example of Riemann double integral

Let us determine the value of the double integral

$I:={\displaystyle {\iint }_D}{\displaystyle \frac{dxdy}{{(1+x^2+y^2)}^2}}$ (1)

where $D$ is the triangle by the lines  $x=0$,  $y=0$  and  $x+y=1$.

Since the triangle is defined by the inequalities  $0\leqq x\leqq 1,\mathrm{\hspace{0.33em}\hspace{0.33em}0}\leqq y\leqq 1-x$,  one can write

$I$ $\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^1}{\displaystyle {\int }_0^{1-x}}{\displaystyle \frac{dxdy}{{(1+x^2+y^2)}^2}}={\displaystyle {\int }_0^1}{\displaystyle \frac{dx}{{(1+x^2)}^2}}{\displaystyle {\int }_0^{1-x}}{\displaystyle \frac{dy}{{\left[1+{\left(\frac{y}{\sqrt{1+x^2}}\right)}^2\right]}^2}}$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^1}{\displaystyle \frac{1}{{(1+x^2)}^2}}\cdot {\displaystyle \frac{\sqrt{1+x^2}}{2}}\underset{y=0}{\overset{1-x}{/}}\left(\arctan {\displaystyle \frac{y}{\sqrt{1+x^2}}}+{\displaystyle \frac{\frac{y}{\sqrt{1+x^2}}}{1+\frac{y^2}{1+x^2}}}\right)dx$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^1}\left({\displaystyle \frac{1}{2}}{(1+x^2)}^{-\frac{3}{2}}\arctan {\displaystyle \frac{1-x}{\sqrt{1+x^2}}}+{\displaystyle \frac{1-x}{(1-x+x^2)(1+x^2)}}\right)𝑑x.$

The last expression seems quite difficult to calculate to a closed form …

Some appropriate substitution (http://planetmath.org/ChangeOfVariablesInIntegralOnMathbbRn)

$$x:=x(u,v),y:=y(u,v)$$

directly to the form (1) could offer a better is

${\displaystyle {\iint }_D}f(x,y)𝑑x𝑑y={\displaystyle {\iint }_{\mathrm{\Delta }}}f(x(u,v),y(u,v))\left|{\displaystyle \frac{\partial (x,y)}{\partial (u,v)}}\right|𝑑u𝑑v.$ (2)

What kind a change of variables would be good?  One idea were to use some “natural substitution”, i.e. such one that would give constant limits (http://planetmath.org/DefiniteIntegral).  For example, the equations

$$x+y:=u,\frac{y}{x}:=v,$$

map the triangular domain (http://planetmath.org/Domain2) $D$ to the “rectangle”

Then we need the Jacobian

$$\frac{\partial (x,y)}{\partial (u,v)}=\frac{u+v^2}{{(v+1)}^3}.$$

By (2), we have

$$I={\int }_0^1{\int }_0^{\mathrm{\infty }}\frac{{(v+1)}^4}{u^2+2v^2+2v+1}\cdot \frac{u+v^2}{{(v+1)}^3}𝑑u𝑑v={\int }_0^{\mathrm{\infty }}(v+1)𝑑v{\int }_0^1\frac{u+v^2}{u^2+2v^2+2v+1}𝑑u.$$

But here after integrating with respect to $u$, one obtains a difficult single integral.  Thus, when the , the integrand may become awkward.

A second idea would be to try to make the integrand simpler.  For this end, the transition to the polar coordinates

$$x:=r\cos \phi ,y:=r\sin \phi$$

in (1) is more suitable.  We have

The Pythagorean theorem gives the equation  $r^2=x^2+y^2={(r\cos \phi )}^2+{(1-r\cos \phi )}^2$,  i.e.

$$r^2\cos 2\phi -2r\cos \phi +1=\mathrm{\hspace{0.33em}0},$$

from which we get the upper limit

$$r=\frac{2\cos \phi \pm \sqrt{4{\cos }^2\phi -4\cos 2\phi }}{2\cos 2\phi }=\frac{\cos \phi \pm \sin \phi }{{\cos }^2\phi -{\sin }^2\phi };$$

this is ${\displaystyle \frac{1}{\cos \phi +\sin \phi }}$, since the “+” alternative can be excluded by choosing e.g.  $\phi =\frac{\pi }{2}$.  Thus

$$\mathrm{\Delta }:\mathit{\hspace{1em}}0\leqq \phi \leqq \frac{\pi }{2},0\leqq r\leqq \frac{1}{\cos \phi +\sin \phi }$$

and

$$I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\int }_0^{\frac{1}{\cos \phi +\sin \phi }}\frac{2rdr}{{(1+r^2)}^2}𝑑\phi =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{d\phi }{2+\sin 2\phi }.$$

Here, the http://planetmath.org/node/9380Weierstrass substitution  $\tan \phi :=t$  easily yields the final result

$I={\displaystyle \frac{2\pi \sqrt{3}}{9}}.$ (3)

Title	example of Riemann double integral
Canonical name	ExampleOfRiemannDoubleIntegral
Date of creation	2013-03-22 19:12:22
Last modified on	2013-03-22 19:12:22
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	11
Author	pahio (2872)
Entry type	Example
Classification	msc 26A42
Classification	msc 28-00
Related topic	SubstitutionNotation
Related topic	ChangeOfVariablesInIntegralOnMathbbRn
Related topic	ExampleOfRiemannTripleIntegral