What's wrong in this expansion of this erfc-like function?

Question

Given the function $$H(x)=\int_{-x}^{+\infty}dy \frac{e^{-y^2/2}}{\sqrt{2\pi}}$$, I would like to write it as a series expansion employing integration by parts: $$\int_{-x}^{+\infty}dy \frac{e^{-y^2/2}}{\sqrt{2\pi}}=y\frac{e^{-y^2/2}}{\sqrt{2\pi}}\vert_{-x}^{+\infty}+\int_{-x}^{+\infty}dy y^2 \frac{e^{-y^2/2}}{\sqrt{2\pi}}=\frac{e^{-x^2/2}}{\sqrt{2\pi}}\left(x+\frac{x^3}{3}+...+\frac{x^{2k+1}}{(2k+1)!!}+...\right)$$. The problem is evident: in $x=0$ the expansion doesn't seem to be equal to $1/2$ as it should. What am I doing wrong?

Answer 1

firstly you did integration by parts correctly (which was useless here) but you cant use series because of infinity.

So you need just to use partition of an interval $$ H(x)=\int_{-x}^{\infty} \frac{e{-\frac{y^2}{2}}}{\sqrt{2\pi}}dy=\int_{0}^{\infty} \frac{e^{-\frac{y^2}{2}}}{\sqrt{2\pi}}dy+\int_{-x}^{0} \frac{e^{-\frac{y^2}{2}}}{\sqrt{2\pi}}dy$$ then for the first integral its Gaussian integral which give us $$ \int_{0}^{\infty} \frac{e^{-\frac{y^2}{2}}}{\sqrt{2\pi}}dy=\frac{1}{2}$$ also for the second $$\int_{-x}^{0} \frac{e^{-\frac{y^2}{2}}}{\sqrt{2\pi}}dy=\int_{-x}^{0} \frac{1}{\sqrt{2\pi}}\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{-y^2}{2}\right)^ndy=\frac{1}{\sqrt{2\pi}} \sum_{n=0}^\infty \frac{-(-x)^{2n+1}}{n!(2n+1)}\left(\frac{-1}{2}\right)^n$$ using the series form for the error function and get $$ H(x)=\frac{1}{2}+\frac{1}{2}\frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty \frac{(-1)^n}{n!(2n+1)}\left(\frac{x}{\sqrt{2}}\right)^{2n+1}$$ finally $$ H(x)=\frac{1}{2} \left( \text{erf} \left(\frac{x}{\sqrt{2}}\right)+1 \right) $$