## Monday Math 148

Find $I=\int_{-1}^{1}\int_{\frac{x^2}2+x-\frac12}^{-\frac{x^2}2+x+\frac12}\frac{x^5-y^3}{x^2+y^2}\,dy\,dx$.

One could integrate over y by using
$\frac{x^5-y^3}{x^2+y^2}=\frac{x^5}{x^2+y^2}-\frac{y^3}{x^2+y^2}=\frac{x^5}{x^2+y^2}-y+\frac{x^2y}{x^2+y^2}$; however, when you put in the limits, the resulting integral over x becomes very difficult.

Note that the region of integration, the region bounded by the parabolas $y=\frac{x^2}2+x-\frac12$ and $y=-\frac{x^2}2+x+\frac12$, has twofold rotational symmetry about the origin. Thus, we consider the change in variables $u=-x,\;\;v=-y$, which represents axes rotated by 180°. We then have
$\begin{array}{rcl}I&=&\int_{-1}^{1}\int_{\frac{x^2}2+x-\frac12}^{-\frac{x^2}2+x+\frac12}\frac{x^5-y^3}{x^2+y^2}\,dy\,dx\\&=&\int_{1}^{-1}\int_{-\frac{u^2}2+u+\frac12}^{\frac{u^2}2+u-\frac12}\frac{(-u)^5-(-v)^3}{(-u)^2+(-v)^2}\,(-dv)\,(-du)\\&=&\int_{-1}^{1}\int_{\frac{u^2}2+u-\frac12}^{-\frac{u^2}2+u+\frac12}\frac{-u^5+v^3}{u^2+v^2}\,dv\,du\\I&=&-\int_{-1}^{1}\int_{\frac{u^2}2+u-\frac12}^{-\frac{u^2}2+u+\frac12}\frac{u^5-v^3}{u^2+v^2}\,dv\,du\end{array}$
and if we rename u and v as x and y, respectively, we see
$\begin{array}{rcl}I&=&-\int_{-1}^{1}\int_{\frac{x^2}2+x-\frac12}^{-\frac{x^2}2+x+\frac12}\frac{x^5-y^3}{x^2+y^2}\,dy\,dx\\I&=&-I\\I&=&{0}\end{array}$
For our integrand $f(x,y)=\frac{x^5-y^3}{x^2+y^2}$, we have $f(-x,-y)=-f(x,y)$, which combined with the rotational symmetry of the region of integration, means that the integral is zero.