Defining the overlapping area of two log-normal distributions with different means, same variance, and different scaling factors that add up to 1 Define
$$
\begin{cases}
X_1\sim Lognormal(ln(\mu_1), \sigma^2) \\
X_2\sim Lognormal(ln(\mu_2), \sigma^2)
\end{cases}
$$
where $\mu_2>\mu_1>0$ and that there is a definite proportion, $\eta\in(0,1)$,  between $X_1$ and $X_2$ such that
$$
\begin{cases}
f_1(x)=\frac{\eta}{x\sigma\sqrt{2\pi}}e^{-{\frac{(ln(x)-ln(\mu_1))^2 \,\,\,\,\,\,}{2\sigma^2}}} \\
f_2(x)=\frac{1-\eta}{x\sigma\sqrt{2\pi}}e^{-\frac{(ln(x)-ln(\mu_2))^2 \,\,\,\,\,\,}{2\sigma^2}}
\end{cases}
$$
where $f_1$ and $f_2$ represent the $\eta$-scaled PDF's of $X_1$ and $X_2$, respectively.
Based on the above definitions, note that $\int_{x=0}^\infty f_1(x)\,dx\,+\int_{x=0}^\infty f_2(x)\,dx=1$.

Given $\mu_1$, $\mu_2$, $\sigma$, and $\eta$, how is the overlapping area of the two probability distribution curves, $OVL=f(\mu_1,\mu_2,\sigma,\eta)$, defined?
Please see a illustrative plot below, where $OVL=f(\mu_1=5,\mu_2=10,\sigma=20\%,\eta=50\%)$ is highlighted in yellow:

I am able to perform numerical approximation for $OVL$ using the trapezoidal rule, but I need to express $OVL$ explicitly and I am not sure how to do so.
 A: With help from a friend we solved the problem ourselves:
Let $\{\tau\in\Bbb{R}^+|f_1(\tau,\mu_1,\sigma,\eta)=f_2(\tau,\mu_2,\sigma,\eta)\}$, then
$$
\begin{align}
\frac{\eta}{\tau\sigma\sqrt{2\pi}}e^{-\frac{(\ln\tau-\ln{\mu_1})^2}{2\sigma^2}}&=\frac{1-\eta}{\tau\sigma\sqrt{2\pi}}e^{-\frac{(\ln\tau-\ln{\mu_2})^2}{2\sigma^2}} \\
e^\frac{(\ln\tau-\ln{\mu_1})^2-(\ln\tau-\ln{\mu_2})^2}{2\sigma^2}&=\frac{\eta}{1-\eta}\quad (\because\tau>0) \\
\frac{2\ln\tau\ln{\tau_2}-2\ln\tau\ln{\tau_1}+(\ln{\mu_1})^2-(\ln{\mu_2})^2}{2\sigma^2}&=\ln\frac{\eta}{1-\eta} \\
\ln\tau&=\frac{\sigma^2[\ln\eta-\ln(1-\eta)]}{\ln{\mu_2}-\ln{\mu_1}}+\frac{\ln{\mu_2}+\ln{\mu_1}}{2} \\
\tau&=e^{\frac{\sigma^2[\ln\eta-\ln(1-\eta)]}{\ln{\mu_2}-\ln{\mu_1}}+\frac{\ln{\mu_2}+\ln{\mu_1}}{2}}
\end{align}
$$
Since $\mu_2>\mu_1$ and $\eta\in(0,1)$, $\exists!\tau\in\Bbb{R}^+|f_1=f_2$

$\forall x\in\Bbb{R}^+$,
$\ln{f_2(x,\mu_2,\sigma,\eta)}-\ln{f_1(x,\mu_1,\sigma,\eta)}$
$=\ln\frac{1-\eta}{x\sigma\sqrt{2\pi}}-\ln\frac{\eta}{x\sigma\sqrt{2\pi}}+\frac{(\ln{x}-\ln{\mu_1})^2}{2\sigma^2}-\frac{(\ln{x}-\ln{\mu_2})^2}{2\sigma^2}$
$=\ln\frac{1-\eta}{\eta}+\frac{(\ln{\mu_1})^2-(\ln{\mu_2})^2}{2\sigma^2}+\frac{\ln{\mu_2}-\ln{\mu_1}}{\sigma^2}\ln{x}$, which is strictly increasing on $x\quad(\because\mu_2>\mu_1)$
$\therefore\frac{f_2(x,\mu_2,\sigma,\eta)}{f_1(x,\mu_1,\sigma,\eta)}$ is strictly increasing on $x$
It follows that:
$\begin{cases}
f_2(x,\mu_2,\sigma,\eta)>f_1(x,\mu_1,\sigma,\eta)\quad\forall x>\tau \\
f_2(x,\mu_2,\sigma,\eta)<f_1(x,\mu_1,\sigma,\eta)\quad\forall x<\tau
\end{cases}\quad\quad\ldots\ldots(*)$

The $\eta$-scaled CDF's of $X_1$ and $X_2$, denoted as $F_1$ and $F_2$, are defined as follows:
$\begin{cases}
F_1(x,\mu_1,\sigma,\eta)=\frac{\eta}{2}[1+erf(\frac{\ln{x}-\ln{\mu_1}}{\sqrt{2}\sigma})] \\
F_2(x,\mu_2,\sigma,\eta)=\frac{1-\eta}{2}[1+erf(\frac{\ln{x}-\ln{\mu_2}}{\sqrt{2}\sigma})]
\end{cases}$, where $erf(z)=\frac{2}{\sqrt\pi}\int_0^z{e^{-t^2}dt}$
By (*),
$\begin{align}
OVL=f(\mu_1,\mu_2,\sigma,\eta)&=F_2(\tau,\mu_2,\sigma,\eta)+[1-F_1(\tau,\mu_1,\sigma,\eta)] \\
&=1+F_2(\tau,\mu_2,\sigma,\eta)-F_1(\tau,\mu_1,\sigma,\eta)
\end{align}$
