b| < a$. Due to periodicity, the inner integral becomes
$$int_0^{2 pi} frac {e^{-i m' phi'}} {a + b cos (phi - phi')} dphi' =
e^{-i m' phi} int_0^{2 pi} frac 1 {sqrt {a^2 - b^2}}
frac {1 - r^2} {1 - 2 r cos phi' + r^2} e^{-i m' phi'} dphi',$$
where $r = (sqrt {a^2 - b^2} - a)/b$. This, up to a constant factor, is the $n$th Fourier series coefficient of a Poisson kernel, which is known and equals $r^{|n|}$. Then
$$int_0^{2 pi} int_0^{2 pi}
frac {e^{i (m phi - m' phi')}} {a + b cos (phi - phi')} dphi' dphi =
frac {2 pi r^{|m'|}} {sqrt {a^2 - b^2}}
int_0^{2 pi} e^{i (m - m') phi} dphi =
frac {4 pi^2 r^{|m|}} {sqrt {a^2 - b^2}} delta_{m, m'}.$$
The $n$th Fourier coefficient for the second integrand is the convolution of $r^{|n|}$ with itself:
$$sum_{k in mathbb Z} r^{|k|} r^{|n - k|} =
left( |n| + frac {1 + r^2} {1 - r^2} right) r^{|n|}, \
int_0^{2 pi} int_0^{2 pi} frac
{e^{i (m phi - m' phi')}}
{(a + b cos (phi - phi'))^2} dphi' dphi =
frac {4 pi^2 r^{|m|}} {a^2 - b^2}
left( |m| + frac {1 + r^2} {1 - r^2} right) delta_{m, m'}.$$






share|cite|improve this answer











$endgroup$



I'm going to assume that $m, m' in mathbb Z$ and $0 < |b| < a$. Due to periodicity, the inner integral becomes
$$int_0^{2 pi} frac {e^{-i m' phi'}} {a + b cos (phi - phi')} dphi' =
e^{-i m' phi} int_0^{2 pi} frac 1 {sqrt {a^2 - b^2}}
frac {1 - r^2} {1 - 2 r cos phi' + r^2} e^{-i m' phi'} dphi',$$
where $r = (sqrt {a^2 - b^2} - a)/b$. This, up to a constant factor, is the $n$th Fourier series coefficient of a Poisson kernel, which is known and equals $r^{|n|}$. Then
$$int_0^{2 pi} int_0^{2 pi}
frac {e^{i (m phi - m' phi')}} {a + b cos (phi - phi')} dphi' dphi =
frac {2 pi r^{|m'|}} {sqrt {a^2 - b^2}}
int_0^{2 pi} e^{i (m - m') phi} dphi =
frac {4 pi^2 r^{|m|}} {sqrt {a^2 - b^2}} delta_{m, m'}.$$
The $n$th Fourier coefficient for the second integrand is the convolution of $r^{|n|}$ with itself:
$$sum_{k in mathbb Z} r^{|k|} r^{|n - k|} =
left( |n| + frac {1 + r^2} {1 - r^2} right) r^{|n|}, \
int_0^{2 pi} int_0^{2 pi} frac
{e^{i (m phi - m' phi')}}
{(a + b cos (phi - phi'))^2} dphi' dphi =
frac {4 pi^2 r^{|m|}} {a^2 - b^2}
left( |m| + frac {1 + r^2} {1 - r^2} right) delta_{m, m'}.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 8:20

























answered Dec 6 '18 at 6:15









MaximMaxim

5,7231220




5,7231220












  • $begingroup$
    $m, m'$ are indeed integers. As far as I understand, for the Poisson kernel, $r in [0, 1)$. I do not think that is guaranteed for the r that you have presented here. Also do you have any suggestions for the second integral?
    $endgroup$
    – monstergroup42
    Dec 6 '18 at 7:32












  • $begingroup$
    Please see the update. It is sufficient to require $|r| < 1$, which holds for $0 < |b| < a$.
    $endgroup$
    – Maxim
    Dec 6 '18 at 8:26


















  • $begingroup$
    $m, m'$ are indeed integers. As far as I understand, for the Poisson kernel, $r in [0, 1)$. I do not think that is guaranteed for the r that you have presented here. Also do you have any suggestions for the second integral?
    $endgroup$
    – monstergroup42
    Dec 6 '18 at 7:32












  • $begingroup$
    Please see the update. It is sufficient to require $|r| < 1$, which holds for $0 < |b| < a$.
    $endgroup$
    – Maxim
    Dec 6 '18 at 8:26
















$begingroup$
$m, m'$ are indeed integers. As far as I understand, for the Poisson kernel, $r in [0, 1)$. I do not think that is guaranteed for the r that you have presented here. Also do you have any suggestions for the second integral?
$endgroup$
– monstergroup42
Dec 6 '18 at 7:32






$begingroup$
$m, m'$ are indeed integers. As far as I understand, for the Poisson kernel, $r in [0, 1)$. I do not think that is guaranteed for the r that you have presented here. Also do you have any suggestions for the second integral?
$endgroup$
– monstergroup42
Dec 6 '18 at 7:32














$begingroup$
Please see the update. It is sufficient to require $|r| < 1$, which holds for $0 < |b| < a$.
$endgroup$
– Maxim
Dec 6 '18 at 8:26




$begingroup$
Please see the update. It is sufficient to require $|r| < 1$, which holds for $0 < |b| < a$.
$endgroup$
– Maxim
Dec 6 '18 at 8:26


















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mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

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0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
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How to change which sound is reproduced for terminal bell?

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3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
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