Is indefinite integration largely a heuristic or it can be mechanical too?

Question

I am sorry for putting multiple questions in the same post, but I think providing answers here will be better.

As far as I know, there is no 'product formula' for integrals, like we have for the derivative. Also, I can be wrong, but I think a general class of functions, differing by a constant, have the same derivative. So, ignoring the constant, one might think that such a product formula might exist. So, my first question:

Can it be proved that there is no 'product formula' for integrals, or is it just that it has not been discovered yet?

Let us reduce our case to just rational functions. Partial fractions for integration is a pretty good technique, but I think it can't be used for all rational functions, because not all polynomials have all real roots. So:

Is there a technique/algorithm to integrate all rational functions?

Another open-ended question, that I think of, is that are there any techniques/formulas for integrals not discovered yet, or are the existence of these techniques/formulas been proven false by some theorem? Answer to any question is suffice.

EDIT FROM HERE
After reading some of the answers, I felt I need to be more precise. The integration by parts formula, as far as I know, is again a heuristic, and not mechanical. So, there is no scope for integration by parts theorem to be such a product formula. Another user answered that I thought of a formula combining functions in an elementary way, and proposed that it is well known that the sinc function does not have a closed form integral, and so such a formula doesn't exist. But, to add to this, there is also a possibility that such a formula may produce an undefined result, or some weird result, which we can relate to the absence of such a closed-form solution. What I am looking for is a theorem or result which clearly proves the non-existence of such a formula, taking into account all possibilities.

YES! There is a technique for all rational functions! Really long though. — Simply Beautiful Art, Dec 26 '16 at 17:10
Previously: Is there a general formula for the antiderivative of rational functions? (See Michael Hardy's answer for the case with only real coefficients.) — , Dec 26 '16 at 18:14
Related: https://math.stackexchange.com/questions/21305/algorithms-for-symbolic-manipulation — Watson, Dec 26 '16 at 19:30
I think your question is reasonably precise; some will downvote a question because it is beyond them. But you could make your words much more readable by adding some white space. Format your question into paragraphs or, in this context, if in doubt, just put a blank line after every full stop. It's nice to be able to breathe! — Borodin, Dec 26 '16 at 23:03
Possible duplicate of How can you prove that a function has no closed form integral? — Bill Dubuque, Dec 28 '16 at 03:41
@BillDubuque, as the asker, I think I am pretty clear in what I have asked, and that your given link is no where near it. — codetalker, Dec 28 '16 at 13:42
@Siddhant The refeences given in my answer in the linked question suffice to answer both of your questions (and, btw, one should ask only one question per post). — Bill Dubuque, Dec 28 '16 at 14:14

abnry · Accepted Answer · 2017-01-01T21:53:45.697

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Due to criticism in the comments, I am limiting the scope of and rewriting my answer. I still think I have something to offer even though I admit my answer is not complete.

The scope of the question itself is not entirely clear. There are possibly three questions: (1) How to integrate any function symbolically? (2) Does a "product rule" for integrals exist? (3) How do you integrate any rational function. I think (2) is a very interesting one.

I'll address (2) and (3) but (1) is outside of my knowledge. I am aware of related terms such as differential algebra and differential galois theory and the risch algorithm which may shed insight but, again, I am not knowledgeable in this.

For (2):

I suspect you are hoping for a formula such as the following. Let us take inspiration from the product rule for derivatives which has the form $$\frac{d}{dx}(fg)=\lambda(f,f',g,g')$$ with $\lambda$ an elementary function defined as $\lambda(w,x,y,z)=wz+xy$. Let $F$ and $G$ be the antiderivatives of $f$ and $g$ respectively. Taking the differentiation case, you are hoping for a formula $$\int fg dx = \lambda(f,F,g,G)$$ where $\lambda$ is some sort of function that's "nice". Implicit also is the "plus C". Specifically, I would say it is made of elementary functions or less precisely, the sort of functions you find in pre-calc. (I admit this does not cover all bases, however, the following will give at least some insight.)

Now take the case of $f=1/x$, $F=\log x$, $g=\sin x$ and $G = -\cos x$. We'd hope for some expression

$$\int \frac{\sin x}{x} dx = \lambda (1/x, \log x, \sin x, -\cos x).$$

However, it is well known that there is no closed form to the integral of the the so-called sinc function $\sin x/x$ in terms of elementary functions. But if $\lambda$ takes in elementary functions and combines them in an elementary way, the above expression would also have to be elementary. So no "nicely expressed", or elementary formula for an "integral product rule" exists.

This tells us at the least it would be a beast of a problem to find such a function. But in a way, this is backwards since the real work of showing $(\sin x)/x$ has no nice indefinite integral is itself a hard problem.

For (3):

Let me remind how you can deal with quadratic factors in partial fractions such as $$r(x) = \sum \frac{p_i(x)}{q_i(x)}$$ where $q_i(x)$ is degree 2 and has complex roots and $p_i$ is degree 1 at most. Now you can always complete the square and add and subtract in the numerator to get a form $$\int \frac{2c(x-a)+d}{(x-a)^2+b^2} dx = \int \frac{2c(x-a)}{(x-a)^2+b^2} + \frac{d}{(x-a)^2+b^2} dx.$$ The first term evaluates to $c\log((x-a)^2+x)$. The second to $d/b \arctan[(x-a)/b]$.

This is not a thorough explanation of how to compute the antiderivatives of rational functions but I want to make you aware of this point.

edited Jan 01 '17 at 21:53

answered Dec 26 '16 at 17:53

abnry

14,664

thnx for the partial fraction part, it is really helpful(+1). But, I doubt that a product formula doesn't exist. I think it may, but needs more math to work out. If you can maybe prove rigorously its non-existence, it would be awesome. – codetalker Dec 27 '16 at 08:38
As far as the example of a sinc function, I think that it is not wise enough to state its non-elementary form as being an argument against a product formula. Maybe that formula evaluates to something 'different' for the sinc function... – codetalker Dec 27 '16 at 08:39
@Siddhart, you need to make what you think a "product rule" for integrals would be precise. I took direct inspiration from the product rule for derivatives. That is, $D(ab)=\lambda(a,a',b,b')$ with $\lambda(w,x,y,z)=wz+xy$. So it is natural to suppose a product rule for integrals would be in a similar form as $I(ab)=\lambda(a,A,b,B)$, with possibly a different $\lambda$, where $A$ and $B$ are antiderivatives instead of derivatives. Turns out this is not possible based on my argument with the sinc function. – abnry Dec 27 '16 at 16:08
@Siddhant, are you trying to argue that there may be a non-elementary function $\lambda$ such that $I(ab)=\lambda(a,A,b,B)$? – abnry Dec 27 '16 at 16:10
It occurs to me you may be flipping the order of your quantifiers. The product rule for derivatives states that there exists a $\lambda(w,x,y,z)$ such that for all functions $a$ and $b$ we have $D(ab)=\lambda(a,a',b,b')$. You may be allowing the possibility in your mind for a product rule for integrals such that for all $a$, $b$, there exists a $\lambda(w,x,y,z)$ (that depends on $a$ and $b$ unlike before!) such that $I(ab)=\lambda(a,A,b,B).$ However, you can almost certainly always do this and without a $\lambda$ that works generally this isn't useful. – abnry Dec 27 '16 at 16:17
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There is no "Differential Galois Theory" used in integration algorithms. This comes into play only for higher-order differential equations. Nor is the product rule (or anything else above) related to the algorithms used in integration. – Bill Dubuque Dec 28 '16 at 03:01
Rosenlitch's paper Integration in Finite Terms which discusses the non-integrability of $\exp(x^2)$ in terms of elementary functions is precisely about differential galois theory. He uses all the language of Galois theory with the addition of a derivation. See paper here: http://www.dm.unipi.it/pages/gianni/public_html/Alg-Comp/rosenlicht.pdf – abnry Dec 28 '16 at 03:09
A proposed product rule for integration in terms of an elementary expression is not possible, as I have shown, because the sine integral is not expressibly in terms of elementary functions (as one example). The fact that the sine integral is not integrable in terms of elementary functions is related to Liouville's theorem, which is proven in Rosenlitch's paper. This is why I brought up differential galois theory. – abnry Dec 28 '16 at 03:12
Any integration algorithm must contend with the question of whether or not a particular function has an elementary integral. Whether or not a function has an elementary integral is a question of Differential Galois Theory. Indeed, Liouville's theorem is listed on Wikipedia as part of differential algebra. https://en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra) – abnry Dec 28 '16 at 03:14
First, you are confusing "differential algebra" with "differential Galois theory". The former is (slightly) used in integration algorithms. The latter is not. fyi: I implemented the latest version of indefinite integration in Macsyma, so I know this subject well. I'm sorry to inform you that almost everything that you wrote above is incorrect or misleading. I highly recommend that you delete the answer so as not to cause any further confusion. – Bill Dubuque Dec 28 '16 at 03:27

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

First of all, there a few pointers. Always check for $u$-substitution, for if that is possible, then by all means please use. Secondly, we have integration by parts, designed for integrating products. And also, we have partial fraction decomposition, as explained in this answer.

There also happens to be this really nice answer that gives the solution to some special cases, which can be modified via substitutions to provide solutions to many hard integrals.

Is indefinite integration largely a heuristic or it can be mechanical too?

2 Answers2