I am trying to prove that the integral $$\int_{\mathbb{R}^2}|\frac{xy}{(x^2+y^2)^2}|\exp(\frac{-(x^2+y^2)}2) \, dx \otimes \, dy$$ does not converge.
I transformed into polar coordinates, with $x^2+y^2=r^2$ and $x=r\cos(\phi)$, $y=r\sin(\phi)$ and it became
$$\int_{[0, 2\pi) \times (0,\infty)}|\frac{r^2\sin(\phi)\cos(\phi)}{r^4}|\exp\Big(\frac{-r^2}{2}\Big) \, d\phi \otimes \, dr=$$
$$\int_{[0, 2\pi) \times (0,\infty)}|\frac{\sin(\phi)\cos(\phi)}{r^2}|\exp\Big(\frac{-r^2}{2}\Big) \, d\phi \otimes \, dr=$$
Using double angle formula and then grouping variables together
$$\int_{[0, 2\pi) \times (0,\infty)}|\frac{\sin(2\phi)}{2r^2}|\exp\Big(\frac{-r^2}{2}\Big) \, d\phi \otimes \, dr=\int_{[0, 2\pi) \times (0,\infty)}\frac{\exp(\frac{-r^2}{2})}{2r^2}|\sin(2\phi)| \, d\phi \otimes \, dr=$$
Transforming $r^2$ by $r^2=t$ and then using $1+x \leq \exp(x)$
$$\int_{[0, 2\pi) \times (0,\infty)}\frac{\exp(\frac{-t}{2})}{2t}|\sin(2\phi)| \, d\phi \otimes \, dt \geq \int_{[0, 2\pi) \times (0,\infty)}\frac{1-\frac{t}{2}}{2t}|\sin(2\phi)| \, d\phi \otimes \, dt=$$
$$\int_{[0, 2\pi) \times (0,\infty)}\frac{2-t}{4t} \, |\sin(2\phi)| \, d\phi \otimes \, dt$$
I feel like this function is simpler than at the start, however I fail to see why this integral does not converge. Any hints are welcome.
Best Answer
Note $$\int_{[0, 2\pi) \times (0,\infty)}|\frac{\sin(2\phi)}{2r^2}|\exp\Big(\frac{-r^2}{2}\Big) \, d\phi \otimes \, dr=\int_{[0, 2\pi)}|\sin(2\phi)| \, d\phi \int_{(0,\infty)}\frac{\exp(\frac{-r^2}{2})}{2r^2}\, dr. \tag1$$ But $$\int_{[0, 2\pi)}|\sin(2\phi)| \, d\phi=const. $$ and $$\int_{(0,\infty)}\frac{\exp(\frac{-r^2}{2})}{2r^2}\, dr=\int_{(0,1)}\frac{\exp(\frac{-r^2}{2})}{2r^2}\, dr+\int_{(1,\infty)}\frac{\exp(\frac{-r^2}{2})}{2r^2}\, dr.$$ Note the second integral converges and the second one diverges and so (1) diverges. This is because $$ \frac{\exp(\frac{-r^2}{2})}{2r^2} \ge e^{-\frac12} \frac{1}{2r^2}$$ and $\int_0^1\frac{1}{2r^2}dr$ diverges and hence $$ \int_{(0,1)}\frac{\exp(\frac{-r^2}{2})}{2r^2}\,dr $$ diverges.