Determining whether $\iint_{|x|+|y| \leq 1} \ln(x^{2}+y^{2}) \,dx\,dy$ is positive or negative I have an integral:
$$\iint_{|x|+|y| \leq 1} \ln(x^{2}+y^{2}) \,dx\,dy$$
So basically it's:
$$\int_{-1}^{0}\,dx \int_{-x-1}^{x+1} \ln(x^{2}+y^{2})\,dy + \int_{0}^{1}\,dx \int_{x-1}^{-x+1} \ln(x^{2}+y^{2})\,dy$$
But it's two huge integrals, and it takes lots of time and calculations to get an answer. So, I wonder maybe there is another easy way to find out whether the answer is positive or negative. Maybe I don't see something.
 A: Observe that
$$
x^2 + y^2 \leq (|x| + |y|)^2 \leq 1
\quad\text{in your domain},
$$
so that integrand is $\leq 0$ in the integration domain.
A: We can compute the value of this integral exactly. By rotational symmetry we have that the integral is equivalent to the integral on the square (and subsequent triangle):
$$I = \iint_{\left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]^2} \log\left(x^2+y^2\right)\:dA = 8\int_0^{\frac{1}{\sqrt{2}}} \int_0^x \log\left(x^2+y^2\right)\:dy\:dx$$
In polar coordinates we get the integral
$$4 \int_0^{\frac{\pi}{4}}\int_0^{\frac{\sec\theta}{\sqrt{2}}}2r\log\left(r^2\right)\:dr\:d\theta = 2\int_0^{\frac{\pi}{4}}\sec^2\theta\left[\log\left(\sec^2\theta\right)-\log 2 - 1\right]\:d\theta$$
which by the substitution $x = \tan\theta$ gives
$$I = 2\int_0^1\log\left(1+x^2\right)\:dx - 2 (1+\log 2)$$
Just the integral piece is taken care of by an integration by parts
$$\int_0^1\log\left(1+x^2\right)\:dx = x\log\left(1+x^2\right)\Bigr|_0^1 - \int_0^1\frac{2x^2}{1+x^2}\:dx = \log 2 - 2 + \frac{\pi}{2}$$
which means our final answer is given by
$$I = \pi-6$$
which is clearly negative.
