Multidimensional angular integrals with complex exponentials
$begingroup$
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
asked Dec 5 '18 at 5:13
monstergroup42monstergroup42
155
$endgroup$
add a comment |
$begingroup$
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
asked Dec 5 '18 at 5:13
monstergroup42monstergroup42
155
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 5 '18 at 5:13
monstergroup42monstergroup42
155
$endgroup$
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
asked Dec 5 '18 at 5:13
monstergroup42monstergroup42
155
asked Dec 5 '18 at 5:13
monstergroup42monstergroup42
155
edited Dec 5 '18 at 5:40
monstergroup42
asked Dec 5 '18 at 5:13
monstergroup42monstergroup42
155
asked Dec 5 '18 at 5:13
monstergroup42monstergroup42
155
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:
$$sum_{k in mathbb Z} r^{|k|} r^{|n - k|} =
left( |n| + frac {1 + r^2} {1 - r^2} right) r^{|n|}, \
int_0^{2 pi} int_0^{2 pi} frac
{e^{i (m phi - m' phi')}}
{(a + b cos (phi - phi'))^2} dphi' dphi =
frac {4 pi^2 r^{|m|}} {a^2 - b^2}
left( |m| + frac {1 + r^2} {1 - r^2} right) delta_{m, m'}.$$
answered Dec 6 '18 at 6:15
MaximMaxim
5,7231220
$endgroup$
$begingroup$
$m, m'$ 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
add a comment |
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1 Answer
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$begingroup$
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:
$$sum_{k in mathbb Z} r^{|k|} r^{|n - k|} =
left( |n| + frac {1 + r^2} {1 - r^2} right) r^{|n|}, \
int_0^{2 pi} int_0^{2 pi} frac
{e^{i (m phi - m' phi')}}
{(a + b cos (phi - phi'))^2} dphi' dphi =
frac {4 pi^2 r^{|m|}} {a^2 - b^2}
left( |m| + frac {1 + r^2} {1 - r^2} right) delta_{m, m'}.$$
answered Dec 6 '18 at 6:15
MaximMaxim
5,7231220
$endgroup$
$begingroup$
$m, m'$ 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
add a comment |
$begingroup$
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:
$$sum_{k in mathbb Z} r^{|k|} r^{|n - k|} =
left( |n| + frac {1 + r^2} {1 - r^2} right) r^{|n|}, \
int_0^{2 pi} int_0^{2 pi} frac
{e^{i (m phi - m' phi')}}
{(a + b cos (phi - phi'))^2} dphi' dphi =
frac {4 pi^2 r^{|m|}} {a^2 - b^2}
left( |m| + frac {1 + r^2} {1 - r^2} right) delta_{m, m'}.$$
answered Dec 6 '18 at 6:15
MaximMaxim
5,7231220
$endgroup$
$begingroup$
$m, m'$ 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
add a comment |
$begingroup$
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:
$$sum_{k in mathbb Z} r^{|k|} r^{|n - k|} =
left( |n| + frac {1 + r^2} {1 - r^2} right) r^{|n|}, \
int_0^{2 pi} int_0^{2 pi} frac
{e^{i (m phi - m' phi')}}
{(a + b cos (phi - phi'))^2} dphi' dphi =
frac {4 pi^2 r^{|m|}} {a^2 - b^2}
left( |m| + frac {1 + r^2} {1 - r^2} right) delta_{m, m'}.$$
answered Dec 6 '18 at 6:15
MaximMaxim
5,7231220
$endgroup$
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:
$$sum_{k in mathbb Z} r^{|k|} r^{|n - k|} =
left( |n| + frac {1 + r^2} {1 - r^2} right) r^{|n|}, \
int_0^{2 pi} int_0^{2 pi} frac
{e^{i (m phi - m' phi')}}
{(a + b cos (phi - phi'))^2} dphi' dphi =
frac {4 pi^2 r^{|m|}} {a^2 - b^2}
left( |m| + frac {1 + r^2} {1 - r^2} right) delta_{m, m'}.$$
answered Dec 6 '18 at 6:15
MaximMaxim
5,7231220
edited Dec 6 '18 at 8:20
answered Dec 6 '18 at 6:15
MaximMaxim
5,7231220
answered Dec 6 '18 at 6:15
MaximMaxim
5,7231220
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
add a comment |
$begingroup$
$m, m'$ 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
add a comment |
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Are $m, m'$ integers or arbitrary reals? For $m, m' in mathbb Z$ and $0 < |b| < a$, the integral is $$frac {4 pi^2} {sqrt {a^2 - b^2}} left( frac {sqrt {a^2 - b^2} - a} b right)^{|m|} delta_{m, m'}.$$
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– Maxim
Dec 5 '18 at 16:05
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@Maxim How do you get that? When I use Mathematica to do the integral I get zero for non-zero integers $m, m'$.
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– monstergroup42
Dec 5 '18 at 23:05