How to find $\lim_{x\to +\infty}{\left(\sqrt{\pi}x-\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+2}}{2n!(2n+1)(2n+2)}\right)}$?

Question

$$\lim_{x\to+\infty}{\left(\sqrt{\pi}x-\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+2}}{2n!(2n+1)(2n+2)}\right)}$$

This limit comes from the following integral, which I expand it into Taylor series. $$S=2\int_0^{+\infty}\left(\frac{\sqrt{\pi}}{2}-\int_0^xe^{-t^2}dt\right)dx$$

My attempt is to solve the integral $S$, but I can't find a proper way, so I expand it into Taylor series and the question I ask is where I stuck.

Answer 1

Hint. You may just integrate your initial integral by parts, $$ \begin{align}S&=2\int_0^{+\infty}\left(\frac{\sqrt{\pi}}{2}-\int_0^xe^{-t^2}dt\right)dx\\ &=2\left[x\left(\frac{\sqrt{\pi}}{2}-\int_0^xe^{-t^2}dt\right)\right]_0^\infty+\underbrace{2\int_0^{+\infty}xe^{-x^2}dx}_{u=x^2\,\,du=2xdx}\\ &=0+\int_0^{+\infty}e^{-u}du\\ &=1. \end{align} $$ The desired limit is then equal to $1$.

Observe that we have used the classic gaussian integral result,

$$ \int_0^\infty e^{-t^2}dt=\frac{\sqrt{\pi}}{2} $$

to obtain, as $x \to +\infty$, $$ \left|x\left(\frac{\sqrt{\pi}}{2}-\int_0^xe^{-t^2}dt\right)\right|=\left|x\int_x^\infty e^{-t^2}dt\right|\leq \left|x\int_x^\infty e^{-xt}dt\right|=e^{-x^2}\to 0. $$