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Why is integration so much harder than differentiation?
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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?
calculus<br>integration
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edited Aug 9, 2020 at 20:12
Rodrigo de Azevedo
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asked Feb 5, 2011 at 20:38
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$\begingroup$<br>@Patrick: This doesn't exactly answer your question but is on related lines. math.arizona.edu/~mleslie/files/integrationtalk.pdf<br>$\endgroup$
user17762
user17762
2011-02-05 20:42:17 +00:00
Commented<br>Feb 5, 2011 at 20:42
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$\begingroup$<br>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<br>$\endgroup$
user856
user856
2011-02-05 21:34:27 +00:00
Commented<br>Feb 5, 2011 at 21:34
$\begingroup$<br>@sivaram: that's an interesting slideshow! The last slide is hilarious.<br>$\endgroup$
Myself
Myself
2011-02-06 02:15:14 +00:00
Commented<br>Feb 6, 2011 at 2:15
$\begingroup$<br>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.<br>$\endgroup$
Matt
Matt
2011-02-06 02:39:19 +00:00
Commented<br>Feb 6, 2011 at 2:39
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$\begingroup$<br>To rephrase Jacques's quote: differentiation is symbolically easy but numerically hard, while integration is numerically easy but symbolically hard.<br>$\endgroup$
J. M. ain't a mathematician
J. M. ain't a mathematician
2011-04-20 11:03:50 +00:00
Commented<br>Apr 20, 2011 at 11:03
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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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edited Jul 30, 2014 at 12:59
user166959
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answered Feb 6, 2011 at 1:28
Qiaochu Yuan
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$\begingroup$<br>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.<br>$\endgroup$
Christian Blatter
Christian Blatter
2011-02-06 13:26:13 +00:00
Commented<br>Feb 6, 2011 at 13:26
$\begingroup$<br>@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.<br>$\endgroup$
Emily
Emily
2014-06-04 20:06:47 +00:00
Commented<br>Jun 4, 2014 at 20:06
$\begingroup$<br>True, without an underlying notion of an integral/derivative, saying $d/dx\ x^2 = 2x$ is nothing more than arbitrary symbol mucking. But we do have an underlying sense of what is and what isn't a function, a derivative, and an integral, so those things do matter. Symbolic manipulations are mathematically meaningless when you elide the mathematical context, so trying to make other mathematical arguments about them is meaningless as well.<br>$\endgroup$
Emily
Emily
2014-06-05 21:07:44 +00:00
Commented<br>Jun 5, 2014 at 21:07
$\begingroup$<br>In...