Maclaurin series of G(x)=$ \int_{0}^{x} \frac{1-e^{-y^2}}{y^2} \ dy $

Question

I have to make Maclaurin series of following function.

$$ G(x)= \int_{0}^{x} \frac{1-e^{-y^2}}{y^2} \ dy $$ Then i need to determine about domain of convergence of that series and and determine if it converges uniformly on that domain.

I did this:

$$ g(x)=\frac{d}{dx}(G(x))= \frac{1-e^{-x^2}}{x^2} = x^{-2}-x^{-2}\sum _{n=0}^{\infty }\frac{(-1)^nx^{2n}}{n!} $$

then,

$$ \int g(x) \ dx = -2x^{-1}-x^{-1}\sum _{n=0}^{\infty }\frac{(-1)^nx^{2n}}{(2n-1)n!} + const$$

But now i am not sure how to deal with case when x=0. I also tried find out domain of convergence by solving of this limit:

$$ L = \lim_{n\to \infty}\frac{2n-1}{2n+1}\frac{n!}{(n+1)!} = 0 $$

hence,

O = R

I do not know if my procedure is correct and like i said how to deal with case when x= 0

Can anybody help me please.

Answer 1

The series is \begin{align*} x-\dfrac{1}{3\cdot 2!}x^{3}+\dfrac{1}{5\cdot 3!}x^{5}-\cdots, \end{align*} let $S_{n}(x)=\displaystyle\sum_{k=1}^{n}(-1)^{k+1}\dfrac{x^{2k+1}}{k!(2k+1)}$, if it were $S_{n}(x)\rightarrow G(x)$ uniformly, then for some $n$ we have \begin{align*} |S_{n+1}(x)-S_{n}(x)|<1,~~~~x\in\mathbb{R}, \end{align*} then \begin{align*} \dfrac{x^{2n+3}}{(n+1)!(2n+3)}<1,~~~~x>0. \end{align*} Now taking $x\rightarrow\infty$ will make the above inequality blow up.