I am reading an optics textbook that uses the following integral to evaluate the power squared in the lower tail of the following Gaussian integral.
$$\frac{1}{{{\sigma _P} \cdot \sqrt {2 \cdot \pi } }} \cdot \int\limits_{ - \infty }^0 {{P^2} \cdot \exp \left( { - \frac{{{{\left( {P - {P_b}} \right)}^2}}}{{2 \cdot \sigma _P^2}}} \right)} \cdot dP $$
The author gives the following approximation for the value of this integral with no explanation. $$\sqrt {\frac{2}{\pi }} \cdot \frac{{\sigma _P^5}}{{P_b^3}} \cdot \exp \left( { - \frac{{P_b^2}}{{2 \cdot \sigma _P^2}}} \right) $$
I have been unable to derive this approximation. Are any of you familiar with it?