Proof for Simplifying Integral involving Gaussian and Error Function How do we simplify this integral?
\begin{eqnarray*}
\int_{-\infty}^{\infty}\left\{ \frac{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right)}{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\mu_{X}}{\sigma_{X}}\right)}\right\} yf\left(y\right)dy
\end{eqnarray*}
Please note $k<0$ here.
\begin{eqnarray*}
Y\sim N\left(\mu_{Y},\sigma_{Y}^{2}\right); k<0
\end{eqnarray*}
Here, $f\left(y\right)$ is the probability density function for $y$, and $\mathbf{\Phi}$ is the standard normal CDF.
STEPS TRIED
Based on other suggestions, please see related link below. It seems one of the two assertions below are valid. But I am not sure if (and which) of these are correct or how we can prove it? Could someone please clarify and provide steps?
I think the second assertion below holds when $\lim_{y\to0^+}$ though I am not sure and hence would appreciate clarifications as well. 
How about other cases? (Can this integral be simplied in some region?) 
1)
\begin{eqnarray*}
\left[\int_{-\infty}^{0}\left\{ \frac{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right)}{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\mu_{X}}{\sigma_{X}}\right)}\right\} yf\left(y\right)dy=\int_{-\infty}^{0}yf\left(y\right)dy\right]
\end{eqnarray*}
2)
\begin{eqnarray*}
\left[\int_{0}^{\infty}\left\{ \frac{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right)}{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\mu_{X}}{\sigma_{X}}\right)}\right\} yf\left(y\right)dy=\int_{0}^{\infty}yf\left(y\right)dy\right]
\end{eqnarray*}
This comes up during the proof for this question. Conditional Expected Value of Product of Normal and Log-Normal Distribution
 A: The equalities never hold.
The following analysis illustrates the benefits of simplifying complicated expressions before attempting to analyze them.

Because $\Phi$ is the CDF of a continuous variable supported everywhere, it is strictly increasing and nonzero.  One way to write this fact is
$$\Phi(z + \lambda) / \Phi(z) \gt 1$$
for any all $z$ and any $\lambda \gt 0$.  Applying this to
$$z = \frac{-\log(-k/y) + \mu_x}{\sigma_x}$$
(assuming $k \gt 0$ in the first integral, where $y\lt 0$, and $k \lt 0$ in the second integral where $y\gt 0$, for otherwise the logarithm is not defined) and
$$\lambda = \sigma_x \gt 0,$$
and observing that the remaining factor $y f(y)$ has a constant sign in either integral, allows us to replace everything in braces $\{\cdots \}$--which is precisely $\Phi(z+\lambda)/\Phi(z)$--by the strictly smaller value $1$, achieving a strict inequality.
Since in (1) $yf(y) \lt 0$ and in (2) $yf(y) \gt 0$, the integrand on the left hand side of (1) is strictly greater than the integrand on the right hand side and the integrand of the lhs of (2) is strictly less than the integrand of its rhs.  Integration over any set of positive measure (which is the case here) will preserve such a strict inequality.
Therefore, there are no possible values of $k$, $\mu_x$, or $\sigma_x$ that make either (1) or (2) true.
