14
$\begingroup$

So it's known that there's no product rule and quotient rule for integration. But is there a reason why they don't exist, and the rules exist for differentiation?

$\endgroup$
  • 19
    $\begingroup$ It's integration by parts. $\endgroup$ – Randall Jan 16 '18 at 3:15
  • $\begingroup$ what do you mean by product rule? $\endgroup$ – Guy Fsone Jan 16 '18 at 16:54
  • $\begingroup$ Regarding the roles of both the product and quotient rules in integration, look at math.stackexchange.com/questions/1653814/… $\endgroup$ – David K Jan 17 '18 at 4:14
  • 1
    $\begingroup$ A small comment on the answers so far: if you write integration by parts as $\int uv'=uv-\int u'v$ then it doesn't really look like a product rule, but if you write $w=v'$ then it becomes $\int uw = u\int w - \int\left(u'\int w\right)$, which is indeed a formula for the integral of a product. $\endgroup$ – Nathaniel Jan 17 '18 at 7:30
  • $\begingroup$ On a similar to vein with what @Nathaniel said: Are you asking for the formula to evaluate $\int f(x)\,g(x)\,dx$ precisely? $\endgroup$ – Chase Ryan Taylor Jan 18 '18 at 0:17
23
$\begingroup$

The product rules is in fact an easy way to derive integration by parts! If we have $$ \frac{d}{dt}\bigl(f(t)\,g(t)\bigr)=f'(t)\,g(t)+f(t)\,g'(t) $$ then the FTC tells us that $$ \int_{a}^x\frac{d}{dt}\bigl(f(t)\,g(t)\bigr)\,\mathrm dt=\int_a^xf'(t)\,g(t)\,\mathrm dt+\int_{a}^xf(t)\,g'(t)\,\mathrm dt\\ \implies f(x)\,g(x)-\int_a^xf'(t)\,g(t)\,\mathrm dt=\int_{a}^xf(t)\,g'(t)\,\mathrm dt+C $$ where $C=f(a)\,g(a)$.

$\endgroup$
  • $\begingroup$ (+1) I hope you don’t mind the little aesthetic edits I made—I think they make the post a bit easier to read. $\endgroup$ – Chase Ryan Taylor Jan 16 '18 at 3:52
  • 1
    $\begingroup$ @ChaseRyanTaylor not a problem, looks great! $\endgroup$ – qbert Jan 16 '18 at 4:04
15
$\begingroup$

First and foremost, the integration rule that's most "analogous" to the product rule is integration by parts: $$ \int u \, dv = uv - \int v \, du $$ This formula can be established by starting with the product rule: $$ d(uv) = v \cdot du + u \cdot dv $$ Now integrate both sides and rearrange a bit. This leaves out some details, but at a high level that's what's going on.

In general, it's just the case that integration is a much more difficult problem than differentiation. You can take the derivative of any elementary function you can write down. But the same is not true of integration. For example, $e^{\sin x}$ is very easy to differentiate with the chain rule. But there is no formula to integrate it.

$\endgroup$
8
$\begingroup$

Edit: In light of Carmeister’s comment, I would like to clarify that (in my opinion) there does exist an integral analogue to the product rule, and that analogue is integration by parts.

Unfortunately you can’t comment an image, but I think this visual would supplement the above answers very nicely:

integration by parts

To see how this relates to the product rule, take qbert’s answer and change $f(t)$ and $g(t)$ with $u$ and $v$.

You can see how this works for definite and indefinite integrals by taking the standard integration by parts formula,

$$\int v \, du = uv - \int u\,dv$$

and evaluating it between the points $(v_1,u_1)$ and $(v_2,u_2$), giving you

$$\int_{u_1}^{u_2} v \, du = u_2v_2 - u_1v_1 -\int_{v_1}^{v_2} u\,dv$$

Side note: I hate doing integration by parts with $u$ and $v$ — they just look too similar! That’s why I prefer to use $w$ and $z$.

$\endgroup$
  • 4
    $\begingroup$ Actually it appears that someone downvoted every single answer, so maybe I shouldn’t be too concerned. $\endgroup$ – Chase Ryan Taylor Jan 16 '18 at 4:30
  • 4
    $\begingroup$ I'm not the downvoter, but it seems like the question is more asking about why there doesn't exist a formula for $\int uv$, which none of the answers address. $\endgroup$ – Carmeister Jan 16 '18 at 6:22
  • 3
    $\begingroup$ Someone downvoted the question too. Lol. $\endgroup$ – tilper Jan 16 '18 at 15:15
  • 1
    $\begingroup$ Well, integration by parts does not solve general integrals of products. It solves only very specific integrals of products. In the other cases, it converts one integral of a product to another one. From this point of view, you fail to answer the question. You claim that there actually is a product rule, but this can be considered misleading or wrong. $\endgroup$ – yo' Jan 17 '18 at 13:44
  • 1
    $\begingroup$ @yo', I never claimed there is a product rule for integration or that IBP is a general solution. I stand by my exact words, "...the integration rule that's most 'analogous' to the product rule is integration by parts." By that same logic I can say your answer is wrong because you said indefinite integration is the inverse of differentiation, which it isn't, because the $+C$ slightly ruins that. The truth is nobody here has answered the general question of "why." I'm not sure how answerable that is to any level of satisfaction. Integration is just harder than differentiation, but... why? $\endgroup$ – tilper Jan 18 '18 at 0:04
2
$\begingroup$

Integration doesn't follow any type of general rule in case of fractions.

However, for integration of product functions, we've the following breakdown -

Let there be two functions $u(x)$ and $v'(x)$ (further denoted by $u$ and $v$), such that $\displaystyle \int v'(x) \; dx = v(x)$,

Then by product rule, we've $$d(uv)=u'v+uv'$$

Integrating both the sides

$$uv=\int u'v+\int uv' $$

$$\implies \color{blue}{\int uv'=uv-\int u'v}$$

This is known as Integration by Parts

$\endgroup$
  • $\begingroup$ I feel its either typo or confused notation, when you write $\int v' = v$ or $d(uv) = u'v+uv'$. $\endgroup$ – samjoe Jan 16 '18 at 3:28
  • $\begingroup$ I meant the common notation is $d(uv) = u dv + v du$ (if you are referring to differentials) $\endgroup$ – samjoe Jan 16 '18 at 3:51
  • $\begingroup$ With respect to what variable are you integrating? I feel like specifying that would encourage more up-votes. $\endgroup$ – Chase Ryan Taylor Jan 16 '18 at 3:55
2
$\begingroup$

I'll try to approach this question in a bit more abstract way. I know I won't be 100% precise in my statements, but I think that this is the point.

First, note that indefinite integration is basically an inverse operation to the derivative, and this is basically the only definition we have. Yes, you can define indefinite integral as $F(x) = \int_a^x f(\xi) d\xi$ and use Riemann (or Lebesgue) integral, but these involve crazy formulas with strange limits of sums and stuff.

The fact that there's an easy formula for the derivative of a product is useless for the inverse derivative of a product. Similarly, note that computing a product of two functions is easy, but computing the inverse function of a product is tough.

Actually, product of invertible functions may not even be invertible! (Example: $e^x\sqrt{1/x}$.) Or, the inverse may not be expressible in elementary functions. (Example: $e^x\sqrt{x}$.) And this is in my opinion similar to the integrals, where a product of two functions with integrals expressible in elementary functions may not have such an integral. (The same two examples work.)

So somehow, this issue is related to the fact that when some operation works well on products, its inverse need not to, implicitly.

$\endgroup$
1
$\begingroup$

I agree with @yo', but I thought it was worth saying more. Because of the chain rule and the product rule, along with the derivative formulas for basic functions, derivatives in general can be calculated mechanically. Algebraically, integration is of course the inverse process, and sometimes inverses can be calculated just as mechanically as the forward operation, but not usually.

I think the idea of entropy is useful here, at least as a metaphor. There are a lot more ways for a formula to get more complicated (one might almost say disorderly, although it is driven by the underlying order of mathematics) than for it to get simpler. So most differentiation operations (especially the product rule) are going to make formulas more complicated. That means the inverse operation is unlikely to have a simple form.

In the specific case of the product rule, it's impossible for there to be a simple product rule for integration, because the product rule for derivatives goes from a product of two functions to a sum of two products. Reversing the operation requires you to carefully cancel the extra product of functions, and that's not always possible. But that "careful cancellation" is just the process of integration by parts that several other answers talked about.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.