How can the following be transformed in to a sum of complete elliptic integrals of the first and second kind
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I have the following, that I known from a numerical implementation of the problem by a third party should be able to be transformed in to elliptic integrals of the first and second kind however I can't find a way of achieving this can any one help.
$$
Aint_{0}^{pi}{frac{1}{left(sqrt{1pm Ccostheta}right)^3}dtheta} + Bint_{0}^{pi}{frac{costheta}{left(sqrt{1pm Ccostheta}right)^3}dtheta}
$$
I need to transform it to the following:
$$
alpha int_{0}^{frac{pi}{2}}{frac{1}{sqrt{1-frac{2C}{1+C}sin^2theta}}dtheta} + beta int_{0}^{frac{pi}{2}}{sqrt{1-frac{2C}{1+C}sin^2theta}dtheta}
$$
I known that $sqrt{1+Ccostheta} = sqrt{1-frac{2C}{1+C}sin^2frac{theta}{2}}$, however I can't work out how to reduce the cube term or remove the $costheta$ from the numerator in the second integral. I also verified that the expression I have for $C$ is equivalent to $frac{2C}{1+C}$ from the numerical solution in the code I have. However $A$ and $B$ aren't related to $alpha$ and $beta$ so I expect both integrals are expressed as some sum of the two elliptic integrals, and this would resolve that disparity.
integral-transforms elliptic-integrals
asked Dec 11 '18 at 23:56
Glen FletcherGlen Fletcher
1314
$endgroup$
add a comment |
$begingroup$
I have the following, that I known from a numerical implementation of the problem by a third party should be able to be transformed in to elliptic integrals of the first and second kind however I can't find a way of achieving this can any one help.
$$
Aint_{0}^{pi}{frac{1}{left(sqrt{1pm Ccostheta}right)^3}dtheta} + Bint_{0}^{pi}{frac{costheta}{left(sqrt{1pm Ccostheta}right)^3}dtheta}
$$
I need to transform it to the following:
$$
alpha int_{0}^{frac{pi}{2}}{frac{1}{sqrt{1-frac{2C}{1+C}sin^2theta}}dtheta} + beta int_{0}^{frac{pi}{2}}{sqrt{1-frac{2C}{1+C}sin^2theta}dtheta}
$$
I known that $sqrt{1+Ccostheta} = sqrt{1-frac{2C}{1+C}sin^2frac{theta}{2}}$, however I can't work out how to reduce the cube term or remove the $costheta$ from the numerator in the second integral. I also verified that the expression I have for $C$ is equivalent to $frac{2C}{1+C}$ from the numerical solution in the code I have. However $A$ and $B$ aren't related to $alpha$ and $beta$ so I expect both integrals are expressed as some sum of the two elliptic integrals, and this would resolve that disparity.
integral-transforms elliptic-integrals
asked Dec 11 '18 at 23:56
Glen FletcherGlen Fletcher
1314
$endgroup$
add a comment |
$begingroup$
I have the following, that I known from a numerical implementation of the problem by a third party should be able to be transformed in to elliptic integrals of the first and second kind however I can't find a way of achieving this can any one help.
$$
Aint_{0}^{pi}{frac{1}{left(sqrt{1pm Ccostheta}right)^3}dtheta} + Bint_{0}^{pi}{frac{costheta}{left(sqrt{1pm Ccostheta}right)^3}dtheta}
$$
I need to transform it to the following:
$$
alpha int_{0}^{frac{pi}{2}}{frac{1}{sqrt{1-frac{2C}{1+C}sin^2theta}}dtheta} + beta int_{0}^{frac{pi}{2}}{sqrt{1-frac{2C}{1+C}sin^2theta}dtheta}
$$
I known that $sqrt{1+Ccostheta} = sqrt{1-frac{2C}{1+C}sin^2frac{theta}{2}}$, however I can't work out how to reduce the cube term or remove the $costheta$ from the numerator in the second integral. I also verified that the expression I have for $C$ is equivalent to $frac{2C}{1+C}$ from the numerical solution in the code I have. However $A$ and $B$ aren't related to $alpha$ and $beta$ so I expect both integrals are expressed as some sum of the two elliptic integrals, and this would resolve that disparity.
integral-transforms elliptic-integrals
asked Dec 11 '18 at 23:56
Glen FletcherGlen Fletcher
1314
$endgroup$
I have the following, that I known from a numerical implementation of the problem by a third party should be able to be transformed in to elliptic integrals of the first and second kind however I can't find a way of achieving this can any one help.
$$
Aint_{0}^{pi}{frac{1}{left(sqrt{1pm Ccostheta}right)^3}dtheta} + Bint_{0}^{pi}{frac{costheta}{left(sqrt{1pm Ccostheta}right)^3}dtheta}
$$
I need to transform it to the following:
$$
alpha int_{0}^{frac{pi}{2}}{frac{1}{sqrt{1-frac{2C}{1+C}sin^2theta}}dtheta} + beta int_{0}^{frac{pi}{2}}{sqrt{1-frac{2C}{1+C}sin^2theta}dtheta}
$$
I known that $sqrt{1+Ccostheta} = sqrt{1-frac{2C}{1+C}sin^2frac{theta}{2}}$, however I can't work out how to reduce the cube term or remove the $costheta$ from the numerator in the second integral. I also verified that the expression I have for $C$ is equivalent to $frac{2C}{1+C}$ from the numerical solution in the code I have. However $A$ and $B$ aren't related to $alpha$ and $beta$ so I expect both integrals are expressed as some sum of the two elliptic integrals, and this would resolve that disparity.
integral-transforms elliptic-integrals
integral-transforms elliptic-integrals
asked Dec 11 '18 at 23:56
Glen FletcherGlen Fletcher
1314
asked Dec 11 '18 at 23:56
Glen FletcherGlen Fletcher
1314
asked Dec 11 '18 at 23:56
Glen FletcherGlen Fletcher
1314
asked Dec 11 '18 at 23:56
Glen FletcherGlen Fletcher
1314
asked Dec 11 '18 at 23:56
Glen FletcherGlen Fletcher
1314
1314
add a comment |
add a comment |
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