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There is a well-defined algorithm (taught in calculus) which, given a function, gives you that function's derivative. Why is there no such thing for integration?


marked as duplicate by Jonas Meyer, user147263, Antonio Vargas, Ahaan S. Rungta, dustin Mar 29 '15 at 21:13

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  • $\begingroup$ Could you please explain your premise? What equation? Do you mean the equation $\displaystyle{f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}}$? Are you referring to definite integrals or indefinite integrals? $\endgroup$ – Jonas Meyer Aug 2 '11 at 0:50
  • $\begingroup$ Yes, that's the equation I mean. I don't know how to type out math like that yet. And either of them. $\endgroup$ – Peter Aug 2 '11 at 1:00
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    $\begingroup$ Differentiation is local; integration is global. That's why Riemann integrals have a more involved limit definition. $\endgroup$ – anon Aug 2 '11 at 1:07
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    $\begingroup$ What about an equation like $\displaystyle{\int_0^1 f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)}$? $\endgroup$ – Jonas Meyer Aug 2 '11 at 1:07
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    $\begingroup$ @Jonas, that's good for definite integrals. Of course, it leaves open the question for indefinite integrals (yes, I know you know this). $\endgroup$ – Gerry Myerson Aug 2 '11 at 2:57

All the functions built up, with the usual tools, from the standard functions have a derivative which is also built up from such functions: the usual differentiation formulas tell you that.

However, many simple functions, such as $e^{-x^2}$, do not have an antiderivative that can be expressed in terms of elementary functions.

However, there are algorithms which, with a few caveats, do the following:

(i) If you input an elementary function $f(x)$ which does not have an elementary antiderivative, the algorithm tells you that it doesn't.

(ii) If $f(x)$ has an elementary antiderivative, the algorithm gives you such an antiderivative.

Unfortunately, the algorithms (there are several) are extremely complicated, and as far as I know, none has ever been fully implemented. The first such algorithm was due to Risch. There are partial implementations of a symbolic integration algorithm in most computer algebra systems, like Maple and Mathematica.

For some detail, please see the Wikipedia article on Symbolic Integration.

Added: From the response of the OP to a question by @Jonas Meyer, the above answer does not seem to be what the OP is asking about. For the definite integral, and indeed for the indefinite integral, of a continuous function $f(x)$, there is in a sense an equation, for one antiderivative of $f(x)$ is given by $$\lim_{n\to \infty}\frac{1}{n}\sum_1^n xf(kx/n).$$ This is certainly not as pleasant as the "formula" that defines the derivative at $x=a$. Part of the reason for the difference is that differentiation is "local," the value of the derivative at $a$ depends only on the values of $f(x)$ very near $x=a$. Integration, for example definite integration, is "global," the value is influenced by all the values of the function in the interval, or at least on a suitably dense collection of the values.

The gap between integration and differentiation also diminishes if we work purely formally with series, or, no longer purely formally, with power series that converge suitably fast. For if $$f(x)=a_0+a_1x+a_2x^2+\cdots+ a_nx^n +\cdots,$$ then derivative and antiderivative are given by "formulas" that have very similar structures.


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