Answer
(a) We show that
$\mathop \smallint \limits_0^1 g\left( t \right){\rm{d}}t = \dfrac{\pi }{4} - \dfrac{1}{2}\ln 2$
(b) We show that
$g\left( t \right) = 1 - t - {t^2} + {t^3} + {t^4} - {t^5} - {t^6} + \cdot\cdot\cdot$
(c) $S = 1 - \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6} - \dfrac{1}{7} + \cdot\cdot\cdot = \dfrac{\pi }{4} - \dfrac{1}{2}\ln 2$
Work Step by Step
(a) We have $g\left( t \right) = \dfrac{1}{{1 + {t^2}}} - \dfrac{t}{{1 + {t^2}}}$.
Take the integral:
(1) ${\ \ \ \ \ }$ $\mathop \smallint \limits_0^1 g\left( t \right){\rm{d}}t = \mathop \smallint \limits_0^1 \dfrac{1}{{1 + {t^2}}}{\rm{d}}t - \mathop \smallint \limits_0^1 \dfrac{t}{{1 + {t^2}}}{\rm{d}}t$
1. Consider the integral: $\mathop \smallint \limits_0^1 \dfrac{1}{{1 + {t^2}}}{\rm{d}}t$
From the Table of Integrals at the end of the book, we know that
$\smallint \dfrac{1}{{{a^2} + {u^2}}}{\rm{d}}u = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{u}{a} + C$
So,
$\mathop \smallint \limits_0^1 \dfrac{1}{{1 + {t^2}}}{\rm{d}}t = {\tan ^{ - 1}}t|_0^1 = \dfrac{\pi }{4}$
2. Consider the integral: $\mathop \smallint \limits_0^1 \dfrac{t}{{1 + {t^2}}}{\rm{d}}t$
Let $u = {t^2}$. So, $du = 2tdt$. The integral becomes
$\mathop \smallint \limits_0^1 \dfrac{t}{{1 + {t^2}}}{\rm{d}}t = \dfrac{1}{2}\mathop \smallint \limits_0^1 \dfrac{1}{{1 + u}}{\rm{d}}u = \dfrac{1}{2}\ln \left( {1 + u} \right)|_0^1 = \dfrac{1}{2}\ln 2$
Substituting $\mathop \smallint \limits_0^1 \dfrac{1}{{1 + {t^2}}}{\rm{d}}t = \dfrac{\pi }{4}$ and $\mathop \smallint \limits_0^1 \dfrac{t}{{1 + {t^2}}}{\rm{d}}t = \dfrac{1}{2}\ln 2$ in equation (1) gives
$\mathop \smallint \limits_0^1 g\left( t \right){\rm{d}}t = \dfrac{\pi }{4} - \dfrac{1}{2}\ln 2$
(b) From Table 2, we have
$\dfrac{1}{{1 + t}} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{t^n} = 1 - t + {t^2} - {t^3} + {t^4} - \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| t \right| \lt 1$
Substituting ${t^2}$ for $t$ in the series above gives
$\dfrac{1}{{1 + {t^2}}} = \mathop \sum \limits_{n = 0}^\infty {\left( { - 1} \right)^n}{t^{2n}} = 1 - {t^2} + {t^4} - {t^6} + {t^8} - \cdot\cdot\cdot$, ${\ \ \ }$ converges for $\left| {{t^2}} \right| \lt 1$ or $\left| t \right| \lt 1$
Thus,
$g\left( t \right) = \dfrac{1}{{1 + {t^2}}} - \dfrac{t}{{1 + {t^2}}}$
$g\left( t \right) = \left( {1 - {t^2} + {t^4} - {t^6} + {t^8} - \cdot\cdot\cdot} \right) - t\left( {1 - {t^2} + {t^4} - {t^6} + {t^8} - \cdot\cdot\cdot} \right)$
$g\left( t \right) = \left( {1 - {t^2} + {t^4} - {t^6} + {t^8} - \cdot\cdot\cdot} \right) - \left( {t - {t^3} + {t^5} - {t^7} + {t^9} - \cdot\cdot\cdot} \right)$
Hence, $g\left( t \right) = 1 - t - {t^2} + {t^3} + {t^4} - {t^5} - {t^6} + \cdot\cdot\cdot$.
(c) From part (b), we obtain
$g\left( t \right) = 1 - t - {t^2} + {t^3} + {t^4} - {t^5} - {t^6} + \cdot\cdot\cdot$
Taking the integral on both sides gives
$\mathop \smallint \limits_0^1 g\left( t \right){\rm{d}}t = \mathop \smallint \limits_0^1 1{\rm{d}}t - \mathop \smallint \limits_0^1 t{\rm{d}}t - \mathop \smallint \limits_0^1 {t^2}{\rm{d}}t + \mathop \smallint \limits_0^1 {t^3}{\rm{d}}t + \mathop \smallint \limits_0^1 {t^4}{\rm{d}}t - \mathop \smallint \limits_0^1 {t^5}{\rm{d}}t - \mathop \smallint \limits_0^1 {t^6}{\rm{d}}t + \cdot\cdot\cdot$
From part (a), we know that
$\mathop \smallint \limits_0^1 g\left( t \right){\rm{d}}t = \dfrac{\pi }{4} - \dfrac{1}{2}\ln 2$
So,
$\dfrac{\pi }{4} - \dfrac{1}{2}\ln 2 = \left( t \right)|_0^1 - \dfrac{1}{2}\left( {{t^2}} \right)|_0^1 - \dfrac{1}{3}\left( {{t^3}} \right)|_0^1 + \dfrac{1}{4}\left( {{t^4}} \right)|_0^1 + \dfrac{1}{5}\left( {{t^5}} \right)|_0^1 - \dfrac{1}{6}\left( {{t^6}} \right)|_0^1 - \dfrac{1}{7}\mathop \smallint \limits_0^1 {t^7}{\rm{d}}t + \cdot\cdot\cdot$
$\dfrac{\pi }{4} - \dfrac{1}{2}\ln 2 = 1 - \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6} - \dfrac{1}{7} + \cdot\cdot\cdot$
Write $S = 1 - \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6} - \dfrac{1}{7} + \cdot\cdot\cdot$, so
$S = 1 - \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6} - \dfrac{1}{7} + \cdot\cdot\cdot = \dfrac{\pi }{4} - \dfrac{1}{2}\ln 2$
