Integration over a formula including arcsin
$begingroup$
When i calulate a probability formula, i met a double integration as follows:
$$int^{rho}_{0}int^{rho}_{0}frac{1}{sqrt{t}} dr_{1}dr_{2}$$
where,
$$t=(1-r_{1}^2)(1-r_{2}^2)+(1-r_{1}^2)(1-rho^2)+(1-r_{2}^2)(1-rho^2)+2(1-r_{1}r_{2})(1-rho^2)+2(1-r_{1}rho)(1-r_{2}^2)+2(1-r_{2}rho)(1-r_{1}^2)$$
To calculate such an integration, i first integrate over $dr_{2}$ and get:
$$int^{rho}_{0}frac{1}{sqrt{-r_{1}^2-2rho r_{1}-rho^2+4}}{(s-t)dr_{1}}$$
where,
$$s = arcsinfrac{-2rho_{1}r_{1}^2+(1-3rho^{2})r_{1}+rho(3-rho^{2})}{2(-r_{1}^2-rho r_{1}+3-rho^{2})}$$
$$t = arcsinfrac{(1-rho^{2})r_{1}+(1-r_{1}^2)rho}{2(-r_{1}^2-rho r_{1}+3-rho^{2})}$$
And i don`t know what to do next to solve such a long integration. So, i just wonder can this double integration finally be solved, even with a long answer. And i really need help for the direction to solve(detail can be ignored). I should integrate over $dr_{1}$ next or just start from double integartion and change of variables.
Thanks a lot!
calculus integration
asked Jan 14 at 12:43
leslie zhangleslie zhang
112
$endgroup$
add a comment |
$begingroup$
When i calulate a probability formula, i met a double integration as follows:
$$int^{rho}_{0}int^{rho}_{0}frac{1}{sqrt{t}} dr_{1}dr_{2}$$
where,
$$t=(1-r_{1}^2)(1-r_{2}^2)+(1-r_{1}^2)(1-rho^2)+(1-r_{2}^2)(1-rho^2)+2(1-r_{1}r_{2})(1-rho^2)+2(1-r_{1}rho)(1-r_{2}^2)+2(1-r_{2}rho)(1-r_{1}^2)$$
To calculate such an integration, i first integrate over $dr_{2}$ and get:
$$int^{rho}_{0}frac{1}{sqrt{-r_{1}^2-2rho r_{1}-rho^2+4}}{(s-t)dr_{1}}$$
where,
$$s = arcsinfrac{-2rho_{1}r_{1}^2+(1-3rho^{2})r_{1}+rho(3-rho^{2})}{2(-r_{1}^2-rho r_{1}+3-rho^{2})}$$
$$t = arcsinfrac{(1-rho^{2})r_{1}+(1-r_{1}^2)rho}{2(-r_{1}^2-rho r_{1}+3-rho^{2})}$$
And i don`t know what to do next to solve such a long integration. So, i just wonder can this double integration finally be solved, even with a long answer. And i really need help for the direction to solve(detail can be ignored). I should integrate over $dr_{1}$ next or just start from double integartion and change of variables.
Thanks a lot!
calculus integration
asked Jan 14 at 12:43
leslie zhangleslie zhang
112
$endgroup$
$begingroup$
May we assume that $0<rhole1$ so as to guarantee $tge0$?
$endgroup$
– David H
Jan 14 at 14:08
$begingroup$
yeah,$rho$is correlation
$endgroup$
– leslie zhang
Jan 14 at 14:13
$begingroup$
I'm mostly sure the value of the integral in the $rho=1$ case is $I=frac{2pi^{2}}{15}$. If you're lucky, perhaps the general case can be reduced to dilogarithms.
$endgroup$
– David H
Jan 15 at 10:57
$begingroup$
Thanks so much. Dilogarithms is quite a possible result. Hope i am lucky enough.
$endgroup$
– leslie zhang
Jan 15 at 11:54
add a comment |
$begingroup$
When i calulate a probability formula, i met a double integration as follows:
$$int^{rho}_{0}int^{rho}_{0}frac{1}{sqrt{t}} dr_{1}dr_{2}$$
where,
$$t=(1-r_{1}^2)(1-r_{2}^2)+(1-r_{1}^2)(1-rho^2)+(1-r_{2}^2)(1-rho^2)+2(1-r_{1}r_{2})(1-rho^2)+2(1-r_{1}rho)(1-r_{2}^2)+2(1-r_{2}rho)(1-r_{1}^2)$$
To calculate such an integration, i first integrate over $dr_{2}$ and get:
$$int^{rho}_{0}frac{1}{sqrt{-r_{1}^2-2rho r_{1}-rho^2+4}}{(s-t)dr_{1}}$$
where,
$$s = arcsinfrac{-2rho_{1}r_{1}^2+(1-3rho^{2})r_{1}+rho(3-rho^{2})}{2(-r_{1}^2-rho r_{1}+3-rho^{2})}$$
$$t = arcsinfrac{(1-rho^{2})r_{1}+(1-r_{1}^2)rho}{2(-r_{1}^2-rho r_{1}+3-rho^{2})}$$
And i don`t know what to do next to solve such a long integration. So, i just wonder can this double integration finally be solved, even with a long answer. And i really need help for the direction to solve(detail can be ignored). I should integrate over $dr_{1}$ next or just start from double integartion and change of variables.
Thanks a lot!
calculus integration
asked Jan 14 at 12:43
leslie zhangleslie zhang
112
$endgroup$
When i calulate a probability formula, i met a double integration as follows:
$$int^{rho}_{0}int^{rho}_{0}frac{1}{sqrt{t}} dr_{1}dr_{2}$$
where,
$$t=(1-r_{1}^2)(1-r_{2}^2)+(1-r_{1}^2)(1-rho^2)+(1-r_{2}^2)(1-rho^2)+2(1-r_{1}r_{2})(1-rho^2)+2(1-r_{1}rho)(1-r_{2}^2)+2(1-r_{2}rho)(1-r_{1}^2)$$
To calculate such an integration, i first integrate over $dr_{2}$ and get:
$$int^{rho}_{0}frac{1}{sqrt{-r_{1}^2-2rho r_{1}-rho^2+4}}{(s-t)dr_{1}}$$
where,
$$s = arcsinfrac{-2rho_{1}r_{1}^2+(1-3rho^{2})r_{1}+rho(3-rho^{2})}{2(-r_{1}^2-rho r_{1}+3-rho^{2})}$$
$$t = arcsinfrac{(1-rho^{2})r_{1}+(1-r_{1}^2)rho}{2(-r_{1}^2-rho r_{1}+3-rho^{2})}$$
And i don`t know what to do next to solve such a long integration. So, i just wonder can this double integration finally be solved, even with a long answer. And i really need help for the direction to solve(detail can be ignored). I should integrate over $dr_{1}$ next or just start from double integartion and change of variables.
Thanks a lot!
calculus integration
calculus integration
asked Jan 14 at 12:43
leslie zhangleslie zhang
112
asked Jan 14 at 12:43
leslie zhangleslie zhang
112
edited Jan 14 at 13:19
leslie zhang
asked Jan 14 at 12:43
leslie zhangleslie zhang
112
asked Jan 14 at 12:43
leslie zhangleslie zhang
112
asked Jan 14 at 12:43
leslie zhangleslie zhang
112
112
$begingroup$
May we assume that $0<rhole1$ so as to guarantee $tge0$?
$endgroup$
– David H
Jan 14 at 14:08
$begingroup$
yeah,$rho$is correlation
$endgroup$
– leslie zhang
Jan 14 at 14:13
$begingroup$
I'm mostly sure the value of the integral in the $rho=1$ case is $I=frac{2pi^{2}}{15}$. If you're lucky, perhaps the general case can be reduced to dilogarithms.
$endgroup$
– David H
Jan 15 at 10:57
$begingroup$
Thanks so much. Dilogarithms is quite a possible result. Hope i am lucky enough.
$endgroup$
– leslie zhang
Jan 15 at 11:54
add a comment |
$begingroup$
May we assume that $0<rhole1$ so as to guarantee $tge0$?
$endgroup$
– David H
Jan 14 at 14:08
$begingroup$
yeah,$rho$is correlation
$endgroup$
– leslie zhang
Jan 14 at 14:13
$begingroup$
I'm mostly sure the value of the integral in the $rho=1$ case is $I=frac{2pi^{2}}{15}$. If you're lucky, perhaps the general case can be reduced to dilogarithms.
$endgroup$
– David H
Jan 15 at 10:57
$begingroup$
Thanks so much. Dilogarithms is quite a possible result. Hope i am lucky enough.
$endgroup$
– leslie zhang
Jan 15 at 11:54
$begingroup$
May we assume that $0<rhole1$ so as to guarantee $tge0$?
$endgroup$
– David H
Jan 14 at 14:08
$begingroup$
May we assume that $0<rhole1$ so as to guarantee $tge0$?
$endgroup$
– David H
Jan 14 at 14:08
$begingroup$
yeah,$rho$is correlation
$endgroup$
– leslie zhang
Jan 14 at 14:13
$begingroup$
yeah,$rho$is correlation
$endgroup$
– leslie zhang
Jan 14 at 14:13
$begingroup$
I'm mostly sure the value of the integral in the $rho=1$ case is $I=frac{2pi^{2}}{15}$. If you're lucky, perhaps the general case can be reduced to dilogarithms.
$endgroup$
– David H
Jan 15 at 10:57
$begingroup$
I'm mostly sure the value of the integral in the $rho=1$ case is $I=frac{2pi^{2}}{15}$. If you're lucky, perhaps the general case can be reduced to dilogarithms.
$endgroup$
– David H
Jan 15 at 10:57
$begingroup$
Thanks so much. Dilogarithms is quite a possible result. Hope i am lucky enough.
$endgroup$
– leslie zhang
Jan 15 at 11:54
$begingroup$
Thanks so much. Dilogarithms is quite a possible result. Hope i am lucky enough.
$endgroup$
– leslie zhang
Jan 15 at 11:54
add a comment |
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$begingroup$
May we assume that $0<rhole1$ so as to guarantee $tge0$?
$endgroup$
– David H
Jan 14 at 14:08
$begingroup$
yeah,$rho$is correlation
$endgroup$
– leslie zhang
Jan 14 at 14:13
$begingroup$
I'm mostly sure the value of the integral in the $rho=1$ case is $I=frac{2pi^{2}}{15}$. If you're lucky, perhaps the general case can be reduced to dilogarithms.
$endgroup$
– David H
Jan 15 at 10:57
$begingroup$
Thanks so much. Dilogarithms is quite a possible result. Hope i am lucky enough.
$endgroup$
– leslie zhang
Jan 15 at 11:54