Let $N(x)$ denote the cdf of standard normal and $n(x)$ denote the pdf of standard normal.

How to evaluate the integral $\int\limits_{-\infty}^\infty N(a+x) n(x) \mathrm{d} x$ ?

Thanks a lot!

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Use $\partial_a N(a+x) = n(a+x)$ and $n(-x) = n(x)$: $$\partial_a \int_{-\infty}^\infty N(a+x) n(x) \mathrm{d} x = \int_{-\infty}^\infty n(a+x) n(x) \mathrm{d} x = \int_{-\infty}^\infty n(-a-x) n(x) \mathrm{d} x = n\left(-\frac{a}{\sqrt{2}} \right)\frac{1}{\sqrt{2}}$$ The last equality reflects that the the sum $Z = X+Y$ of two standard normal random variates $X$ and $Y$ is another random variable with zero mean and variance $2$. The pdf of $Z$ is a convolution of pdfs of $X$ and $Y$: $$f_Z(z) = \int_{-\infty}^\infty f_X(z-y) f_Y(y) \mathrm{d} y$$ Thus: $$\int_{-\infty}^\infty N(a+x) n(x) \mathrm{d} x = \int_{-\infty}^a n\left(-\frac{b}{\sqrt{2}} \right) \frac{1}{\sqrt{2}} \mathrm{d} b = 1 - N\left(-\frac{a}{\sqrt{2}} \right) = N\left(\frac{a}{\sqrt{2}} \right)$$

Check:

In[15]:= With[{a = 2/3.},
{NIntegrate[
CDF[NormalDistribution[], a + x] PDF[NormalDistribution[],
x], {x, -Infinity, Infinity}],
CDF[NormalDistribution[], a/Sqrt[2]]}]

Out[15]= {0.681324, 0.681324}

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