# Why are some non elementary integrals defined and others are not?

Why are some integrals that cannot be integrated in elementary terms defined and given names, while others aren’t? Based on what criteria are they chosen? Applicability to real life? And what is the point if we cannot solve them?

For example:

$$\int\frac{\sin(x)}{x}\,dx=\text{Si}(x), \quad -\int_{-x}^\infty \frac{e^{-t}}{t}\,dt=\text{Ei}(x), \quad \int \cos\left(x^2\right)\,dx = \sqrt{\frac\pi2} \text{C}\left( \sqrt{\frac2\pi}x\right)$$ while others like $$\int x^x \,dx$$ are not defined.

• Just because they turn out to be widely useful, yep; usually because they turn up in more places than just "the result of the integral". The exponential integral turns out to be part of the solution to several physically-motivated differential equations, for example. – Patrick Stevens Feb 5 '18 at 9:07
• Since there are infinitely many non-elementary antiderivatives and useful technical terminology is finite, how could it be otherwise? – John Coleman Feb 5 '18 at 14:56
• Also, what do you mean by saying that " we cannot solve them"? A definition of a function in terms of an a definite integral is a perfectly useful definition. You can plot such functions, differentiate them, evaluate them to any desired degree of accuracy, etc. Elementary functions are just a small subset of all functions. They shouldn't constrain your intuition about what a function really is. Most people don't encounter non-elementary functions in high school, but so what? – John Coleman Feb 5 '18 at 16:49

I can define one right now. I hereby declare that $$\mathrm{Xi}(x) := \int_0^xt^tdt$$ Now, the only thing that remains is to see whether other mathematicians pick it up and start using it.

As with any other mathematical notation, there isn't really a committee somewhere who sits and weighs criteria to decide what mathematical notation is correct and should be accepted. What decides whether a piece of notation gets accepted and conventional is simply whether enough other mathematicians find it useful and start using it.

• Good job keeping up the cover story. Remember, TINC (there is no committee). – Yakk Feb 5 '18 at 15:30
• A function is only as important as it is useful. – omegadot Feb 5 '18 at 23:10

And what is the point if we cannot solve them?

That's actually a big part of the motivation for giving some integral a name. An integral that can be solved exactly doesn't really need a name – you can just go ahead, solve it right now and thus avoid having to write out the integral anymore. Or else, keep writing it out in verbatim, and anybody can calculate the results themselves later.

OTOH, an integral whose result we can't write in elementary form can be a pain in the back, if you have to keep carrying around. You may still be able to obtain numerical results, derive differential relationships etc., but all that is tedious work that'll need to be done over and over again by everybody who encounters the integral.

Much better is if you realise “ah, that's the Sine Integral”, then you can invoke some standard formulas / implementations / tables to look up values.

"Cannot solve them" might not be the right point of view. Example: consider $$\tag1 \text{erf}(x)=\frac2{\sqrt\pi}\,\int_{0}^xe^{-t^2}\,dt.$$ This is the error function, the prototypical function that cannot be expressed in terms of elementary functions. Now, if you had a formula, it would probably involve the exponential; and any time you want to obtain numbers from the exponential, you need to use its Taylor series. So, approximations of the series $$\tag2 \text{erf}\,(x)=\frac2{\sqrt\pi}\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{n!(2n+1)}$$ are actually the mosta direct way of calculating it (and fairly efficient for $x$ not too big).

More importantly, just the fact that a function has a name, doesn't really make it "easier": compare $(2)$ with $$\tag3 \sin x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}.$$ Do you think that having $\sin x$ in a formula is much better than having $\text{erf}\,(x)$?

• What's the bounds on that Taylor series for erf? – Joshua Feb 6 '18 at 6:27
• @Joshua Do you mean the radius of convergence? Just as with $e^x$ and $\sin x$, for any $x$ the $n!$ in the denominator is eventually dominating, and the radius is infinite. – Arthur Feb 6 '18 at 7:22
• @Arthur correct, still in practice you wouldn't want to evaluate either of these Taylor series very far from 0, because it gets numerically unstable. In case of the trig and exp functions, that's fortunately easy to avoid by exploiting periodicity etc.; in case of the error function it's a bit less straightforward. – leftaroundabout Feb 6 '18 at 10:18