Define $B:=\{(x,y) \in \mathbb R^{2}: x^2+y^2\leq1\}$, determine whether
$$ \int_{B}\frac{1}{(x^2+y^2)^{3/2}}d\lambda^2(x,y) \quad \text{and} \quad \int_{\mathbb R^2-B}\frac{1}{(x^2+y^2)^{3/2}}d\lambda^2(x,y) $$ are integrable.
My ideas:
Set $C:=\{(x,y) \in \mathbb R_{+}^{2}: x^2+y^2\leq1\}$ to get $$ \begin{split} I &= \int_{B}\frac{1}{(x^2+y^2)^{3/2}}d\lambda^2(x,y) \\ &= 2\int_{C}\frac{1}{(x^2+y^2)^{3/2}} d\lambda^2(x,y) \quad \text{and using polar coordinates} \\ &= 2\int_0^1 \int_0^\pi \frac{rd\phi dr}{r^3} \\ &= 2\pi \left[\left.-\frac{1}{r}\right|_{0}^{1}\right] =\infty \end{split} $$
I do not know what to say on the case $$\int_{\mathbb R^2-B}\frac{1}{(x^2+y^2)^{3/2}}d\lambda^2(x,y).$$ It is clear that it is not integrable but how do I show this?
Questions:
In the above case finding $$ \int_B \left|\frac{1}{(x^2+y^2)^{3/2}}\right|d\lambda^2(x,y) $$ was easy due to the symmetry of on $B$, but how can I solve a problem like when there is no symmetry involved? Do I simply divide the functions in positive and negative parts?
Our professors are very pedantic on the use of correct reasoning. I need to reason why I can write $$\int_{B}|\frac{1}{(x^2+y^2)^{3/2}}|d\lambda^2(x,y),$$ in other words I need to reason why $$\left|\frac{1}{(x^2+y^2)^{3/2}}\right|$$ is measurable but surely I cannot show this as ($f$ measurable $\Rightarrow |f|$ measurable ) is not true in general is it?