Integration by parts: the variance of a standard normal I am calculating the variance of a standard normal, but I stuck with the following part (the answer is different from what I know). What is wrong with my calculation?
$$
\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} y^2 e^{- y^2 / 2} = \left[ y^2 \cdot \left( - \frac{1}{y} \right) \cdot e^{- y^2 / 2} \right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} 2y \cdot \left( - \frac{1}{y} \right) \cdot e^{- y^2 / 2} dy = 2?({\rm Should \ be\ 1})
$$
I used integration by parts:
$$
\int_{a}^{b} f(x) g'(x) dx = \left[f(x)g(x) \right]_{a}^{b} - \int_{a}^{b} f'(x) g(x) dx
$$
I thought $f(y) = y^2$, so $f'(y) = 2y$, and $g'(y) = e^{- y^2 / 2}$, so $g(y)= (- 1/y) \cdot e^{- y^2 / 2}$
I also used the result of Gaussian integral:
$$ \int_{-\infty}^{\infty} e^{-a x^2} dx = \sqrt{\frac{\pi}{a}}$$

\begin{align}
   \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} y^2 e^{- y^2 / 2}dy
  &=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} y^2 e^{- y^2 / 2} dy\\
 &\text{put $\frac{y^2}{2}=t$,we have $ydy=dt$}\\
&=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- t} \sqrt{2t}dt\\
&=\frac{2\sqrt 2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- t} t^{\frac12}dt\\
&=\frac{2\sqrt 2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- t} t^{\frac32-1}dt\\
&=\frac{2\sqrt 2}{\sqrt {2\pi}}\Gamma(\frac32)\\
&=\frac{2\sqrt 2}{\sqrt {2\pi}}\times\frac{\sqrt \pi}{2}\\
&=1
\end{align}
