In this derivation of Black's formula for puts, we have that $\mathbb{E}[e^X 1_{e^X \leq K/S_0}]$ somehow equals $S_0 e^{\mu + 0.5 \sigma^2} N$ (as above in the formula).
I tried breaking apart the formula into
$$\mathbb{E}[e^X 1_{e^X \leq K/S_0}]) = \mathbb{E}(e^X)\mathbb{E}(1_{e^X \leq K/S_0}) + \operatorname{Cov}(e^X,1_{e^X \leq K/S_0})$$
and ended up with
$$S_0 e^{\mu + 0.5\sigma^2} N((\ln(K/S_0) - \mu)/\sigma) + \operatorname{Cov}(e^X, 1_{e^X \leq K/S_0})$$
by applying the MGF and standardizing the inside of the Normal
and now I'm stuck on what to do next to simplify it into the form given in step 3.
My question is how to calculate away the Covariance term to make it match last line of the circled formula, aka how do you get from step 2 to step 3 of the circled.
Consider the new random variable $Y:=\exp(X)$. Then you are asked to calculate $\mathbb{E}(Y1_{Y\leq c})$ for some constant c. This boils down to evaluating the integral $\int_0^c y f_Y(y)\mathrm{d}y$