Let $\phi(\cdot)$ and $\Phi(\cdot)$ be the probability and cumulative density functions, respectively, of a random variable with distribution $\text{N}(0,\,1)$. That is,
$$\forall x\in\mathbb{R}:\,\phi(x)=\frac{1}{\sqrt{2\cdot\pi}}\cdot\text{e}^{x^2/2}$$
and
$$\forall x\in\mathbb{R}:\,\Phi(x)=\int_{\infty}^{x}\phi(u)\,\text{d}u.$$
I was wondering if you could help me to compute
$$\int_{\infty}^{w}\phi(x)\cdot\Phi(a+b\cdot x)\,\text{d}x\mbox{,}$$
where $(a,\,b,\,w)\in\mathbb{R}^3$ and $b\neq 0$, please.
Thanks a lot for your help.

1$\begingroup$ How did this integral come up? Do you expect it to have some kind of closedform solution? $\endgroup$ – theHigherGeometer Oct 20 '17 at 6:42

$\begingroup$ Hi, @DavidRoberts. This integral is part of an algorithm I am programming in Rsoftware. My purpose is to avoid the use of integrals in Rsoftware because they reduce the efficiency of my algorithm. For this reason, I need to find a closedform solution for this integral. $\endgroup$ – Student1981 Oct 20 '17 at 10:09

$\begingroup$ Both Maple and Mathematica fail with it. Wanting is not harmful, harmful not want to. $\endgroup$ – user64494 Oct 20 '17 at 10:51
This integral can be found in D. B. Owen (1980) A table of normal integrals, Communications in Statistics  Simulation and Computation, 9:4, 389419:
BvN
denotes the bivariate normal probability function.
Check in R:
> a < 2
> b < 3
> w < 5
> f < function(x) dnorm(x)*pnorm(a+b*x)
> integrate(f, lower=Inf, upper=w)
0.7364551 with absolute error < 1.3e06
>
> rho < b/sqrt(1+b^2)
> Sigma < cbind(c(1,rho),c(rho,1))
> mvtnorm::pmvnorm(upper=c(a/sqrt(1+b^2), w), sigma=Sigma)
[1] 0.7364551
attr(,"error")
[1] 1e15
attr(,"msg")
[1] "Normal Completion"
Alternatively, you can express this integral with the Owen $T$function:
> library(OwenQ)
> 1/2*(pnorm(a/sqrt(1+b^2)) + pnorm(w)  2*OwenT(w, (b*w+a)/w)  2*OwenT(a/sqrt(1+b^2), (a*b+w*(1+b^2))/a)  (a <= 0))
[1] 0.7364551
Benchmark:
> library(mvtnorm)
> library(OwenQ)
> library(microbenchmark)
>
> a < 2
> b < 3
> w < 1
>
> microbenchmark(
+ integral = integrate(function(x) dnorm(x)*pnorm(a+b*x), lower=Inf, upper=w),
+ mvtnorm = {rho < b/sqrt(1+b^2); pmvnorm(upper=c(a/sqrt(1+b^2), w), sigma=cbind(c(1,rho),c(rho,1)))},
+ OwenT = 1/2*(pnorm(a/sqrt(1+b^2)) + pnorm(w)  2*OwenT(w, (b*w+a)/w)  2*OwenT(a/sqrt(1+b^2), (a*b+w*(1+b^2))/a)  (a <= 0))
+ )
Unit: microseconds
expr min lq mean median uq max neval cld
integral 80.677 83.5860 116.97275 90.0240 93.0625 2878.062 100 b
mvtnorm 320.550 327.0625 339.22625 330.3975 336.0315 595.829 100 c
OwenT 22.682 24.6360 28.89006 29.2685 31.9955 51.015 100 a

1$\begingroup$ @Student1981 I've just added a speed comparison in my answer.
OwenT
is the way to go. $\endgroup$ – Stéphane Laurent Oct 21 '17 at 10:18 
$\begingroup$ These are efficient reductions of the integral to standard libraries. But I'd say a solution with the bivariate normal CDF or OwenT is not in closed form; I would restrict the term "closedform" to quantities that can be calculated with at most a single integral of elementary functions. $\endgroup$ – Matt F. Oct 22 '17 at 1:03

$\begingroup$ I appreciate your observation, @MattF. I believe I did not choose well my words when I wrote "closedform solution". I will be more careful next time. However, for my purposes, the approach of Stéphane Laurent's answer is good enough. $\endgroup$ – Student1981 Oct 22 '17 at 2:21

$\begingroup$ FYI, the condition in the OwenT derivation is wrong. It should be
(a/w<0)
and not(a<=0)
. The conclusion can be derived from 10,010.3 of Owen's paper and using the property $T(u,v) + T(uv,1/v) = \frac{1}{2} (\Phi(u) + \Phi(uv))  \Phi(u) \Phi(uv)  \frac{1}{2} [v<0]$. $\endgroup$ – jvdillon Feb 4 at 5:34
There is no known closedform solution for this integral, though many have asked for it. However, for the case of $w=\infty$ you can use Geller and Ng (http://nvlpubs.nist.gov/nistpubs/jres/73B/jresv73Bn1p1_A1b.pdf), integral 4.3.13.