# Why is integration so much harder than differentiation?

If a function is a combination of other functions whose derivatives are known via composition, addition, etc., the derivative can be calculated using the chain rule and the like. But even the product of integrals can't be expressed in general in terms of the integral of the products, and forget about composition! Why is this?

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@Patrick: This doesn't exactly answer your question but is on related lines. math.arizona.edu/~mleslie/files/integrationtalk.pdf – user17762 Feb 5 '11 at 20:42
This question reminded me of a tangentially related post on MathOverflow: "... on formulas differentiation is nice and integration is hard, but on computable functions differentiation is hard and integration is nice." -- Jacques Carette – Rahul Narain Feb 5 '11 at 21:34
@sivaram: that's an interesting slideshow! The last slide is hilarious. – Myself Feb 6 '11 at 2:15
I think most analysts out there would say integration is much easier than differentiation...but of course they have a different thing in mind than your question. – Matt Feb 6 '11 at 2:39
To rephrase Jacques's quote: differentiation is symbolically easy but numerically hard, while integration is numerically easy but symbolically hard. – J. M. Apr 20 '11 at 11:03

Here is an extremely generic answer. Differentiation is a "local" operation: to compute the derivative of a function at a point you only have to know how it behaves in a neighborhood of that point. But integration is a "global" operation: to compute the definite integral of a function in an interval you have to know how it behaves on the entire interval (and to compute the indefinite integral you have to know how it behaves on all intervals). That is a lot of information to summarize. Generally, local things are generally much easier than global things.

On the other hand, if you can do the global things, they tend to be useful because of how much information goes into them. That's why theorems like the fundamental theorem of calculus, the full form of Stokes' theorem, and the main theorems of complex analysis are so powerful: they let us calculate global things in terms of slightly less global things.

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Patrick asked about the integration of of "function terms", i.e. finite expressions. In your answer you omitted to say that such an expression is a global object to begin with. – Christian Blatter Feb 6 '11 at 13:26

The family of functions you generally consider (e.g., elementary functions) is closed under differentiation, that is, the derivative of such function is still in the family. However, the family is not in general closed under integration. For instance, even the family of rational functions is not closed under integration because you $\int 1/x = \log$.

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Answering an old question just because I saw it on the main page. From Roger Penrose (Road To Reality):

... there is a striking contrast between the operations of differentiation and integration, in this calculus, with regard to which is the ‘easy’ one and which is the ‘difficult’ one. When it is a matter of applying the operations to explicit formulae involving known functions, it is differentiation which is ‘easy’ and integration ‘difficult’, and in many cases the latter may not be possible to carry out at all in an explicit way. On the other hand, when functions are not given in terms of formulae, but are provided in the form of tabulated lists of numerical data, then it is integration which is ‘easy’ and differentiation ‘difficult’, and the latter may not, strictly speaking, be possible at all in the ordinary way. Numerical techniques are generally concerned with approximations, but there is also a close analogue of this aspect of things in the exact theory, and again it is integration which can be performed in circumstances where differentiation cannot.

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Just for the record: Penrose discusses these matters on p. 103-120. Yet he gives no good reason why this is so (esp. for the symbolic case). – vonjd May 31 '11 at 9:22

Think about what the two represent. If you have some function (it need not be too complicated) such as $f(x) = x^2 \sin(x)$, what does $\frac{df}{dx}(3)$ and $\int_0^3 f(s) ds$ mean?

Suppose you don't know any theory, and you are given the graph of the function, which will be easier to find?

Then it's not surprising that you can calculate the derivative of complicated functions whereas the integral might not even have a closed form.

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