# 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 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. –  Guess who it is. 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 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 question is about symbolic derivation and integration. Nothing to do with the local/global behavior of the underlying functions. This answer pertains to numerical derivation and integration. –  Yves Daoust May 12 '14 at 14:37
@YvesDaoust This is an old answer, for one. For another, that's not true. The symbolic integrability/differentiability is absolutely dependent on local/global contexts. In fact, the definitions of those things depend on those local/global characteristics, and the symbolic results follow from those definitions. In particular, the Risch algorithm requires the identification of terms that are identically zero; this cannot be done locally. –  Arkamis Jun 4 '14 at 20:06
In fact, if we're going to just talk general abstract (nonsensical) symbol manipulation, then I declare the symbolic derivative of $x^2$ to be $4x$. You can't prove me wrong. –  Arkamis Jun 5 '14 at 21:08
@user161825: you also seem to ignore that the OP is explicitly about symbolic derivation/integration. –  Yves Daoust Oct 16 '14 at 13:07

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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It seems to be hard to find families of functions that are closed under integration. –  lhf Oct 12 '13 at 2:55

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

In the MIT lecture 6.001 "Structure and Interpretation of Computer Programs" by Sussman and Abelson this contrast is briefly discussed in terms of pattern matching. See the lecture video (at 3:56) or alternatively the transcript (p. 2 or see the quote below). The book used in the lecture does not provide further details.

Edit: Apparently, they discuss the Risch algorithm. It might be worthwhile to have a look at the same question on mathoverflow.SE: Why is differentiating mechanics and integration art?

And you know from calculus that it's easy to produce derivatives of arbitrary expressions. You also know from your elementary calculus that it's hard to produce integrals. Yet integrals and derivatives are opposites of each other. They're inverse operations. And they have the same rules. What is special about these rules that makes it possible for one to produce derivatives easily and integrals why it's so hard? Let's think about that very simply.

Look at these rules. Every one of these rules, when used in the direction for taking derivatives, which is in the direction of this arrow, the left side is matched against your expression, and the right side is the thing which is the derivative of that expression. The arrow is going that way. In each of these rules, the expressions on the right - hand side of the rule that are contained within derivatives are subexpressions, are proper subexpressions, of the expression on the left - hand side.

So here we see the derivative of the sum, with is the expression on the left - hand side is the sum of the derivatives of the pieces. So the rule of moving to the right are reduction rules. The problem becomes easier. I turn a big complicated problem it's lots of smaller problems and then combine the results, a perfect place for recursion to work.

If I'm going in the other direction like this, if I'm trying to produce integrals, well there are several problems you see here. First of all, if I try to integrate an expression like a sum, more than one rule matches. Here's one that matches. Here's one that matches. I don't know which one to take. And they may be different. I may get to explore different things. Also, the expressions become larger in that direction. And when the expressions become larger, then there's no guarantee that any particular path I choose will terminate, because we will only terminate by accidental cancellation. So that's why integrals are complicated searches and hard to do.

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I believe the aythors name is Sussman rather than Susskind. –  Spencer Apr 14 at 0:03
@spencer Haha, you are right. Thanks for the correction. –  mcb Apr 15 at 9:07

I will try to bring this to you in another way .Let us start by thinking in terms of something as simple as a straight line . If I give you the equation of a line y = mx + c , it's slope can be easily determined which in this case is nothing but m . .Now let me make the question a bit trickier .Let me say that the line given above intersects the x and y axis at some points .I ask you to give me the area between the line,the abcissa and the ordinate

This is obviously not as easy as finding the slope .You shall have to find the intersection of the line with the axis and get two points of intersection and then taking the origin as a third point find the area . This is not the only method of finding the area as we know there are loads of formulas for finding the area of a triangle . Let us now view this in terms of curves .If the simple process of finding the slope in case of a line is translated to curves we get differential calculus which is a bit more complicated than the method of finding slopes of straight lines .

Add finding the area under the curve to that and you get integral calculus which by our experience from straight lines we know should be much harder than finding the slope ie differentiation .Also there is no one fixed method for finding the area of a figure .hence the Many methods of. Integration.

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Reading this is unbearable...1/ Please space up by putting a double return to the line for new paragraphs. 2/ The full stop "." has a space after and not before. 3/ Sentences start with capital letters. –  user88595 Jun 1 '14 at 10:19
I apologize for the errors I have caused .I typed the entire thing out on my phone and was in a hurry.I will make the required changes asap . –  dhruv Jun 4 '14 at 19:32