1.30. Investigate the absolute convergence of improper multiple integrals:

a) $\iint_{\Omega} \frac{\varphi(x, y)}{\left(x^{2}+y^{2}\right)^{p}} d x d y$, where $\Omega: x^{2}+y^{2} \leqslant 1$ and $\Omega: x^{2}+y^{2} \geqslant 1$; $\varphi(x, y)$ - a bounded function on $\Omega$;

b) $\iiint_{\Omega} \frac{\varphi(x, y, z)}{\left(x^{2}+y^{2}+z^{2}\right)^{p}} d x d y d z$, where $\Omega: x^{2}+y^{2}+z^{2} \leqslant 1$ and $\Omega:$ $x^{2}+y^{2}+z^{2} \geqslant 1 ; \varphi(x, y, z)$ - a bounded function on $\Omega$.
Solution.

a) Due to the boundedness of $\varphi(x, y)$ on $\Omega (|\varphi(x, y)| \leqslant C$, $(x, y) \in \Omega$), the absolute convergence of the studied integral is determined by the convergence of the integral

$$
\iint_{\Omega} \frac{d x d y}{\left(x^{2}+y^{2}\right)^{p}}
$$

To investigate this convergence, we introduce polar coordinates $x=\rho \cos \varphi ; y=\rho \sin \varphi$ and obtain

$$
\iint_{\Omega} \frac{d x d y}{\left(x^{2}+y^{2}\right)^{p}}=\iint_{D} \frac{\rho d \rho d \varphi}{\rho^{2 p}}=\left\{\begin{array}{l}
2 \pi \int_{1}^{x} \rho^{(1-2 p)} d \rho \\
2 \pi \int_{0}^{1} \rho^{(1-2 p)} d \rho
\end{array}\right.
$$

If $\Omega: x^{2}+y^{2} \leqslant 1$, then the original integral converges when $2 p - 1 < 1$, i.e., $p > 1$.

If $p = 1$, in both cases there is no absolute convergence, and therefore, by Theorem 1.6, the improper integral diverges.

b) Similarly to part a), the absolute convergence of the studied integral depends on the convergence of the integral

$$
\iiint_{\Omega} \frac{d x d y d z}{\left(x^{2}+y^{2}+z^{2}\right)^{p}}
$$

To investigate this convergence, we introduce spherical coordinates $x=\rho \sin \theta \cos \varphi ; y=\rho \sin \theta \sin \varphi ; z=\rho \cos \theta$, and obtain

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-079.jpg?height=245&width=823&top_left_y=1332&top_left_x=240)

If $\Omega: x^{2}+y^{2}+z^{2} \leqslant 1$, we arrive at the improper integral $\int_{0}^{1} \rho^{2(1-p)} d \rho$, which converges when $2 p - 2 < 1$, i.e., $p > \frac{3}{2}$.

If $p = \frac{3}{2}$ in both cases, there is no absolute convergence, and therefore, according to Theorem 1.6, the improper integral diverges.

## 1.7 .  PROBLEMS FOR CHAPTER 1

## Compute the repeated integrals.