User:Simfish/Math2

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

By theorem V we have

$$ \int_{0}^{x} \frac{sin(t)}{t} dt = \int_0^{x} \Sigma_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{(2n+1)!} = \Sigma_{n=0}^{\infty} \int_{0}^{x} \frac{(-1)^n t^{2n}}{(2n+1)!} dt = \Sigma_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)^2 (2n)!} So R = infinity$$