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I am two weeks into my first stats course and already I have noticed that, because my class ignores measure theory, the instructors are being sloppy about explaining which kinds of functions are Riemann integrable and which ones require more advanced tools. In general, I feel a lot of handwaving is going on and it seems like this is a generally accepted practice amongst the stat teachers.

For example of (not necessarily Riemann) handwaving,

$$\int e^{-x^2} dx$$ cannot be expressed in terms of elementary functions but the definite integral

$$\int_{-\infty}^{\infty}e^{-x^2} dx$$

can be.

Another example would be the Dirac delta function, which isn't a function but formed through "distributions" or something.

What are some examples of frequently used rules like these whose calculus is actually much trickier than one might expect?

Stan Shunpike
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  • The asymptotic behaviour of the function already tells you alot, that's definitely something you should always look out for. – noctusraid Jan 19 '16 at 07:11
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    There's nothing particularly "tricky" about $\int e^{-x^2}; dx$. The antiderivative just happens to be non-elementary (but nevertheless very useful!). It's certainly not a Riemann integrability issue. – Robert Israel Jan 19 '16 at 07:15
  • @RobertIsrael How does one just know intuitively the indefinite integral can't be put in elementary functions? Like on a test, I'd have to know that in advance. That's what I call "tricky", something I can't figure out in a reasonable amount of time without googling that would be easily solved if a simple fact were known. – Stan Shunpike Jan 19 '16 at 07:56
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    @StanShunpike The fact that the antiderivative of $e^{-x^2}$ is definitely something what you could reasonable expect to be shown in some introductory calculus course. (But it definitely should be mentioned.) See How can one prove the impossibility of writing $ \int e^{x^{2}} , \mathrm{d}{x} $ in terms of elementary functions?. Maybe also other posts linked there might be of interest in connection with this particular topic. – Martin Sleziak Jan 19 '16 at 09:09
  • Unfortunately, calculus courses often give students the false impression that elementary functions have elementary antiderivatives. The reality is that if you write down a more-or-less random elementary function that does not fall in one of the "easy" classes that students are taught how to integrate, it is very likely not to have an elementary antiderivative. Whether you can prove that it does not is another matter: in principle this can be decided using the Risch algorithm, but actually doing that can be hard. – Robert Israel Jan 20 '16 at 01:12
  • As far as tests are concerned: in my experience as an instructor, most students in probability courses have forgotten much of the calculus they learned a year or two ago, and putting complicated integrations on tests is a recipe for disaster. On any of my probability tests, any integral I'd expect the student to do would be very basic (polynomials, powers, exponentials, maybe an easy integration by parts for something like $x e^{-x}$). – Robert Israel Jan 20 '16 at 01:30
  • $\int e^{-x^2/2}; dx$ is a very important special case because it comes up in the normal distribution: it's not elementary, but we have a special function (the standard normal CDF) to handle it. – Robert Israel Jan 20 '16 at 01:32

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Perhaps you're looking for something like this. In elementary probability/statistics courses one deals with random variables that are either "discrete" or "continuous". The discrete ones have countably many possible values with positive probabilities; the continuous ones have a density (piecewise continuous in all examples, but easily generalized to Lebesgue integrable once you have that concept). However, these (and mixtures of these) do not by any means exhaust all the possibilities. There is also a class of singular continuous distributions.

A fairly natural example of a random variable with a singular continuous distribution is this. Consider a sequence of independent tosses of a fair coin, and let $X_n$ be $1$ if the $n$'th toss results in heads, $0$ if tails (thus $X_n$ are independent identically-distributed Bernoulli-$1/2$ random variables), and let $$X = 2 \sum_{n=1}^\infty 3^{-n} X_n $$ Then $X$ has a singular continuous distribution supported on the Cantor set. It's not discrete, because $\mathbb P(X=a) = 0$ for every single point $a$, but $\mathbb P(X \in E) = 1$ where the Cantor set $E$ has measure $0$.

Robert Israel
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  • Where does that formula come from? I don't follow where the 2 and the 3^{-n} come from – Stan Shunpike Jan 19 '16 at 07:50
  • @Stan Are you familiar with the Cantor set? A canonical way of constructing it is as the set of real numbers in $[0,1]$ whose ternary expansion contains no $1$ digit. That formula simply defines a random real number with this form. – A.P. Jan 19 '16 at 09:03