What happened to $\frac{1}{\sqrt{2\pi}}$? I'm doing the following exercise: 

I understand it until they do the substitution. Since $\varphi(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}$, where did the constant go in the exercise when they did the substitution? Also, how did they do the "jump" to the last equation?
I would be very grateful if someone could help me understand it.
 A: They dropped the constant when they did the substitution because if you can integrate $ \int \; z e^{-\frac{z^2}{2}} \; \; dz $ then you can integrate any constant multiple of it which is what $ \int \; z \phi(z) \; \; dz$ would be.
For the first part of the last inequality, they are using the fact that if $A$ is any event and I is the indicator random variable for $A$, then $P(A) = E( I(A) )$.
In the last inequality, they are using the above fact for the event $Z > t$:
$P(Z > t) = E(I(Z>t))$
The second part of the last inequality come from the inequality they establish earlier:
$E(I(Z>t)) \leq \frac{1}{t} \int_t^\infty z \phi(z) \; dz$
Then you would just need to compute the integral $\frac{1}{t} \int_t^\infty z \phi(z) \; dz$.
$$\frac{1}{t} \int_t^\infty z \phi(z) \; dz$$
$$\frac{1}{t} \lim_{b \to \infty} \int_t^b z \phi(z) \; dz$$
$$\frac{1}{t} \lim_{b \to \infty} \int_t^b z \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} \; dz$$
$$\frac{1}{t} \lim_{b \to \infty} \frac{1}{\sqrt{2\pi}} \int_t^b z e^{-\frac{z^2}{2}} \; dz$$
$$\frac{1}{t} \lim_{b \to \infty} \frac{1}{\sqrt{2\pi}} \left [ -e^{-\frac{z^2}{2}} \right ]_t^b$$
$$\frac{1}{t} \lim_{b \to \infty} \frac{1}{\sqrt{2\pi}} \left [ -e^{-\frac{b^2}{2}} - \left ( -e^{-\frac{t^2}{2}} \right ) \right ]$$
$$\frac{1}{t} \lim_{b \to \infty} \frac{1}{\sqrt{2\pi}} \left [ -e^{-\frac{b^2}{2}} + e^{-\frac{t^2}{2}} \right ]$$
$$\frac{1}{t} \frac{1}{\sqrt{2\pi}} e^{-\frac{t^2}{2}} $$
$$\frac{1}{t} \phi(t)$$
