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I am currently trying to evaluate this Integral: $$\int\limits_{u_0}^{u_1} \exp\left[-\angle(H(u),N)^2\right]du$$ Where $H(u)=:\overrightarrow{a}+(\overrightarrow{b}-\overrightarrow{a})*u-\overrightarrow{x}+\overrightarrow{v}$,

($\overrightarrow{a},\overrightarrow{b},\overrightarrow{x}\in\mathbb{R}^3$ constant)

$\angle(\overrightarrow{u},\overrightarrow{v}):=\cos^{-1}\left(\frac{\overrightarrow{u}\cdot \overrightarrow{v}}{\|\overrightarrow{u}\|\;\|\overrightarrow{v}\|}\right)$

The Problem is that this integral is rather complicated and yields no result in mathematica. For that reason, I am looking for ways to simplify it.

Most importantly, you could modify the underlying coordinate system such that $H(u)=\left(\begin{smallmatrix}0\\0\\u*\alpha\end{smallmatrix}\right)$. This would require every other vector to be transformed in the same manner however, and therefore not improve the Integral whatsoever. The shortest form yet was translating and rotating everything to make $\overrightarrow{b}-\overrightarrow{a}=\left(\begin{smallmatrix}0\\0\\u*\beta\end{smallmatrix}\right)$, however I did not receive a closed form aswell. While trying I found that mathematica does not find a closed form for $\int e^{-cos^{-1}(x^2)}dx$, however it finds one for $\int e^{-cos^{-1}(x)}dx$ (This is a different question however)

Now, my question is, is there a closed form to this? Can $e^{cos^{-1}(x^2)}$ be simplified/modified to become easier to integrate? (Highly doubt that, mathematica does not simplify that aswell)

For reference, here is a sketch of the underlying Idea: A sketch.

We Integrate over the Gaussian specular lighting value for all points on the line $\overline{ab}$ (See Gaussian Distribution Specular Highlight). The specular value is given by $k_{spec}=Exp\left(-\left(\frac{\angle(H,N)^2}{m}\right)\right)$ (H is the half-view-vector, N the normal vector). H is variable as we modify the point on the line. Therefore $H(u)=\underbrace{a+(b-a)*u-x}_{\text{L(U)}}+v$

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