Question 1062026
.
Compute for the value of 

{{{int(int((1/(1+4x^2+9Y^2)^2), dx,-infinity, infinity),dy, -infinity, infinity)}}}
~~~~~~~~~~~~~~~~~~~~~~ 


<pre>
1.  Let me show first how to calculate more simple integral 


    {{{int(int((1/(1+4x^2+4y^2)^2), dx,-infinity, infinity),dy, -infinity, infinity)}}}.      (1)


    Change variables for polar coordinates x = {{{r*cos(phi)}}},  y = {{{r*sin(phi)}}}.

    Then dx*dy = {{{r*dr*d(phi)}}}, and the integral (1) becomes 

    {{{int(int((r/(1+4r^2)^2), dr,0, infinity),d(phi), 0, 2pi)}}} = {{{(1/2)*int(int((1/(1+4r^2)^2), d(r^2),0, infinity),d(phi), 0, 2pi)}}} = (introduce new variable z = {{{r^2}}}) = {{{(1/2)*int(int((1/(1+36z)^2), dz,0, infinity),d(phi), 0, 2pi)}}}.


    The internal integral is 
                                                          |{{{infinity}}}
    {{{int((1/(1+4z)^2), dz,0, infinity)}}} = {{{(1/4)*(int((1/(1+4z)^2), d(4z),0, infinity)))}}}  = {{{(-1/4)*(1/(1 + 4z))}}} |         =  {{{1/4}}}.
                                                          |0

    Then the entire double integral (1) is  {{{(1/2)*(1/4)*2pi}}} = {{{pi/4}}}.



2.  Now, let us start with the original integral  

    {{{int(int((1/(1+4x^2+9y^2)^2), dx,-infinity, infinity),dy, -infinity, infinity)}}}.      (2)

To calculate this integral, let me introduce the new coordinate system 

   u = x,
   v = {{{(3/2)*y}}}.

Then

   x = u,  y = {{{(2/3)*v}}};  dx = du, dy = {{{(2/3)*dv}}}, and the integral (2) becomes


   {{{(2/3)*int(int((1/(1+4u^2+4v^2)^2), du,-infinity, infinity),dv, -infinity, infinity)}}}.

   The integral of "u" and "v" after the factor {{{2/3)}}} was just calculated in the section #1.

   Hence, the final answer is {{{(2/3)*(pi/4)}}} = {{{pi/6}}}.
</pre>

<U>Answer</U>. The integral (2) is equal to {{{pi/6}}}.