How much does symbolic integration mean to mathematics?

(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?


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”.

Source : Link , Question Author : Sangchul Lee , Answer Author : J. M. ain’t a mathematician

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