In Vershynin's book High Dimensional Probability, there is an easy exercise I am doing for self-study:
Let $g \sim N(0,1)$. Show that for all $t\geq 1$, we have
$$\mathbb E g^2 \mathbb 1_{\{g>t\}} = t \cdot \frac{1}{\sqrt{2\pi}}e^{-t^2/2} + \mathbb P \{g > t\}.$$
The left hand side is the integral
$$\frac{1}{\sqrt{2\pi}} \int_t^\infty x^2 e^{-x^2/2}dx$$
So, I integrate by parts:
$$ u= x^2 \qquad du = 2x dx$$ $$ v= \frac{e^{-x\frac{x^2}{2}}}{-x} \qquad dv =e^{-x^2/2}dx$$
\begin{align*} \frac{1}{\sqrt{2\pi}} \int_t^\infty x^2 e^{-x^2/2}dx &= \frac{1}{\sqrt{2\pi}} \int u dv \\ &=\frac{1}{\sqrt{2\pi}} \left( uv - \int v du \right) \\ &= \frac{1}{\sqrt{2\pi}} \left(te^{-\frac{t^2}{2}}+ 2\int_t^\infty e^{-x^2/2} dx\right) \\ &=\frac{1}{\sqrt{2\pi}} te^{\frac{-t^2}{2}}+2 \mathbb P\{ g>t\} \end{align*}
I can't figure out how to get rid of the 2 in that second term. Is the book mistaken? Or did I make some stupid error somewhere?
Better choice for u and v. $u=x,\ dv=xe^{-\frac{x^2}{2}}dx, \ v=e^{-\frac{x^2}{2}}$. Integral is the expression you are looking for.