# Math: Intergals harder than derivatives, why? **URL:** \ **Category:** Factual Questions\ **Created:** [March 31, 2003, 3:37am UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356 "2003-03-31T03:37:16Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [March 31, 2003, 4:47pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/21 "2003-03-31T16:47:03Z")
> [@](#): > > \*Originally posted by RyanD004 \* > \*\*Heres the thing. When you take the antiderivitive, you’re almost trying to ‘undo’ a previous process. The problem is that you might not be able to clearly see the process that was used. > > For instance. > If we have a function such that: y = xe^x > > Then the derivitive is: y’= e^x + xe^x > If you think about taking the integral of y’ you’ll probably be considering two integrals- the first one is the easy e^x, and the second could be an integration by parts. It’s not obvious that the function was the derivitive of xe^x. True- that relationship is still there- the information is there- but it’s harder to infer. Now there could be any number of possible equivilant answers. If you do it the way I suggested, you will get the original xe^x +c. \*\* I don’t tend to think of differentiation or integration as a process. For my purposes, it’s generally enough to know that the derivative or integral of a function exists. And in that sense, all the information about the integral and all the information about the derivative are contained in the original function. \*\* > [@](#): > > The second part of your argument makes no sense to me. Saying that integration is harder because it’s more complicated doesn’t answer the question- why is it more complicated?\*\* If you know what a limit is, the definition of a derivative can be given in half a line. The definition of an integral takes a paragraph or so. That’s all I mean by “more complicated”. \*\* > [@](#): > > As for the whole “global” vs “local” argument- I thought about it, but I have one gripe. The slope of the fuction is determined by the other points around it. If you have one point in space,it doesn’t have a slope. There has to be a change in something- which can only exist relative to something else. Just a thought. IANA Mathematician. \*\* Yes, and if you change a function very slightly on a very small interval around a point, the derivative may change quite a bit. Besides, as **Achernar** pointed out, “global” and “local” don’t mean anything precise. Integrals can be carried out over very small intervals. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [March 31, 2003, 5:33pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/22 "2003-03-31T17:33:21Z")
> [@](#): > > A function f is related to its antiderivative in exactly the same way that it’s related to its antiderivative, in the sense that f contains enough information to uniquely determine either related function. I assume one of those "antiderivative"s was supposed to be “derivative”. With an antiderivative, you’ve still got that constant of integration which is _not_ determined by the function. It’s true that you can get _an_ antiderivative locally, and set the constant of integration to whatever you like, but there’s no guarantee that the constant of integration you use at one place will match the one for an antiderivative you get locally elsewhere. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 31, 2003, 5:51pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/23 "2003-03-31T17:51:20Z")
Quoth **Orbifold** : > [@](#): > > Functions which can’t be integrated at all are in a sense unusual; for example, most non-mathematicians would be hard-pressed to think of a non-integrable function. Care to give us an example? Other than divergence to infinity (int(1/x, x=0…1 , for example), I can’t think of any non-integrable functions. And I’m usually pretty good at finding pathological counterexamples, too. ---
**Author:** ![Orbifold](https://avatars.discourse-cdn.com/v4/letter/o/779978/32.png) [@Orbifold](https://boards.straightdope.com/u/Orbifold)\ **Post date:** [March 31, 2003, 7:05pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/24 "2003-03-31T19:05:45Z")
Let f(x) be a function defined on the interval [0,1], such that f(x)=0 if x is irrational and f(x)=1 if x is rational. Such a function is not Riemann integrable: the “upper Riemann sums” are always \>= 1, while the “lower Riemann sums” are always \<=0, so the Riemann sums never converge. The function f(x) is, in fact, _Lebesgue_ integrable; it’s Lebesgue integral is 0. There are functions which aren’t even Lebesgue integrable, but they’re harder to construct. The short version is, let g(x) be a function such that g(x)=1 if x lies in a given set A, and g(x)=0 otherwise, where A is a non-measurable subset of the real line. So you need a “non-measurable subset”. I _think_ the following is non-measurable: define the equivalence relation x~y if x-y is rational, and let A be a subset of [0,1] containing exactly one element from each equivalence class of ~. I’d have to spend some time going over my old analysis textbooks to verify that, however. (I’d also have to go back to my textbooks to even define “measurable” properly, to be honest.) ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [March 31, 2003, 8:12pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/25 "2003-03-31T20:12:42Z")
That’s the standard example of a non-measurable set. I was told that it requires the axiom of choice, but now I’m not so sure. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 31, 2003, 8:13pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/26 "2003-03-31T20:13:43Z")
I guess I was considering functions which are “physicist integrable”, because I was considering the function in your first example to integrate to zero (a physicist function or operation is one which has whichever unspecified properties it needs in order to make the problem interesting, well-defined, and/or doable :)) Similarly, you don’t need to dig up the official mathematician’s definition of “measure”, since I, as a physicist, know exactly what that means, without needing a definition. But y’know, I think that your second example works, so long as you restrict yourself to positive numbers, and you allow the axiom of choice (I think that the equivalence classes might not be well-defined if you allow negative numbers, and you need to choose one element from each class). Rather an interesting set, I must say. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [March 31, 2003, 8:15pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/27 "2003-03-31T20:15:20Z")
I wrote: > [@](#): > > there’s no guarantee that the constant of integration you use at one place will match the one for an antiderivative you get locally elsewhere As an example, which I actually ran into once, I had some nasty integral involving Bessel functions. I could get expansions for the integrand for the limits X --\> 0 and also for X --\> infinity, and could thus find expressions for the integral in both limits. I wanted to say that the integral was zero (IIRC), but I had no way of showing that the constant of integration was the same in the two cases. Looking at a simple case, let f(X) = -J[sub]1/sub, and use the approximation f(x) = -X/2 for X small, and use f(x) = -J[sub]1/sub for X large. Now try to get the integral from 0 to infinity of f(x). Both integrations are easily performed “locally”: F(t) = integral{-X/2} = -t[sup]2[/sup]/4 (t small) F(t) = integral{-J[sub]1/sub} = J[sub]0/sub (t large) So (naively) integral{[from 0 to infinity] f(X)} = F(infinity) - F(0) = 0 - 0 = 0. But that’s wrong; it’s known that the integral{[from 0 to infinity] -J[sub]1/sub} = -1. ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [March 31, 2003, 8:21pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/28 "2003-03-31T20:21:52Z")
It requires the axiom of choice because you’re constructing a set by _choosing_ an element from each equivalence class. ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [March 31, 2003, 8:25pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/29 "2003-03-31T20:25:15Z")
(My post, by the way, was referring to **Orbifold** ’s example of a non-measurable set.) ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [March 31, 2003, 8:25pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/30 "2003-03-31T20:25:31Z")
> [@](#): > > \*Originally posted by Cabbage \* > \*\*It requires the axiom of choice because you’re constructing a set by _choosing_ an element from each equivalence class. \*\* You’re right. I was thinking that each equivalence class had a least member under the standard ordering. :smack: ---
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [March 31, 2003, 9:50pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/31 "2003-03-31T21:50:52Z")
> [@](#): > > \*Originally posted by RM Mentock \* > … finding the cube of a number is a lot easier (it’s just multiplication) than finding the cube root. Although some inverse functions may seem just as easy (I dunno, division always seemed harder to me), the inverse problem in mathmatics is notorious. Actually, multiplication, division and constant roots all take the _same_ amount of time computationally. In fact, they are _equivalent_ problems. That is, a better bound (upper or lower) on one applies to the rest automatically. This is discussed (or given as homework questions) in Aho, Hopcroft and Ullman’s _Algorithm_ book and no doubt others. The standard techniques for such problems given in grade school are not very efficient. Current methods for very large numbers use FFT based algorithms. For some strange reason, FFT is not taught in the 3rd grade. Some problems in modular arithmetic haven’t been mentioned. Finding squares or doing exponentiation in modular fields is easy. Doing roots or logs appears to be hard. Proving that they are in fact hard would make one instantly world famous overnight. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [March 31, 2003, 10:25pm UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/32 "2003-03-31T22:25:17Z")
> [@](#): > > Actually, multiplication, division and constant roots all take the same amount of time computationally. In fact, they are equivalent problems. The analogy I first thought of was multiplying and finding roots of polynomials. Using the quadratic formula to find the roots of a quadratic equation is more complicated than multiplying linear terms, and it rapidly [goes](http://mathworld.wolfram.com/CubicEquation.html) [downhill](http://mathworld.wolfram.com/QuarticEquation.html) [from](http://mathworld.wolfram.com/QuinticEquation.html) [there](http://mathworld.wolfram.com/SexticEquation.html). > [@](#): > > For some strange reason, FFT is not taught in the 3rd grade. Just as well, they’d just teach them the DFT anyway. 😉 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 1, 2003, 12:23am UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/33 "2003-04-01T00:23:05Z")
> [@](#): > > \*Originally posted by ZenBeam \* > \*\*Looking at a simple case, let f(X) = -J[sub]1/sub, and use the approximation f(x) = -X/2 for X small, and use f(x) = -J[sub]1/sub for X large. Now try to get the integral from 0 to infinity of f(x). Both integrations are easily performed “locally”: > > F(t) = integral{-X/2} = -t[sup]2[/sup]/4 (t small) > F(t) = integral{-J[sub]1/sub} = J[sub]0/sub (t large) > > So (naively) integral{[from 0 to infinity] f(X)} = F(infinity) - F(0) = 0 - 0 = 0. But that’s wrong; it’s known that the integral{[from 0 to infinity] -J[sub]1/sub} = -1. \*\* All that that shows is that you have to be careful when you use approximations. Define f ~ g iff f(x) - g(x) = c, a constant, for all x. ~ is an equivalence relation, so that’s why I feel comfortable saying that a function uniquely determines its antiderivative. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [April 1, 2003, 2:44am UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/34 "2003-04-01T02:44:06Z")
> [@](#): > > All that that shows is that you have to be careful when you use approximations. So use the whole series for X and t small. > [@](#): > > Define f ~ g iff f(x) - g(x) = c, a constant, for all x. ~ is an equivalence relation, so that’s why I feel comfortable saying that a function uniquely determines its antiderivative. For any x, I can set the value of the antiderivative to any value I want, confident that one function satisfying your equivalence relation will match that value. That’s an interesting definition of “uniquely” you must be using. ---
**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\ **Post date:** [April 1, 2003, 3:15am UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/35 "2003-04-01T03:15:19Z")
> [@](#): > > You’re right. I was thinking that each equivalence class had a least member under the standard ordering. They do…provided that you assume the axiom of choice ;). ---
**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\ **Post date:** [April 1, 2003, 3:16am UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/36 "2003-04-01T03:16:39Z")
> [@](#): > > \*Originally posted by Tyrrell McAllister \* > \*\*They do…provided that you assume the axiom of choice ;). \*\* D’oh; not with respect to the standard ordering, but with respect to _some_ ordering. ---
**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\ **Post date:** [April 1, 2003, 3:55am UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/37 "2003-04-01T03:55:33Z")
> [@](#): > > \*Originally posted by Achernar \* > Your concepts of global and local are not well-defined enough for you to make this assertion. Ok, fair enough; I have not provided a precise definition of what I mean by “local” and “global”, but I took the following definitions to be understood. Given a map [symbol]F[/symbol] that maps a set of functions defined on some domain _X_ into another set of functions defined on _X_, I say that [symbol]F[/symbol] is **local** iff, for each point _x_ in _X_ and every neighborhood _N_ of _x_, if functions _f_ and _g_ in the domain of [symbol]F[/symbol] agree on _N_, then [symbol]F/symbol and [symbol]F/symbol agree at _x_. I say that [symbol]F[/symbol] is **global** iff [symbol]F[/symbol] is not local. By this definition, if [symbol]F[/symbol] returns the derivative of continuous functions, then it is local; whereas if [symbol]F[/symbol] returns an antiderivative of continuous functions, then it is global. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [April 1, 2003, 4:12am UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/38 "2003-04-01T04:12:43Z")
That may work. But I’m slightly confused on (at least) one thing. Isn’t it true that the antiderivative of a function is not itself a function, because of the constant of integration? Isn’t it instead a set of functions? If so, then wouldn’t antiderivative be an invalid value for F, given its definition? ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 1, 2003, 4:51am UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/39 "2003-04-01T04:51:46Z")
> [@](#): > > \*Originally posted by ZenBeam \* > \*\*For any x, I can set the value of the antiderivative to any value I want, confident that one function satisfying your equivalence relation will match that value. That’s an interesting definition of “uniquely” you must be using. \*\* You’re an engineer or a scientist, right? When I think of a function, I’m not thinking of it as a collection of values–honestly, I’d never though of evaluation until you brought it up–but rather as a point in some space. Turning a space into a “smaller” space whose points are equivalence classes of points in the original space is not unusual. I should say that given a function f, the set of antiderivatives of f, and the set of derivatives of f are both uniquely determined. The fact that one’s a singleton and the other uncountably infinite is of little import. 😃 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 1, 2003, 4:54am UTC](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356/40 "2003-04-01T04:54:31Z")
> [@](#): > > \*Originally posted by Achernar \* > \*\*That may work. But I’m slightly confused on (at least) one thing. Isn’t it true that the antiderivative of a function is not itself a function, because of the constant of integration? Isn’t it instead a set of functions? If so, then wouldn’t antiderivative be an invalid value for F, given its definition? \*\* If you make the range of [symbol]F[/symbol] into the powerset of X -\> A (whatever the range of the original class of functions is), you can get around that. [Previous page](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356.md?page=1) [Next page](https://boards.straightdope.com/t/math-intergals-harder-than-derivatives-why/165356.md?page=3)