Answer
(a) using Fubini’s Theorem, we show that ${I^2} = J$
(b) evaluate $J$ in polar coordinates:
$J = \pi $
(c) we prove that $I = \sqrt \pi $
Work Step by Step
(a) We have $I = \mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {x^2}}}{\rm{d}}x$.
Write $J = \mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {x^2} - {y^2}}}{\rm{d}}x{\rm{d}}y$.
Since ${{\rm{e}}^{ - {x^2} - {y^2}}} = {{\rm{e}}^{ - {x^2}}}{{\rm{e}}^{ - {y^2}}}$, so
$J = \mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {x^2}}}{{\rm{e}}^{ - {y^2}}}{\rm{d}}x{\rm{d}}y$
Using Theorem 3 Fubini's Theorem for the double integral (Section 16.1), we get
$J = \mathop \smallint \limits_{ - \infty }^\infty \left( {\mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {x^2}}}{\rm{d}}x} \right){{\rm{e}}^{ - {y^2}}}{\rm{d}}y$
$ = \left( {\mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {x^2}}}{\rm{d}}x} \right)\left( {\mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {y^2}}}{\rm{d}}y} \right)$
Since $I = \mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {x^2}}}{\rm{d}}x = \mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {y^2}}}{\rm{d}}y$, so $J = {I^2}$.
Hence, ${I^2} = J$.
(b) We evaluate $J = \mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {x^2} - {y^2}}}{\rm{d}}x{\rm{d}}y$ in polar coordinates.
Write $f\left( {x,y} \right) = {{\rm{e}}^{ - {x^2} - {y^2}}}$.
In polar coordinates, $f\left( {r\cos \theta ,r\sin \theta } \right) = {{\rm{e}}^{ - {r^2}}}$.
The infinite region in rectangular coordinates: $\left( {x,y} \right) = \left( { - \infty ,\infty } \right) \times \left( { - \infty ,\infty } \right)$ is equivalent to $\left( {r,\theta } \right) = [0,\infty ) \times \left[ {0,2\pi } \right]$ in polar coordinates.
Using Eq. (4) of Theorem 1, we have
$J = \mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {x^2} - {y^2}}}{\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^\infty r{{\rm{e}}^{ - {r^2}}}{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\mathop {\lim }\limits_{R \to \infty } \mathop \smallint \limits_{r = 0}^R r{{\rm{e}}^{ - {r^2}}}{\rm{d}}r} \right){\rm{d}}\theta $
$ = - \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\mathop {\lim }\limits_{R \to \infty } \left( {{{\rm{e}}^{ - {r^2}}}|_0^R} \right)} \right){\rm{d}}\theta $
$ = - \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\mathop {\lim }\limits_{R \to \infty } \left( {{{\rm{e}}^{ - {R^2}}} - 1} \right)} \right){\rm{d}}\theta $
Since $\mathop {\lim }\limits_{R \to \infty } {{\rm{e}}^{ - {R^2}}} = 0$, so
$J = \mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {x^2} - {y^2}}}{\rm{d}}x{\rm{d}}y = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = \pi $
(c) From part (b) we obtain $J = \pi $.
Since ${I^2} = J$, so $I = \sqrt \pi $.
