Conditional Expectation of the minimum of two identical log-normal distributions I'd like to compute the closed form mean of the minimum of two truncated log-normal distribution (on another interval than its truncation). 
I have: 
$\int_{a}^{\infty} \int_{a}^{\infty} min(v, v') \ f(v_{| v > b})f(v'_{| v' > b}) dv dv'$
Where $v' $ and $v \sim LogNormal(0, \sigma)$, and $f(v | v>b)$ represents the truncated log-normal density on $[b, +\infty]$.
I already computed the density $f$ (and its mean): I have:
$f(v|v>b) =  \frac{\frac{1}{v} \frac{1}{\sigma} \phi(log(v)/\sigma)}{1-\Phi(log(b)/\sigma)}$
where $\phi$ and $\Phi$ are the standard normal density and distribution functions. 
The expectation of this process, on a truncated interval (from $c$ to $d$, where $c > b$) is given by:
$\mathbb{E}[v|b< c < v < d] = \int^d_c v \ f(v | v > b) = exp(\sigma^2/2) \frac{\Phi(\sigma - log(c)/\sigma) - \Phi(\sigma - log(d)/\sigma)}{1 - \Phi(log(b)/\sigma)}$  
What I would like to obtain now is the expectation of the min of two draws from $f(v|v>b)$.
I rewrote the integral:
$\int_{a}^{\infty} \int_{a}^{\infty} min(v, v') \ f(v_{| v > b})f(v'_{| v' > b}) dv dv' = \int_a^\infty v_{min}f_{min}(v_{min} | v_{min} > b)$ 
(I m not sure this is correct but numerical trials seem to indicate that it was correct). 
Since $Pr(v_{min} < x) = Pr(v < x) + Pr(v' < x) - Pr(v < x, v' < x) = 2 Pr(v < x) - Pr(v < x)^2$
(where the last equality comes from the fact that two identical truncated log-normal distributions are independent... I did not do the proof but I think it s correct)
I computed
$f_{min}(v | v > b) = 2 f(v | v > b) [1-F(v | v > b)]$
So now, I would like to compute the expectation of this, from a given $c> b$ to $\infty$, and I am stuck because I have:
$\begin{align*} \int^\infty_c v_{min} f_{min} (v_{min} |v_{min} > c) d v_{min}  &= \int^\infty_c v \ 2 f(v | v > b) [1-F(v | v > b)] dv \\
&= \underbrace{2 \int^\infty_c f(v | v > b) dv}_{\text{OK, expectation formula above}} - 2 \ \underbrace{\int^\infty_c v f(v | v>b) F(v|v>b) dv}_{\text{Problem here}} \end{align*}$
I need to compute the second term, which I know can be rewritten: 
$\begin{align*} 
\int^\infty_c v f(v | v>b) F(v|v>b) dv = \frac{1}{(1-\Phi(log(b)/\sigma))^2} \int_c^\infty v  \frac{1}{v} \frac{1}{\sigma} \phi(log(v)/\sigma) \Phi(log(v)/\sigma) dv - \underbrace{\frac{\Phi(log(b)/\sigma}{1-\Phi(log(b)/\sigma)} \int^\infty_c v f(v|v>b) dv}_{\text{OK, known using expectation formula}}
\end{align*}$

So basically (if all the rest is correct), I need to compute:
$\begin{align*}\int_c^\infty v  \frac{1}{v} \frac{1}{\sigma} \phi(log(v)/\sigma) \Phi(log(v)/\sigma) dv\end{align*}$
I'm not sure integrating by part work or anything... Does anyone has an idea?
Thanks in advance  

Remark:
I also know that $\mathbb{E}[min(X_1, X_2)] = 2 exp(\sigma^2/2) \Phi(-\sigma/\sqrt{2})$ where $X_1$ and $X_2$ are $LN(0, \sigma)$;
I need to extend this to the truncated log normal case.

My whole computation now comes down to this: 
I know that $\int^\infty_{-\infty} v f(v) F(v) dv = \Phi(\sigma/\sqrt{2}) exp(\sigma^2/2)$ 
I want to find : $\int^\infty_{c} v f(v) F(v) dv$. I think the result must not be too far from here, I just don't find it..
 A: Partial answer:
As said, before extending the work, lets work out the ordinary case first. Let $X_1 = e^{\sigma Z_1}, X_2 = e^{\sigma Z_2}$ where $Z_1, Z_2$ are i.i.d. standard normal. So
$$ \begin{align} &~ E[\min\{e^{\sigma Z_1}, e^{\sigma Z_2}\}] \\
=&~ E[e^{\sigma Z_1}\mathbf{1}_{Z_1 < Z_2}] + 
E[e^{\sigma Z_2}\mathbf{1}_{Z_2 < Z_1}]\\
=&~ 2E[e^{\sigma Z_1}\mathbf{1}_{Z_1 < Z_2}] \\
=&~ 2\int_{-\infty}^{\infty}E[e^{\sigma Z_1}\mathbf{1}_{Z_1 < Z_2}|Z_1 = z] \phi(z)dz\\
=&~ 2\int_{-\infty}^{\infty}e^{\sigma z}\Phi(-z)\frac {1} {\sqrt{2\pi}}\exp\left\{-\frac {z^2} {2}\right\}dz\\ 
=&~ 2\exp\left\{\frac {\sigma^2} {2}\right\}\int_{-\infty}^{\infty}\Phi(-z)\frac {1} {\sqrt{2\pi}}\exp\left\{-\frac {z^2-2\sigma z + \sigma^2} {2}\right\}dz\\ 
=&~ 2\exp\left\{\frac {\sigma^2} {2}\right\}\int_{-\infty}^{\infty}\Phi(-z)\phi(z-\sigma)dz\\
=&~ 2\exp\left\{\frac {\sigma^2} {2}\right\}\Pr\{Z_1 + \sigma < Z_2\} \\
=&~ 2\exp\left\{\frac {\sigma^2} {2}\right\}\Pr\left\{\frac {Z_1 - Z_2} {\sqrt{2}} < -\frac {\sigma} {\sqrt{2}}\right\} \\
=&~ 2\exp\left\{\frac {\sigma^2} {2}\right\}\Phi\left(-\frac {\sigma} {\sqrt{2}}\right)\\
\end{align}$$
Here the key trick should be exploiting the normal pdf by completing the square and thus we have changed the measure. To extend it to truncated case, note that
$$ e^{\sigma z} > b \iff z > \frac {\ln b} {\sigma} \triangleq a$$
Now we try to modify the calculation accordingly:
$$ \begin{align} &~ E[\min\{e^{\sigma Z_1}, e^{\sigma Z_2}\}|Z_1 > a, Z_2 > a] \\
=&~ 2\int_a^{\infty}E[e^{\sigma Z_1}\mathbf{1}_{Z_1 < Z_2}|Z_1 = z, Z_2 > a] \frac {\phi(z)} {\Phi(-a)}dz\\
=&~ 2\exp\left\{\frac {\sigma^2} {2}\right\}\int_{a}^{\infty}\frac {\Phi(-z)\phi(z-\sigma)} {\Phi(-a)^2}dz\\
=&~ 2\exp\left\{\frac {\sigma^2} {2}\right\}\frac {1} {\Phi(-a)}\Pr\{Z_1 + \sigma < Z_2|Z_1 + \sigma > a\} \\
\end{align}$$
So in the last step it involve a bivariate normal CDF, which cannot be in terms of the univarate normal CDF as before.
