Help solving the following integral with a CDF Here is my problem
Let $H(x)=\Phi(\frac{Ax+b}{\sqrt{2}})$ where $A$, $b$ are some non-negative constant and $\Phi$ is the CDf of a standard normal distribution.
I want to find
$$\int x \, dH(x)$$
I thought maybe integration by parts which yields:
$$x H(x)-\int H(x) \, dx$$
And now I'm not sure how to proceed or if it's possible to take the integral of a CDf
 A: Let us do some probability here. Let $Z$ denote a standard normal random variable, with CDF $\Phi$, and $X$ a random variable with CDF $H$. Then, for every $x$, $H(x)=\Phi((Ax+b)/\sqrt2)$ implies that $\mathrm P(X\leqslant x)=\mathrm P(Z\leqslant(Ax+b)/\sqrt2)=\mathrm P((\sqrt2Z-b)/A\leqslant x)$. That is, $X$ is distributed as $(\sqrt2Z-b)/A$. In particular, $Z$ is centered hence
$$
\int\limits x\mathrm dH(x)=\mathrm E(X)=\mathrm E((\sqrt2Z-b)/A)=(\sqrt2\mathrm E(Z)-b)/A=-b/A.
$$
A: Since $H$ is an everywhere differentiable function, we have
$$
\int x \, dH(x) = \int x H'(x)\,dx = \int x \Phi'\left(\frac{Ax+b}{\sqrt{2}}\right) \frac{A}{\sqrt{2}}\, dx.
$$
Let $u=\dfrac{Ax+b}{\sqrt{2}}$, so that $du = \dfrac{A\,dx}{\sqrt{2}}$ and $x= \dfrac{\sqrt{2}\; u - b}{A}$.  Then the integral becomes
$$
\int \frac{\sqrt{2}\;u-b}{A} \Phi'(u) \, du.
$$
Then make this into
$$
\frac{\sqrt{2}}{A} \int u \Phi'(u)\,du - \frac{b}{A} \int \Phi'(u)\,du.
$$
The second integral above becomes $\Phi(u)+\text{constant}$.  The first is
$$
\int u \Phi'(u)\,du = \int u e^{-u^2/2} \, du = \int e^{-w}\,dw = -e^{-w}+\text{constant}.
$$
Then convert back to an expression in $u$, then back to an expression in $x$.
If you had in mind a definite integral from $-\infty$ to $\infty$, then
$$
\int_{-\infty}^\infty \Phi'(u)\,du = 1
$$
and
$$
\int_{-\infty}^\infty u \Phi'(u) \, du = 0.
$$
The second integral is $0$ because you're integrating an odd function over an interval that is symmetric about $0$.  The first is $1$ since you're integrating a probability density function.
