I am trying to Integrate the following Integral (all of the variables are reals) $$\int Exp(-(\cos^{-1}\left[\frac{\text{n}_0 (\text{v}_0-\text{x}_0)+\text{n}_1 (\text{v}_1-\text{x}_1)+\text{n}_2 (u \text{v}+\text{v}_2-\text{x}_2)}{\sqrt{\text{n}_0^2+\text{n}_1^2+\text{n}_2^2} \sqrt{(\text{v}_0-\text{x}_0)^2+(\text{v}_1-\text{x}_1)^2+(u \text{v}+\text{v}_2-\text{x}_2)^2}}\right]/m)^2) \, du$$

The formula, simplified, in Mathematica input:

Integrate[Exp[-(ArcCos[(a + b*u)/(c*Sqrt[d + (e + b*u)^2])]/m)^2], u]

If I simply enter that into Mathematica, it instantly returns the same expression. If I enter it as an Integral with set limits (0 and t), it calculates on and on and on....

How do I go about this? Is there any tricks that can be applied to

  1. Find out if there is a closed Form for this (obviously including Erf terms)
  2. Get an Approximation for the Integral (For example using series)
  3. Make Mathematica take more time trying to find a closed form for the indefinite Integral/antiderivative?
  • 3
    $\begingroup$ Actual Mathematica code would be helpful here. Also note that the integration variable shows up only in two places. You can thus simplify the parameters and that might help speed things. For example the numerator of the arccos argument might become a+b*u with a,b assumed to be real. $\endgroup$ – Daniel Lichtblau Jan 22 '13 at 14:47
  • $\begingroup$ Are you sure there is no imaginary unit inside of exponent? Because then it looks more like some integral from optics or something and simplifies to pretty integratable one. $\endgroup$ – swish Jan 22 '13 at 15:15
  • $\begingroup$ @swish I am absolutely sure. This is an integral over specular lighting (Gauss Specular lighting) along a part of the z-axis. The formula is $e^{-\angle(H,L)^2}$ - I just realized I forgot the ^2 >_< $\endgroup$ – CBenni Jan 22 '13 at 15:22
  • $\begingroup$ @DanielLichtblau I will try this. However it will make integrating way harder, as there will be a few new variables? That makes simplifying not easier... $\endgroup$ – CBenni Jan 22 '13 at 15:24
  • $\begingroup$ Should be slightly easier as there are fewer new variables than old. Also your a+b*u and e+b*u appear to be not quite right. I think it should be n2*(a+b*u) in the numerator and a+b*u in the denominator (n2 being the constant in the original formulation). $\endgroup$ – Daniel Lichtblau Jan 22 '13 at 15:41

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