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(Before reading, I apologize for my poor English ability.)

I have enjoyed calculating some symbolic integrals as a hobby, and this has been one of the main source of my interest towards the vast world of mathematics. For instance, the integral below $$ \int_0^{\frac{\pi}{2}} \arctan (1 - \sin^2 x \; \cos^2 x) \,\mathrm dx = \pi \left( \frac{\pi}{4} - \arctan \sqrt{\frac{\sqrt{2} - 1}{2}} \right). $$ is what I succeeded in calculating today.

But recently, as I learn advanced fields, it seems to me that symbolic integration is of no use for most fields in mathematics. For example, in analysis where the integration first stems from, now people seem to be interested only in performing numerical integration. One integrates in order to find an evolution of a compact hypersurface governed by mean curvature flow, to calculate a probabilistic outcome described by Ito integral, or something like that. Then numerical calculation will be quite adequate for those problems. But it seems that few people are interested in finding an exact value for a symbolic integral.

So this is my question: Is it true that problems related to symbolic integration have lost their attraction nowadays? Is there no such field that seriously deals with symbolic calculation (including integration, summation) anymore?

RegDwight
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Sangchul Lee
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    Manipulationally, an analytic (closed) form would be terribly convenient, in the sense that we know so much about (special) functions and the identities they satisfy that we would like to be able to exploit that whole body of knowledge for the integral at hand. For numerical work, a closed form may or may not be the best thing to have, depending on the circumstances. – J. M. ain't a mathematician May 11 '11 at 15:29
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    If you broaden the question, allowing multiparameter (in)definite integrals and sums, then the question is not about only (transcendental) constants but, more generally, about the utility of closed form functions solving differential / difference equations, i.e. special functions. Do you intend to ask only about special constants or, more generally, special functions? – Bill Dubuque May 11 '11 at 15:52
  • @J.M. : Thanks for your comment. But if you won't mind, can you be a little more specific? – Sangchul Lee May 11 '11 at 15:56
  • @Bill Dubuque : Oh, that's my point. I am also interested in special functions. – Sangchul Lee May 11 '11 at 15:57
  • As I said, if you can express some integral in terms of previously known functions/quantities, that is a good thing in the sense that you can apply the properties derived for those previously known functions/quantities for your particular integral. But for numerical purposes, sometimes numerical integration is more efficient (and it certainly requires less thought if somebody already wrote a quadrature routine for you), and sometimes knowing the properties of special functions allows you to compute more efficiently than merely performing quadrature. – J. M. ain't a mathematician May 11 '11 at 16:01
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    Since you find symbolic integration interesting, you may find useful the references I mention in this answer: http://math.stackexchange.com/questions/37088/integration-doubt/37093#37093 – Andrés E. Caicedo May 12 '11 at 16:39

4 Answers4

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I think it would be appropriate at this point to quote Forman Acton:

...at a more difficult but less pernicious level we have the inefficiencies engendered by exact analytic integrations where a sensible approximation would give a simpler and more effective algorithm. Thus

$$\begin{align*}\int_0^{0.3}\sin^8\theta\,\mathrm d\theta&=\left[\left(-\frac18\cos\,\theta\right)\left(\sin^4\theta+\frac76\sin^2\theta+\frac{35}{24}\right)\sin^3\theta+\frac{105}{384}\left(\theta-\sin\,2\theta\right)\right]_0^{0.3}\\ &=(-0.119417)(0.007627+0.101887+1.458333)(0.0258085)+0.004341\\ &=-0.0048320+0.0048341=0.0000021\end{align*}$$

manages to compute a very small result as the difference between two much larger numbers. The crudest approximation for $\sin\,\theta$ will give

$$\int_0^{0.3}\theta^8\,\mathrm d\theta=\frac19\left[\theta^9\right]_0^{0.3}=0.00000219$$

with considerably more potential accuracy and much less trouble. If several more figures are needed, a second term of the series may be kept.

In a similar vein, if not too many figures are required, the quadrature

$$\int_{0.45}^{0.55}\frac{\mathrm dx}{1+x^2}=\left[\tan^{-1}x\right]_{0.45}^{0.55}=0.502843-0.422854=0.079989\approx 0.0800$$

causes the computer to spend a lot of time evaluating two arctangents to get a result that would have been more expediently calculated as the product of the range of integration ($0.1$) by the value of the integrand at the midpoint ($0.8$). The expenditure of times for the two calculations is roughly ten to one. For more accurate quadrature, Simpson's rule would still be more efficient than the arctangent evaluations, nor would it lose a significant figure by subtraction. The student that worships at the altars of Classical Mathematics should really be warned that his rites frequently have quite oblique connections with the external world.

It may very well be that choosing the closed form approach would still end up with you having to (implicitly) perform a quadrature anyway; for instance, one efficient method for numerically evaluating the zeroth-order Bessel function of the first kind $J_0(x)$ uses the trapezoidal rule!

On the other hand, there are also situations where the closed form might be better for computational purposes. The usual examples are the complete elliptic integrals $K(m)$ and $E(m)$; both are more efficiently computed via the arithmetic-geometric mean than by using a numerical quadrature method.

But, as I said in the comments, for manipulational work, possessing a closed form for your integral is powerful stuff; there is a whole body of results that are now conveniently at your disposal once you have a closed form at hand. Think of it as "standing on the shoulders of giants".

In short, again, "it depends on the situation and the terrain".

J. M. ain't a mathematician
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    For the original problem, things are similar. After realizing that $1-\sin^2 x, \cos^2 x = [7+\cos(4x)]/8$, i.e., $7/8$ plus something oscillating, one is tempted to replace the integrand by $\arctan(7/8)$. Then one gets $\pi \arctan(7/8)/2 \approx 1.129$ for the integral which coincides with the exact expression $1.126$ up to $3\times 10^{-3}$. I don't know how much time the OP spend for his answer, but Pareto's principle seems to apply. – Fabian May 11 '11 at 20:36
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    Thanks to everyone, especially J.M., for giving insightful answers. It's really convincing that distinction between numerical one and symbolic one is subject not to a particular classification of research area, but rather to a situation. – Sangchul Lee May 12 '11 at 10:24
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    @sos440: You're very much welcome. :) Don't let my answer deter you from the fun you seem to have in teasing out analytic solutions; what I'm merely saying is that in real-world applications, one must eventually develop a "feel" for choosing the "right tool for the job". – J. M. ain't a mathematician May 12 '11 at 10:28
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I don't think your point of view is the right one. To compute an integral analytically and to compute an integral numerically are different things. A numerical analysis professor of mine once said that, in applications (engineering, physics...) it is often more convenient to directly evaluate integrals by numerical means, even if they are integrable analytically! For example, suppose that you need

$$\int_{0}^{\frac{\pi}{2}} \arctan (1 - \sin^2 x \; \cos^2 x) \,\mathrm dx$$

meters of conducting wire. You make a phone call to the wire factory and ask for what? For $\pi \left( \frac{\pi}{4} - \arctan \sqrt{\frac{\sqrt{2} - 1}{2}} \right)$ meters of wire? More realistically you will ask for something like $1.13$ meters of wire.

To obtain this number $1.13$ you performed an approximation over the non-rational quantity $\pi \left( \frac{\pi}{4} - \arctan \sqrt{\frac{\sqrt{2} - 1}{2}} \right)$. In doing so you wasted information. It would have been more convenient (and, maybe, even more accurate) to perform this approximation on the first integral directly, that is, to evaluate it numerically.

Of course this does not render analytical methods useless. You could have a family of integrals depending on a parameter, for example. Numerical methods tell you nothing here. You could run across an integral in the middle of a proof, and need its exact value for theoretical purposes. The possibilities are countless.

J. M. ain't a mathematician
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Giuseppe Negro
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  • Also, if you don't need that much accuracy (which is often in applications), even the simple-minded methods like trapezoidal or Simpson's might end up being faster to compute with than having to use a special routine for some exotic transcendental. – J. M. ain't a mathematician May 11 '11 at 16:04
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    In short, if I may borrow the usual piece of military wisdom: "it depends on the situation and the terrain". – J. M. ain't a mathematician May 11 '11 at 16:12
  • Surely you mean something like 1.126 instead of 1.49... Also I'm curious if it is sure or just very likely that this number is not rational, as you state. – Myself May 12 '11 at 02:58
  • @Myself: Oh yes, sorry about that, I must have done some mistake in typing that number into Maple. Regarding the irrationality of $\pi \left( \frac{\pi}{4} - \arctan \sqrt{\frac{\sqrt{2} - 1}{2}} \right)$, I haven't checked it. Looks like a safe bet, though... Don't you agree? – Giuseppe Negro May 12 '11 at 17:26
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    Yes, I completely agree. But for all I know these things are typically hard to prove, so I thought perhaps I had missed something. – Myself May 12 '11 at 17:33
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If you're talking about practical engineering applications, then really only numerical approximations are used (and studied in computer science as 'numerical analysis' or more recently 'scientific computing').

As to an academic mathematical field nowadays that deals with symbolic integration, first some perspective. Newton/Leibniz invented integral calculus in ...hm...late 1600's and was popularized (as much as you can say that) in the 1700's. Some basic symbolic integration even occurred (without that name and system) before then. So let's just say there's been at least 300 years of work there.

Also, there's more to inverting derivatives than just integrals. Solving systems of partial differential equations seem to be the big thing (both numerically and symbolically) for almost as long as simple single variable integrals.

That said, there is a small academic group of people working in 'symbolic computation' (with their own journals), and one subarea is symbolic integration. There are proofs of impossibility (i.e. proving that given certain restrictions there is no 'closed form' for a particular integral), and there are algorithms for computing integrals given other certain restrictions (the Risch algorithm). The latter are often implemented in computer algebra packages (Mathematica, Maple, etc.).

There is surely room for solving particular integrals (in the AMM there don't seem to be many integrals though in the Problems section) and for finding patterns. I'd look at those journals to see what particular interest there is for integrals.

Mitch
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    Back in the day, SIAM Reviews used to maintain a "problems column" where people submitted various requests for simple proofs, and yes, integrals/sums/whatever to evaluate. With the computing environments now, I suppose there is now less reason to submit those sorts of problems. – J. M. ain't a mathematician May 11 '11 at 17:25
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    @J.M.: It's been a longstanding (15 years?) challenge by Doron Zeilberger that he (and his computer) can solve automatically any summation sent in to the AMM Problems section. Surely there are summations (and integrals) that are people-solvable and currently not computer-solvable, but as the technology progresses, these will be harder for people to come by. – Mitch May 11 '11 at 17:44
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    I would also note that it may very well be that the symbolic output of current computing environments may be less than optimal, and some further human massaging may be needed. For instance, Mathematica sucks at producing optimal elliptic integral expressions... – J. M. ain't a mathematician May 12 '11 at 01:24
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Symbolic integration becomes less popular indeed and most researchers prefer numerical integration. However, significance of the symbolic integration should not be underestimated. This can be shown by using the integration formula: $$\int_0^1 f(x)dx = 2\sum_{m=1}^M{\sum_{n=0}^\infty{\frac{1}{{\left(2M\right)}^{2n+1}\left({2n+1}\right)!}\left.f^{(2n)}\left(x\right)\right|_{x=\frac{m-1/2}{M}}}}\,\,,$$ where the notation $f(x)^{(2n)}|_{x=\frac{m-1/2}{M}}$ implies 2n-th derivative at the points $x=\frac{m-1/2}{M}$. Once the integral is expanded as a series, we can use it either numerically or analytically (i.e. symbolically). It may be tedious to find by hand 2n-th derivatives. However, with powerful packages supporting symbolic programming like Maple, Mathematica or MATLAB this can be done easily. For example, by taking $f(x) = \frac{\theta}{1 + \theta^2 x^2}$ even at smallest $M = 1$ we can find a rapidly convergent series for the arctangent function: $$\tan^{-1}(\theta)=i\sum_{n=1}^{\infty}\frac{1}{2n-1}\left(\frac{1}{\left(1+2i/\theta\right)^{2n-1}}-\frac{1}{\left(1-2i/\theta\right)^{2n-1}}\right),$$ where $i=\sqrt{-1}$. This example shows that symbolic integration may be highly efficient in many numerical applications.

JWalter
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