Let $f(x_1,x_2)=\frac{1}{(x_1+ix_2)^2} \chi_{ \{ (-1,1) \times (-1,1) \}}$
Part A: Evaluate the $L^p(R^2)$ norm of $f(x_1,x_2)\chi_{ \{ (x_1,x_2): x_1^2+x_2^2 \leq 1 \}}$ for every $1 \leq p \leq \infty$ (this function is =f inside the disk of radius 1 and is =0 outside the disk)
$||f||_{L^p(R^2)}=(\int\int_{\{ (x_1,x_2): x_1^2+x_2^2 \leq 1 \}} |f(x_1,x_2)|^p dx_1dx_2)^{1/p} =(\int\int_{\{ (x_1,x_2): x_1^2+x_2^2 \leq 1 \}} (\sqrt{x_1^2+x_2^2})^{-2p} dx_1dx_2)^{1/p} =(\int_0^{2 \pi}\int_0^1 r^{-2p} rdr d \theta)^{1/p} =(\int_0^{2 \pi}d \theta \times \int_0^1 r^{-2p+1}dr)^{1/p} =(2 \pi \times r^{-2p+2}]_0^1)^{1/p} =(2 \pi \times \frac{1}{2-2p})^{1/p} =(\frac{\pi}{1-p})^{1/p}$
Please check that the work is correct.
Part B: Evaluate $L^2(R^2)$ norm of $f(x_1,x_2)$
$||f||_{L^2(R^2)}=(\int\int_{\{ (-1,1) \times (-1,1) \}} |f(x_1,x_2)|^2 dx_1dx_2)^{1/2} =(\int_{-1}^!\int_{-1}^1 (\sqrt{x_1^2+x_2^2})^{-4} dx_1dx_2)^{1/2} =(\int_{-1}^!\int_{-1}^1 \frac{dx_1dx_2}{(x_1^2+x_2^2)^2} dx_1dx_2)^{1/2}$
If I continue down this path I end up using multiple trig and u substitutions, how can I use Part A to solve this problem?
Since $p\ge 1$, we have $-2p+1\le -1$. Therefore, the integral $\int_0^1 r^{-2p+1}\,dr$ is ... not what you wrote.
Divergence of integral in Part B follows from the divergence of integral in Part A.
Frankly, these are strange questions. Re-check for typos.