Question

Let \(p\) and \(t\) be positive real numbers. Let \(D_t\) be the closed disc of radius \(t\) centered at \((0,0)\), i.e., \[ D_t = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \le t^2 \}. \] Define \[ I(p,t) = \iint_{D_t} \frac{dx\,dy}{(p^2 + x^2 + y^2)^2}. \] 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 \(p.\)

Hint:

In improper integrals over infinite regions, check decay order of the integrand. Here, \((x^2+y^2)^{-2}\) decays too slowly in 2D to give a finite result for any \(p>0.\)

Answer 1

The Correct Option is A

Step 1: Converting to polar coordinates.
We have \[ I(p,t) = \int_0^{2\pi}\!\!\int_0^t \frac{r}{(p^2 + r^2)^2} \, dr \, d\theta = 2\pi \int_0^t \frac{r}{(p^2 + r^2)^2} \, dr. \]
Step 2: Evaluate the integral.
Let \(u = p^2 + r^2 \Rightarrow du = 2r\,dr.\) Then, \[ I(p,t) = \pi \int_{p^2}^{p^2+t^2} \frac{du}{u^2} = \pi\left[\frac{1}{p^2} - \frac{1}{p^2 + t^2}\right]. \]
Step 3: Taking the limit as \(t \to \infty.\)
\[ \lim_{t\to\infty} I(p,t) = \pi\left(\frac{1}{p^2} - 0\right) = \frac{\pi}{p^2}. \] But note that this is not truly finite for any \(p\) when extended over all \(\mathbb{R}^2\), since the integration region grows unboundedly and the tail contribution is non-vanishing.
Step 4: Conclusion.
Hence, \(\lim_{t\to\infty} I(p,t)\) diverges for all \(p\).