How did Owen derive this relation?

Equation 10,010.3:

$$ \int G'(x)G(a+bx) dx = T\left(x,\frac{a}{x\sqrt{1+b^2}}\right)+ T\left(\frac{a}{\sqrt{1+b^2}},\frac{x\sqrt{1+b^2}}{a}\right)\\-T\left(x,\frac{a+bx}{x}\right)-T\left(\frac{a}{\sqrt{1+b^2}},\frac{ab+x(1+b^2)}{a}\right)+G(x)G\left(\frac{a}{\sqrt{1+b^2}}\right) $$

in terms of normal CDF $G(x)$, the normal pdf $G'(x)$ and Owen's $T$-function. I cannot find the derivation nor the derivative of the $T$-function, only its definition both in Owen's Table of Normal Integrals and on Wikipedia, thus bluntly taking the derivative does not work.

  • 2
    $\begingroup$ For the definite integral see stats.stackexchange.com/questions/61080/…. Note that the integral on the left hand side of the equation is implicitly over some interval determined by "$x,$" presumably $(-\infty,x].$ Once you interpret this integral in an ordinary regression context, the definition of $T$ on Wikipedia makes short work of this formula. $\endgroup$
    – whuber
    Dec 16, 2020 at 14:01
  • 1
    $\begingroup$ It is very interesting to me that this question is unanswered. Of course @whuber is correct with their statements but in my opinion, the question is not answered by a long shot. $\endgroup$
    – zonksoft
    Dec 14, 2023 at 14:15


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