Expressing a classic integral having no elementary primitive. Some functions do not have primitive functions which are easy to express with elementary functions.
A famous example is $$\int_a^be^{-x^2}dx$$
But what about if we can write 
$$=\int_a^b1\cdot e^{-x^2}dx = \int_a^b|e^{ix}|^2\cdot e^{-x^2}dx$$
and maybe even:
$$=\int_{-\infty}^\infty B_{a,b}(x)\cdot |e^{ix}|^2\cdot e^{-x^2}dx$$
Where $B_{a,b}(x)$ is box-function ($=1$ on $[a,b]$ and $=0$ otherwise)
Could this help us somehow in expressing this integral in a nicer more easy-to-calculate way?
 A: I'm not sure what question you're asking, but the answer is probably 'no'.
First, $B_{a,b}(x)$ almost surely won't work. I say this because $B_{a,b}(x)$ is already an "elementary" function, or at least, it is "elementary" for the purpose of integration, if the absolute value function is allowed in "elementary". (This is because $B_{a,b}$ is the difference of two step functions, and $|x|/x$ is a step function.) 
It's true that $|\cdot|$ is usually not considered elementary, but I also feel like ordinary calculus books play pretty fast and loose with these more technical definitions. So if the "impossibleness" were really just a matter of absolute value bars, then the error function formula would be famous, and we'd all be telling our students they have to know it for their exam.
As an aside, your verbiage was a little careless at times, so I just want to take a minute to make sure we all are talking about the same thing: it's not just that the integral isn't "easy to express with elementary functions". You can prove that there is no elementary antiderivative of $e^{-x^2}$.
Second, as far as getting an answer is concerned, the exact value of
$\displaystyle \int_a^be^{-x^2}dx $ is already known: it is $\frac{\sqrt\pi}{2}[\text{erf}(b)-\text{erf}(a)]$. You might say that using the error function is cheating, and I agree. But if you're willing to allow $\log(x)$— which is already either defined by an integral or relies on an auxiliary construction, for instance of the number $e$— then I don't see why $B_{a,b}$ is any more natural a function than the error function.
Third, Count Iblis' comment runs along the same lines. Translating into symbols, they are saying that
$$ \int_a^be^{-x^2}dx = k \int_{-\infty}^{\infty}\frac{\sin((b−a)\pi x) \,e^{-x^2}}{x}dx. $$
To me, this is either disallowed by whatever rules are forbidding the previous answer, or it is a less elegant solution than the previous answer.
Finally, if "easy-to-calculate" includes easy-to-estimate, then there's a simpler solution than any of this, which is just to use, say, Simpson's Rule. If you want something more elegant, you can try using the power series, but I suspect that Simpson's rule will get an accurate answer with less computation.
