Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based at 0. a.Give your answer using summation notation. b.Give the interval on which the series converges.

In this question () I derived the Taylor series for [tex]\mathrm{sinc}\,x[/tex] about [tex]x=0[/tex]:[tex]\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}[/tex]Then the Taylor series for[tex]f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt[/tex]is obtained by integrating the series above:[tex]f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}[/tex]We have [tex]f(0)=0[/tex], so [tex]C=0[/tex] and so[tex]f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}[/tex]which converges by the ratio test if the following limit is less than 1:[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}[/tex]Like in the linked problem, the limit is 0 so the series for [tex]f(x)[/tex] converges everywhere.