Help with the specific improper integral for finding galaxy potential
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I'm working on finding the potential of a razor-thin disk for a galaxy. I have encountered a specific improper integral for which I'm looking for an analytical expression. The integral is given below:
$$
Phi(R, z) = -2pi G Sigma_0alphaint_0^infty e^{-k|z|} J_0(kR)left(alpha^2 + k^2 right)^{(-3/2)} dk,,
$$
where $J_0$ is the Bessel function of the first kind and $alpha$ is some constant.
I'm looking for a solution expressed in terms of any special functions that might appear, no matter how complex they are, but not in terms of power series. I tried using WolframAlpha and couple of other online tools, but none of them find any solution. Could somebody help or point me out to some powerful tool for evaluating integrals like this?
Thanks in advance
integration analysis improper-integrals indefinite-integrals potential-theory
asked Dec 6 '18 at 15:59
Djordje SavicDjordje Savic
62
$endgroup$
add a comment |
$begingroup$
I'm working on finding the potential of a razor-thin disk for a galaxy. I have encountered a specific improper integral for which I'm looking for an analytical expression. The integral is given below:
$$
Phi(R, z) = -2pi G Sigma_0alphaint_0^infty e^{-k|z|} J_0(kR)left(alpha^2 + k^2 right)^{(-3/2)} dk,,
$$
where $J_0$ is the Bessel function of the first kind and $alpha$ is some constant.
I'm looking for a solution expressed in terms of any special functions that might appear, no matter how complex they are, but not in terms of power series. I tried using WolframAlpha and couple of other online tools, but none of them find any solution. Could somebody help or point me out to some powerful tool for evaluating integrals like this?
Thanks in advance
integration analysis improper-integrals indefinite-integrals potential-theory
asked Dec 6 '18 at 15:59
Djordje SavicDjordje Savic
62
$endgroup$
$begingroup$
With binomial expansion of $(a^2+k^2)^{-frac32}$ and the series form of $J_0$, you may find the Cauchy of two series and then apply Laplace transform each term $x^k$.
$endgroup$
– Nosrati
Dec 7 '18 at 10:23
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This integral does not seem to have a valid solution for any $R$ at $z=0$. Is that what you are expecting from a physical point of view?
$endgroup$
– James Arathoon
Dec 9 '18 at 12:00
add a comment |
$begingroup$
I'm working on finding the potential of a razor-thin disk for a galaxy. I have encountered a specific improper integral for which I'm looking for an analytical expression. The integral is given below:
$$
Phi(R, z) = -2pi G Sigma_0alphaint_0^infty e^{-k|z|} J_0(kR)left(alpha^2 + k^2 right)^{(-3/2)} dk,,
$$
where $J_0$ is the Bessel function of the first kind and $alpha$ is some constant.
I'm looking for a solution expressed in terms of any special functions that might appear, no matter how complex they are, but not in terms of power series. I tried using WolframAlpha and couple of other online tools, but none of them find any solution. Could somebody help or point me out to some powerful tool for evaluating integrals like this?
Thanks in advance
integration analysis improper-integrals indefinite-integrals potential-theory
asked Dec 6 '18 at 15:59
Djordje SavicDjordje Savic
62
$endgroup$
I'm working on finding the potential of a razor-thin disk for a galaxy. I have encountered a specific improper integral for which I'm looking for an analytical expression. The integral is given below:
$$
Phi(R, z) = -2pi G Sigma_0alphaint_0^infty e^{-k|z|} J_0(kR)left(alpha^2 + k^2 right)^{(-3/2)} dk,,
$$
where $J_0$ is the Bessel function of the first kind and $alpha$ is some constant.
I'm looking for a solution expressed in terms of any special functions that might appear, no matter how complex they are, but not in terms of power series. I tried using WolframAlpha and couple of other online tools, but none of them find any solution. Could somebody help or point me out to some powerful tool for evaluating integrals like this?
Thanks in advance
integration analysis improper-integrals indefinite-integrals potential-theory
integration analysis improper-integrals indefinite-integrals potential-theory
asked Dec 6 '18 at 15:59
Djordje SavicDjordje Savic
62
asked Dec 6 '18 at 15:59
Djordje SavicDjordje Savic
62
edited Dec 6 '18 at 18:35
Djordje Savic
asked Dec 6 '18 at 15:59
Djordje SavicDjordje Savic
62
asked Dec 6 '18 at 15:59
Djordje SavicDjordje Savic
62
asked Dec 6 '18 at 15:59
Djordje SavicDjordje Savic
62
62
$begingroup$
With binomial expansion of $(a^2+k^2)^{-frac32}$ and the series form of $J_0$, you may find the Cauchy of two series and then apply Laplace transform each term $x^k$.
$endgroup$
– Nosrati
Dec 7 '18 at 10:23
$begingroup$
This integral does not seem to have a valid solution for any $R$ at $z=0$. Is that what you are expecting from a physical point of view?
$endgroup$
– James Arathoon
Dec 9 '18 at 12:00
add a comment |
$begingroup$
With binomial expansion of $(a^2+k^2)^{-frac32}$ and the series form of $J_0$, you may find the Cauchy of two series and then apply Laplace transform each term $x^k$.
$endgroup$
– Nosrati
Dec 7 '18 at 10:23
$begingroup$
This integral does not seem to have a valid solution for any $R$ at $z=0$. Is that what you are expecting from a physical point of view?
$endgroup$
– James Arathoon
Dec 9 '18 at 12:00
$begingroup$
With binomial expansion of $(a^2+k^2)^{-frac32}$ and the series form of $J_0$, you may find the Cauchy of two series and then apply Laplace transform each term $x^k$.
$endgroup$
– Nosrati
Dec 7 '18 at 10:23
$begingroup$
With binomial expansion of $(a^2+k^2)^{-frac32}$ and the series form of $J_0$, you may find the Cauchy of two series and then apply Laplace transform each term $x^k$.
$endgroup$
– Nosrati
Dec 7 '18 at 10:23
$begingroup$
This integral does not seem to have a valid solution for any $R$ at $z=0$. Is that what you are expecting from a physical point of view?
$endgroup$
– James Arathoon
Dec 9 '18 at 12:00
$begingroup$
This integral does not seem to have a valid solution for any $R$ at $z=0$. Is that what you are expecting from a physical point of view?
$endgroup$
– James Arathoon
Dec 9 '18 at 12:00
add a comment |
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$begingroup$
With binomial expansion of $(a^2+k^2)^{-frac32}$ and the series form of $J_0$, you may find the Cauchy of two series and then apply Laplace transform each term $x^k$.
$endgroup$
– Nosrati
Dec 7 '18 at 10:23
$begingroup$
This integral does not seem to have a valid solution for any $R$ at $z=0$. Is that what you are expecting from a physical point of view?
$endgroup$
– James Arathoon
Dec 9 '18 at 12:00