Answer
a) $\pi$
b) Only finite numbers are allowed.
Work Step by Step
(a) $I=\int_1^\infty \pi (\dfrac{1}{x})^2 dx= \pi \lim\limits_{a \to \infty} \int_{1}^{a} x^{-2} dx$
or, $=\pi \lim\limits_{a \to \infty}[-x^{-1}]_{1}^{a} $
or, $= \pi \lim\limits_{a \to \infty}[-\dfrac{1}{x}]_{1}^{a}$
So, $I= \pi [0-(-1)] \\=\pi$
(b) From the previous part (a), we can conclude that we cannot take the limit to infinity since only finite numbers are allowed.
