Does the improper integral $\int_{R^2} \frac{\ln(x^2+y^2)}{x^2+y^2}$ converge?
So I know to solve such integrals when the function is all positive, but here it can be also negative.
I tried using the definition $\int f =\int f^+ - \int f^-$ and calculate for each part.
When I try for $f^-$, that means for { $x^2+y^2 <1 $ } I get to the integral (after using polar coordinates) $$\int_{0}^{1} \frac{\ln(r^2)}{r}dr$$ which goes to $- \infty$.
While the same integral for $f^+$ goes to $\infty$.
Does that mean the the original integral goes to $\infty$ ? From what I remember either the integral for $f^-$ or $f^+$ has to converge for the integral to be defined.
Another direction which I thought would be calculating the integral for $$\int_{R^2} \frac{|\ln(x^2+y^2)|}{x^2+y^2}$$ and since it is less than $$\int_{x^2+y^2 > 1} \frac{|\ln(x^2+y^2)|}{x^2+y^2}$$ then it must not converge. Am I right here?
Any help would be appreciated.
You are almost right. Let us do the calculation in a bit detailed way
$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{\ln(x^2+y^2)}{x^2 + y^2}dxdy = 2\pi\int_{0}^{\infty}\frac{\ln r^2}{r}dr = 4\pi\int_{0}^{\infty}\frac{\ln r}{r}dr$
Now, using variable transformation of $\ln r = u$, we obtain $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{\ln(x^2+y^2)}{x^2 + y^2}dxdy = 4\pi \int_{-\infty}^{\infty}u du$$
If you are strictly speaking in the context of elementary calculus, then yes the integral does not converge. However, the Cauchy Principle Value of the integral exists and it tends to zero. You might want to look for Cauchy Principle Value.