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How to solve a double integral using Gaussian integral method? $$\iint _\Omega \frac{xy}{1 + x^2 + y^2}\,\mathrm{d}x\,\mathrm{d}y$$ Where $\Omega$: $0\le x\le 1$ and $0\le y\le 1$.

Jonathan
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    Hey Johnathon, welcome to math.SE! I've submitted an edit to put the maths in your question into MathJax, here's a reference for how to use the system in future. – Lemmon Jun 11 '23 at 03:16

2 Answers2

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$$\int_0^1\int_0^1\frac{xy}{1+x^2+y^2}dxdy=\int_0^1\int_0^1xy\left(\int_0^{\infty}e^{-s-sx^2-sy^2}ds\right)dxdy=\int_0^{\infty}e^{-s}\left(\int_0^1xe^{-sx^2}dx\right)^2ds=\frac{1}{4}\int_0^{\infty}e^{-s}\left(\int_0^1e^{-su}du\right)^2ds$$ etc.

Gérard Letac
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  • Can you please explain the sub you have made? – Jonathan Jun 11 '23 at 08:16
  • what is a 'sub'? – Gérard Letac Jun 11 '23 at 10:04
  • The integral you inserted…how so? – Jonathan Jun 11 '23 at 10:27
  • Oh, sorry I missed the $ds$ in the first integral $\int_0^{\infty} $ – Gérard Letac Jun 11 '23 at 20:28
  • @GérardLetac sub means substitution (in this case replacing one expression with an equivalent one, rather than an integration by substitution). – J.G. Jun 11 '23 at 20:30
  • Oh, I realise that finally $\int_0^{\infty}e^{-s}\frac{(1-e^{-s})^2}{s^2}ds$ is not so trivial and deserve more than an etc. I suggest to add a parameter $F(t)=\int_0^{\infty}e^{-st}\frac{(1-e^{-s})^2}{s^2}ds$ and to compute $F''(t)=\frac{1}{t}-\frac{2}{1+t}+\frac{1}{2+t}$ leading to $F(t)=t\log t-2(1+t)\log(1+t)+(2+t)\log (2+t)$ and $F(1)=\log\frac{27}{16}.$ – Gérard Letac Jun 11 '23 at 20:52
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You need the zeroes of the orthogonal Legendre polynomials on the unit intervall $(0,1)$
$$x\in (0,1) \to L_n(2x+1):\ \ L_n(x_k)=0 $$ lets say for n=4. Solve for weights $w_k$ in the exact integrals of the first 4 powers $$\int_0^1 dx \ x^n = \frac{1}{n} =\sum_1^4 w_k x_k^n $$ and define e.g. in Mathematica

     GaussIntegrate[f_, { x_, w_} ]:= (f/@x) . w

This integration approximation is exact up to polynomials of degree 7 (check) by the orthogonality of the Legendre polynomials. $$ \int x^7 dx = \int x^3 . x^4 dx $$

For checks: The numerical value of the integrals is $\ \ \frac{3}{4}\log 3 - \log 2$

It's quite intersting how difficult it is to generate a symmetric double integral. By introduction of polar coordinates

$$ \frac{x y}{1 + x^2 + y^2} dxdy ,\ : \ \left( x \to r \cos\phi, y \to r \sin \phi, dxdy \to r dr d\phi\right) \longrightarrow \frac{r^3\ dr}{r^2+1} \ d\phi \ \sin \phi \ \cos \phi $$

$$p= \left[\left(\int \frac{r^3}{r^2+1} \, dr\right) \int \sin (\phi ) \cos (\phi ) \, d\phi \text{//.}\, \left\{\cos (\phi ):\to \frac{x}{r},r\to \sqrt{x^2+y^2}\right\}\right]$$

$$\frac{1}{4} x^2 \left(\frac{\log \left(x^2+y^2+1\right)}{x^2+y^2}-1\right)$$ This expression is difficult to reduce to the integrand by $\partial_{x,y}$. Symmetrizing yields a trivial antiderivative

$$\frac{\left(x^2+y^2\right) \log \left(x^2+y^2+1\right)}{2 \left(x^2+y^2\right)}-\frac{1}{2} \left(x^2+y^2\right)$$

The problem with indefinite multiple integrals is the fact, that the constants of integration are free functions of all the other inactive variables.

Roland F
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