minimization of integral of multi-variable function Let $\alpha \in (0,1) $ be some parameter which we can choose and $Y$ be some random variable (e.g. standard normal distribution), then we can define a function as below (where $v,e < 0$ are two variables):
$S(Y,v,e; \alpha) =  -\frac{1}{\alpha e}\cdot(v-Y) \cdot \mathbf{1}_{(Y \le v)} + \frac{v}{e} - log(-e) - 1 $
If we minimized the expected value with respect to $v$ and $e$, we should get
$$ v=\Phi^{-1}_Y(\alpha)$$
and
$$
e = \frac{1}{\alpha}*\phi_Y( \Phi^{-1}_Y(\alpha) )
$$
 A: Let's first get an analytical expression for $E_Y[S]$ with $Y \sim \mathcal{N}(0,1)$:
\begin{align}
E_Y[S] &= -\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} \frac{1}{\alpha e} (v - y) \mathbf{1}_{\{ y \leq v\}} \exp(-\frac{1}{2} y^2)dy + \frac{v}{e} - \log(-e) -1 \\ 
&= -\frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{v} \frac{1}{\alpha e} (v - y)  \exp(-\frac{1}{2} y^2)dy  + \frac{v}{e} - \log(-e) -1 \\  
& = -\frac{v}{\sqrt{2 \pi}} \int_{- \infty}^{v} \frac{1}{\alpha e}\exp(-\frac{1}{2} y^2)dy +\frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{v} \frac{y}{\alpha e}\exp(-\frac{1}{2} y^2)dy + \frac{v}{e} - \log(-e) -1 \\ 
&= \frac{-v}{ \alpha e} \Phi_Y(v) - \frac{1}{\alpha e \sqrt{2\pi}}\exp(-v^2/2)+ \frac{v}{e} - \log(-e) -1 
\end{align}
Now we are set to get your first equation:
\begin{align}
\frac{\delta}{\delta v} E[ S(Y,v,e; \alpha) ] &= -\frac{1}{\alpha e}  \left( \Phi_Y(v) + \frac{v}{\sqrt{2\pi}}\exp(-v^2 / 2) \right) + \frac{v}{\alpha e \sqrt{2 \pi}} \exp(-v^2 / 2)+\frac{1}{e}\\
&= -\frac{1}{\alpha e} \Phi_Y(v) + \frac{1}{e}
\end{align}
By computing $\frac{\delta}{\delta v} E[ S(Y,v,e; \alpha) ] = 0$ we get the result.
For the second equation:
\begin{align}
\frac{\delta}{\delta e} E[ S(Y,v,e; \alpha) ] &=  \frac{v}{\alpha e^2} \Phi_Y(v) + \frac{1}{\alpha e^2 \sqrt{2\pi}}\exp(-v^2/2) - \frac{v}{e^2} - \frac{1}{e} 
\end{align}
We write $\frac{\delta}{\delta e} E[ S(Y,v,e; \alpha) ] =0$ to get:
\begin{equation}
e = \frac{v}{\alpha} \Phi_Y(v) + \frac{1}{\alpha}\phi_Y(v) - v
\end{equation}
To conclude, we use the optimun of $v = \Phi_Y^{-1}(\alpha)$ found in the first equation and the fact that $\Phi_Y(\Phi_Y^{-1}(\alpha)) = \alpha$ to get:
\begin{equation}
e = \frac{v}{\alpha} \alpha + \frac{1}{\alpha}\phi_Y(v) - v = \frac{1}{\alpha} \phi_Y(\Phi_Y^{-1}(\alpha))
\end{equation}
It's not exactly what you are asking for, but I think it's right; I cannot find any typo in my procedure. Maybe the $\frac{1}{1-\alpha}$ it's just $\frac{1}{\alpha}$ in your question.
