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I am trying to compute the weigth of an ellipse with the density: $\exp(-r^2/2)$ or:

$M_2=\int_{U} \frac{1}{(2\pi)}\exp\Big(-\frac{(x^2+y^2)}{2}\Big)\; dx dy \quad\text{et}\quad U=\Big\{ \frac{x^2}{a^2}+\frac{y^2}{b^2}<1\Big\}$ and more generally:

$M_n=\int_{U}\frac{1}{{(2\pi)}^{n/2}}\exp\Big(-\frac{1}{2}\Big(x_1^2+\cdots+x_n^2\Big)\Big)\; dx_1\cdots dx_n\quad\text{et}\quad U=\{ \frac{x_1^2}{a_1^2}+\cdots+\frac{x_n^2}{a_n^2}<1\}$

Simple variable substitutions (polar, rational) don't work. I believe that one needs use improper convergent integrals and use the residue theorem. This seems to be a delicat matter.

Thanks for your help.

JP

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    $\begingroup$ These are not "elliptic integrals"... elliptic integrals arise from the calculation of arc lengths of an ellipse. See: en.wikipedia.org/wiki/Elliptic_integral $\endgroup$
    – Stanley Yao Xiao
    Feb 14, 2018 at 13:22
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    $\begingroup$ I took the liberty of changing the title in response to this comment. $\endgroup$
    – Carlo Beenakker
    Feb 14, 2018 at 20:00

1 Answer 1

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The integral $M_2$ was studied in a 1961 US military report, to calculate the probabability of a missile hitting an elliptical target. See equation 31, where the integral $M_2=p(a,b)$ is given as a sum over Bessel functions $I_n$. (There does not seem to be a closed form solution.)

The answer can be seen as a two-dimensional analogue of the error-function, defined as an integral over $I_0$,

$$E(R,r)=e^{-r^2/2}\int_0^R e^{-t^2/2}I_0(rt)tdt$$

Then the desired integral over the ellipse is

$$M_2=\frac{1}{2\pi}\int\int_{x^2/a^2+y^2/b^2\leq 1} e^{-(x^2+y^2)/2}dxdy$$ $$=E[(a+b)/2,(a-b)/2]-E[(a-b)/2,(a+b)/2]$$

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  • $\begingroup$ Thank you very much Carlo for this information and solution. It does not bode well for a simple solution in the case n>2. $\endgroup$
    – JP M
    Feb 19, 2018 at 10:34
  • $\begingroup$ I tried to calculate E(R,r) with[I_0(x)=\frac{1}{\pi}\int_0^\pi \cos(-x\sin \theta)d\theta=\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{ix\sin \theta}d\theta] $\endgroup$
    – JP M
    Feb 19, 2018 at 10:36
  • $\begingroup$ but did not get anywhere. JP $\endgroup$
    – JP M
    Feb 19, 2018 at 10:37

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