# What would qualify as a valid reason to believe there is a closed form?

I noticed that almost every non-homework-level integral posted on this site prompts somebody to ask "Do you have any reason to believe there is a closed form?" (some recent examples here and here)

I actually never fully understood either meaning or purpose of such questions.

• What would qualify as a valid reason to believe there is a closed form?
• Why should a lack of such reason inhibit an interest in an integral?

Personally, I hold the following view at this subject: we should always believe there is a closed form. Even if eventually somebody manages to prove there is no closed form in a certain class of functions, we might introduce a new special function or constant to represent a value of the integral/product/sum in question, study its properties, find other integrals that can be represented using it, discover interesting connections with other functions, etc. This happened many times in the past (think of $\operatorname{erf}(x)$, $\Gamma(x)$, $\zeta(x)$, $\operatorname{Li}_s(x)$, $K$, $A$, etc) and always helped to move math forward.

• The problem is that all those questions are badly stated. You certainly don't want the answer of "Evaluate $\int f(x)dx$" to be "$I_f(x)+C$, where $I_f(x):=\int_{0}^{x}f(t)dt$". – OR. Nov 11 '13 at 18:03
• @ABC I do not see how they are badly stated. Most problems in books or in homework assignments stated in the same way. Everybody understands that the answer should not be just a repetition of the question. It is expected to be in terms of constants/functions that we have better understanding and intuition about. – Vladimir Reshetnikov Nov 11 '13 at 18:10
• @ABC, indeed that is the purpose of those comments. Some utterly arbitrary-looking integral appears without any context to indicate why it matters and someone wants a closed form that most likely doesn't exist (because most integrals don't have them!) – dfeuer Nov 11 '13 at 18:28
• @dfeuer "most integrals don't have them" - most in what sense? Can this claim be proved? – Vladimir Reshetnikov Nov 11 '13 at 18:30
• @VladimirReshetnikov This (download.springer.com/static/pdf/565/…) (Look in A7) could be a place to find the answers to those questions. So, if for every finite collection of 'basic functions' we can find an indefinite integral for which Khovanskii's geometric obstructions prevent it from being expressible in terms of those 'basic functions', then we have more integrals that we cannot express in closed form. – OR. Nov 11 '13 at 18:55

The question of whether a given integral has a closed form does not have a consistent answer from one practitioner to the next. In my experience, however, there's a plausibility that there is one when the integration limits include a branch point or an otherwise integrable singularity.

But what is a closed form? That said, we can debate until we turn blue as to what constitutes a closed form. In my humble opinion, a closed form implies a means of computing the value of the integral that results in fewer operations that simply computing the integral by some numerical scheme. Knowing when this condition has been satisfied takes a certain level of familiarity with the special functions we use to express our integration results.

For example, erf is considered a closed form, and rightly so, because we have amazingly accurate schemes to compute the error function using very few operations. These schemes are much more often then not much more efficient than any numerical integration scheme, so we may call it a closed form.

On the other hand, consider the following answer to a question posed yesterday:

$$-{\frac {\,{\mbox{_3F_2}(1/6,1/2,1/2;\,7/6,3/2;\,1)}\Gamma \left( 5/6 \right) \Gamma \left( 2/3 \right) -{\pi }^{3/2}}{6 \Gamma \left( 5/6 \right) \Gamma \left( 2/3 \right) }}$$

Is this considered a closed form? It depends. Is there a scheme that computes such a hypergeometric faster than the original (double) integral, which has an integrable singularity? As I am not very familiar with such hypergeometrics, I'd have a hard time answering that question. My rule of thumb is that, when I see any hypergeometric higher than $_2F_1$ in any result, I do not consider it closed form because it looks so...icky. People who are better versed than me in such matters may feel free to disagree with me.

(NB The actual answer to the question, by the way, is $\pi/24$, which is a whole other can of worms.)

Of course, we could go on about the notion of computability in general, but I'd rather just stick with the notion of tolerance and counting operations.

In any case, I hope these observations won't temper the OP's enthusiasm with working with integrals like the ones he poses and solves here, but will further challenge him to provide more beautiful results that can be useful as well.