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Let $a,b,c \ge 0$.

In the course of solving gaussian integral of power of cdf : $\int_{-\infty}^{+\infty} \Phi(x)^n \cdot \phi(a+bx) \cdot dx$ we came across a following integral: \begin{eqnarray} T(a,b,c):= \int\limits_a^\infty \phi(\xi) T(\xi b,c) d\xi \end{eqnarray} where $T(.,.)$ is the Owen's T function https://en.wikipedia.org/wiki/Owen%27s_T_function. As the Wikipedia page tells us this function is related to likelihoods of a bivariate Gaussian distribution and furthermore the integral itself is related to likelihoods of a triple-variate Gaussian distribution.

Now, by going to spherical coordinates we managed to find our integral in case $a=0$. We have:

\begin{equation} T(0,b,c) = \frac{1}{4\pi} \arcsin\left( \frac{c}{\sqrt{(1+b^2)(1+c^2)}}\right) \end{equation}

My question is therefore what is the value of the integral for generic $a>0$ ?

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It turns out that the answer is actually quite simple and can be obtained using elementary methods. We have: \begin{eqnarray} T(a,b,c)= \int\limits_{{\mathbb R}^3} 1_{\xi_0 > a} 1_{\xi_1 > \xi_0 b} 1_{c \xi_1 > \xi_2 >0} \prod\limits_{p=0}^2 \phi(\xi_p) d \xi_p \end{eqnarray} Now we go to spherical coordinates as follows: \begin{eqnarray} \xi_0&=&r \sin(\theta) \cos(\phi)\\ \xi_1&=&r \sin(\theta) \sin(\phi)\\ \xi_2&=& r \cos(\theta) \end{eqnarray} where $\theta \in[0,\pi/2)$ and $\phi\in[-\pi/2,\pi/2)$ which follows from the fact that all three coordinates are positive. Now we carefuly analyze the inequality conditions: \begin{eqnarray} 1_{\xi_0 > a} &:& r \sin(\theta) \cos(\phi) > a \quad \Longrightarrow \quad \theta > \frac{a}{r \cos(\phi)} \quad \Longrightarrow \quad \theta > \arcsin(\frac{a}{r \cos(\phi)})\\ 1_{\xi_1>\xi_0 b} &:& r \sin(\theta) \sin(\phi) > b r \sin(\theta) \cos(\phi) \quad \Longrightarrow \quad \tan(\phi) > b\\ 1_{c \xi_1> \xi_2 >0} &:& c r \sin(\theta) \sin(\phi) > r \cos(\theta) \quad \Longrightarrow \quad c \sin(\phi) > \cot(\theta) \quad \Longrightarrow \quad \theta > arccot(c \sin(\phi)) \end{eqnarray}

Now, in the last equation on the right in the top line above we must have $r > a/\cos(\phi) = a \sqrt{1+\tan(\phi)^2} > a \sqrt{1+b^2}$. Therefore we write down the integral in spherical coordinates as follows: \begin{eqnarray} (2\pi)^{3/2}T(a,b,c)&=& \int\limits_{a\sqrt{1+b^2}}^\infty e^{-1/2 r^2} r^2 \int\limits_{\arctan(b)}^{\arccos(a/r)} \int\limits_{\arcsin(\frac{a}{(r \cos(\phi))})\vee arccot(c \sin(\phi))}^{\pi/2} \sin(\theta) d\theta \quad d\phi \quad dr\\ &=& \int\limits_{a\sqrt{1+b^2}}^\infty e^{-1/2 r^2} r^2 \int\limits_{\arctan(b)}^{\arccos(a/r)} % \left[\cos(\arcsin(\frac{a}{r \cos(\phi)})) \wedge \cos(arccot(c \sin(\phi))) \right] % \quad d\phi \quad dr\\ &=& \int\limits_{a\sqrt{1+b^2}}^\infty e^{-1/2 r^2} r^2 \int\limits_{\arctan(b)}^{\arccos(a/r)} % \left[ \frac{c \sin(\phi)}{\sqrt{1+c^2 \sin(\phi)^2}} \wedge \frac{\sqrt{r^2 \cos(\phi)^2-a^2}}{r \cos(\phi)} \right] % \quad d\phi \quad dr \\ &=& \int\limits_{a\sqrt{1+b^2}}^{a\sqrt{1+b^2(1+c^2)}} e^{-1/2 r^2} r^2 \int\limits_{\arctan(b)}^{\arccos(a/r)} \frac{\sqrt{r^2 \cos(\phi)^2-a^2}}{r \cos(\phi)} d\phi dr +\\ && \int\limits_{a\sqrt{1+b^2(1+c^2)}}^\infty e^{-1/2 r^2} r^2 \left( \int\limits_{\arctan(b)}^{\arccos(\frac{a \sqrt{1+c^2}}{\sqrt{a^2 c^2 + r^2}})} \frac{c \sin(\phi)}{\sqrt{1+c^2 \sin(\phi)^2}} d\phi + % \int\limits_{\arccos(\frac{a \sqrt{1+c^2}}{\sqrt{a^2 c^2 + r^2}})}^{\arccos(a/r)} \frac{\sqrt{r^2 \cos(\phi)^2-a^2}}{r \cos(\phi)} d\phi \right) =\\ &=& \int\limits_{a\sqrt{1+b^2}}^{a\sqrt{1+b^2(1+c^2)}} e^{-1/2 r^2} r^2 \left( \frac{a}{r} \arctan[\frac{a b}{\sqrt{r^2-a^2(1+b^2)}}]- \arctan[\frac{r b}{\sqrt{r^2-a^2(1+b^2)}}] \right)dr +\\ && \frac{\pi ^{3/2}}{2 \sqrt{2}} \left(\sqrt{\frac{2}{\pi }} a \left(\sqrt{b^2+1}-1\right) e^{-\frac{1}{2} a^2 \left(b^2+1\right)}-\frac{\sqrt{2} a e^{-\frac{1}{2} a^2 \Delta^2} \left(\Delta \left(\pi -2 \arctan\left(\frac{c}{\Delta}\right)\right)-2 \arctan\left(\frac{1}{c}\right)\right)}{\pi ^{3/2}}-\text{erf}\left(\frac{a \sqrt{b^2+1}}{\sqrt{2}}\right)+\frac{\text{erf}\left(\frac{a \Delta}{\sqrt{2}}\right) \left(\pi -2 \arctan\left(\frac{c}{\Delta}\right)\right)}{\pi }+\frac{2 \arctan\left(\frac{c}{\Delta}\right)}{\pi }\right) \end{eqnarray} where $\Delta:=\sqrt{1+b^2(1+c^2)}$. As a sanity check we look at the case $a=0$. In here only the very last term survives and we have: \begin{eqnarray} (2\pi)^{3/2} T(0,b,c)&=& \sqrt{\frac{\pi}{2}} \arctan(\frac{c}{\Delta})\\ \Longrightarrow\\ T(0,b,c)&=& \frac{1}{4\pi} \arctan(\frac{c}{\Delta})= \frac{1}{4\pi} \arcsin(\frac{c}{\sqrt{(1+b^2)(1+c^2)}}) \end{eqnarray} as it should be.

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