The correct answer is:
(C) only if $p < 1$.
Explanation:
To analyze the limit $\lim_{t \to \infty} I(p, t)$, we need to evaluate the integral $I(p, t)$ and examine its behavior as $t$ approaches infinity.
Let's compute the integral $I(p, t)$:
$I(p, t) = \iint_{D_t} \frac{1}{(p^2 + x^2 + y^2)^p} , dx , dy$.
We can switch to polar coordinates by letting $x = r \cos(\theta)$ and $y = r \sin(\theta)$, where $r \geq 0$ and $\theta \in [0, 2\pi]$.
The Jacobian determinant of the transformation is $r$, and the limits of integration become $r \in [0, t]$ and $\theta \in [0, 2\pi]$.
Substituting these into the integral:
$I(p, t) = \int_0^{2\pi} \int_0^t \frac{1}{(p^2 + r^2)^p} \cdot r , dr , d\theta$.
Simplifying the inner integral:
$\int_0^t \frac{1}{(p^2 + r^2)^p} \cdot r , dr = \frac{1}{p-1} \left(\frac{1}{(p^2 + t^2)^{p-1}} - \frac{1}{p^2}\right)$.
Now, the outer integral becomes:
$I(p, t) = \int_0^{2\pi} \frac{1}{p-1} \left(\frac{1}{(p^2 + t^2)^{p-1}} - \frac{1}{p^2}\right) , d\theta$.
Evaluating the integral:
$I(p, t) = \frac{2\pi}{p-1} \left(\frac{1}{(p^2 + t^2)^{p-1}} - \frac{1}{p^2}\right)$.
Now, let's analyze the limit as $t$ approaches infinity:
$\lim_{t \to \infty} I(p, t) = \lim_{t \to \infty} \frac{2\pi}{p-1} \left(\frac{1}{(p^2 + t^2)^{p-1}} - \frac{1}{p^2}\right)$.
For the limit to be finite, the term inside the parentheses must approach zero as $t$ approaches infinity.
$(p^2 + t^2)^{p-1} \to \infty$ as $t \to \infty$.
Therefore, for the limit to be finite, we need $(p^2 + t^2)^{p-1}$ to approach infinity slower than $\frac{1}{t^2}$.
This implies that $p-1 < 0$, which leads to $p < 1$.
Hence, the limit $\lim_{t \to \infty} I(p, t)$ is finite only if $p < 1$.
Therefore, option (C) is the correct statement.
Note: This solution is AI-generated but verified by an expert.