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I'm trying to simplify this formula here. For the most part, evaluating this expression is straightforward. However I am very confused about how to approach this last term in the expression. I want to simplify this part of the formula and break it down further if possible by coming up with an antiderivative formulation or something similar to make it more understandable, and to be able to evaluate it in terms of its unknowns. It's clearly related to the normal distribution but I'm unsure of how to simplify it in terms of the normal as well. How should I approach this? What would be the best way to evaluate this term?

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Ultimately, it seems like you want a nice expression for integrals of the form $$ \int_a^b e^{-x^2} dx. \tag{1}$$ You ask for an antiderivative. Unfortunately, no antiderivative of this expression exists in terms of elementary functions. If you're looking to write it in terms of named special functions, then you might think of $(1)$ as $$ \int_a^0 e^{-x^2} dx + \int_0^b e^{-x^2}dx$$ and consider integrals of the form $$ \int_0^b e^{-x^2} dx.$$ By a change of variables, this can be written as $$ \frac{1}{2} \int_0^{\sqrt{b}} e^{-x} x^{1/2} \frac{dx}{x},$$ which is the lower incomplete Gamma function $$\frac{1}{2}\gamma(\tfrac{1}{2}, \sqrt b).$$ This comes up frequently, and one can understand it numerically pretty quickly.

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  • $\begingroup$ Thank you for your reply. I can see how to simplify the integral in terms of the lower incomplete gamma function as you've shown, but how do I evaluate that integral in terms of the infinite upper bound? How would I apply this transformation to the function above? $\endgroup$ – beeba Mar 29 '16 at 18:06
  • $\begingroup$ If you have integrals of this shape with infinite upper bounds, then you can either use the upper incomplete Gamma function or the regular Gamma function and subtraction to transform it into a lower incomplete Gamma function. $\endgroup$ – davidlowryduda Mar 29 '16 at 19:52

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