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Let $p$ and $t$ be positive real numbers. Let $D_t$ be the closed disc of radius $t$ centered at $(0, 0)$, i.e., $D_t = {(x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq t^2}$. Define

$I(p, t) = \iint_{D_t} \frac{1}{(p^2 + x^2 + y^2)^p} , dx , dy$.

Then $\lim_{t \to \infty} I(p, t)$ is finite

(A) only if $p > 1$.

(B) only if $p = 1$.

(C) only if $p < 1$.

(D) for no value of $p$.
in MA2021 | 101 views

(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.

by Professor