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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026663%2fmultidimensional-angular-integrals-with-complex-exponentials%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026663%2fmultidimensional-angular-integrals-with-complex-exponentials%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
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
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
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:"
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more