Let $B=\{(x,y): x^2+y^2\leq 1\}$. Decide if the following integral exists and calculate it if it does.
$$\int_B \frac{1}{(x^2+y^2)^{3/2}}\,d\lambda^2(x,y)$$
I wanted to use the transformation formula to solve this integral, using polar coordinates.
Without transformation we get the integral:
$$\int_{0}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{(x^2+y^2)^{3/2}}\, dy\,dx$$
After the transformation with $x=r\cos(\varphi)$ and $y=r\sin(\varphi)$, I am not sure if my boundaries are correct. From the picture (we integrate over the circle with radius $1$) it should be correct.
$$\int_{0}^1\int_0^{2\pi} r\cdot\frac{1}{r^3}\,d\varphi\,dr$$
Where the factor $r$ comes from the determinant, and $\dfrac{1}{r^3}$ after substituting the polar coordinates.
Then the calculation shows, that this integral does not exist.
Doing the same for the integral
$$\int_{\mathbb{R}^2\setminus B} \frac{1}{(x^2+y^2)^{3/2}}\, d\lambda^2(x,y)$$
I should get to the transformed integral:
$$\int_1^\infty\int_{0}^{2\pi}\frac{1}{r^2}\,d\varphi\,dr=2\pi$$
Where the lower and upper bound $1$ and $\infty$ respectively are again clear from the picture. The set $\mathbb{R}^2\setminus B$ contains all the points with distance greater then $1$ from the origin.
Can you confirm this calculation, or point out some eventual flaws?
Thanks in advance.