User:Simfish/Math2

${\displaystyle {\frac {sin(t)}{t}}=\Sigma _{n=0}^{\infty }{\frac {(-1)^{n}t^{2n}}{(2n+1)!}}.}$

By theorem V we have

${\displaystyle \int _{0}^{x}{\frac {sin(t)}{t}}dt=\int _{0}^{x}\Sigma _{n=0}^{\infty }{\frac {(-1)^{n}t^{2n}}{(2n+1)!}}=\Sigma _{n=0}^{\infty }\int _{0}^{x}{\frac {(-1)^{n}t^{2n}}{(2n+1)!}}dt=\Sigma _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)^{2}(2n)!}}SoR=infinity}$