# Analogous of $\int 1.999 x^{0.999} \mathrm dx \approx x^2$ for $\int e^{-x^2} \mathrm dx$

So everyone knows that $\int e^{-x^2} \mathrm dx$ can not be evaluated in terms of elementary functions. So is there a way to have something approximate? for example although the following is not well defined yet it is not too hard to allow :

$\int 1.999 x^{0.999} \mathrm dx \approx x^2$ but in this case $1.999 x^{0.999}$ can be integrated in terms of elementary functions, but if it was not possible, we could have replaced it with $2x$.

My question is can we do something similar for $e^{-x^2}$ or other functions that have no elementary integrals and substitute in something that is close enough to the original function that has an integral in terms of elementary functions?

By "close enough" I have no formalised definition to say what I mean, please mention things that can make 'close enough' in this context more meaningful.

• You can expand the power series for $e^{-x^2}$ to an arbitrarily large number of terms. – Potato Sep 1 '13 at 8:22
• @Potato : Yes, but also could have done that for $1.999x^{0.999}$, but instead using $x$ somehow seemed preferable. can we swap $e^{-x^2}$ with some other function ( even if that other function depended on the interval? – Arjang Sep 1 '13 at 8:27

There are some possibilities, but they are valid only in a limited range of values for $x$. The same holds for your $1.999x^{0.9999}$ actually :-)
Consider the following $$D(\frac1{2x}e^{-x^2})=\frac{-2x}{2x}e^{-x^2}-\frac1{2x^2}e^{-x^2}=-e^{-x^2}(1+\frac1{2x^2}).$$ So for very large $x$ the derivative of $-e^{-x^2}/(2x)$ is close to $e^{-x^2}$. You can repeat the dose and use $$D\left(e^{-x^2}(-\frac1{2x}+\frac1{4x^3})\right)=e^{-x^2}(1+\frac3{4x^4})$$ or go even further.
These approximations are useful only for relatively large values of $x$, e.g. estimating the tail probability of a normal random variable having a value, say $+10$SD or above. For smaller values of $x$ undoubtedly other types of tricks work better, but I'm not an expert.