Show that: $$\int_{0}^\infty 8t^2\sigma^2\phi(t)\Phi(t)\ dt\ -\int_{0}^\infty 4t^2\sigma^2\phi(t)\ dt\ =\left(1+\frac{2}{\pi}\right)\sigma^2 $$ where $\phi(.)$ and $\Phi(.)$ i.e. normal pdf and cdf.
I've tried as the answer has been given on here
this what I've tried
$$8\sigma^2\left(\int_{0}^\infty t^2\phi(t)\left(\Phi(t)-\tfrac12\right)\ dt\right)$$ Suppose: $$I=\int_{0}^\infty t^2\phi(t)\left(\Phi(t)-\tfrac12\right)\ dt$$ By definition of $\phi$ as the PDF of a standard normal random variable $X$, $$I=E\left(X^2\left(\Phi(X)-\tfrac12\right):X>0\right)$$
For every $x>0$, for some standard normal random variable $Y$ independent of $X$, $$\Phi(x)-\tfrac12=P(x>Y>0)$$ hence $$I=E\left(X^2:X>Y>0\right)$$
for every $y$, $$E(X^2:X>y)=\int_y^\infty x^2\phi(x)dx=\int_y^\infty -x\phi'(x)dx=\int_\infty^y x\phi'(x)dx=x\phi(x)\rvert_\infty^{y}-\int_\infty^y\phi(x)dx = y\phi(y)-\Phi(y)$$ which yields $$I=E(y\phi(y)-\Phi(y):Y>0)=\int_0^\infty \left(y\phi(y)-\Phi(y)\right)\phi(y)dy =\int_0^\infty y\phi^2(y)dy-\int_0^\infty \Phi(y)\phi(y)dy $$
I'm stuck here. Please help me.
Thank you