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Computing antiderivatives is more challenging than computing derivatives, in part due to the lack of a ``product formula''; namely, while $(fg)'$ can be expressed in terms of $f,f',g,g'$, there seems to be no way to express $\int fg$ in terms of $f, \int f, g \int g$ and related quantities. Is there any intuitive reason or heuristic explanation for why no such formula exists? I'm looking for a non-rigorous explanation which can be understood by first-year calculus students.

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See also… . – Qiaochu Yuan Jul 10 '11 at 18:52
Maybe use the notion of "elementary function". There are elementary functions $A,B$ such that both $\int A$ and $\int B$ are elementary, but $\int(AB)$ is not elementary. So there can be no simple "product rule" for integration, since any simple combinations of elementary functions is again elementary. – GEdgar Jul 10 '11 at 20:26
Because there isn't a good product formula for derivatives? – André Nicolas Jul 10 '11 at 20:30
@GEdgar - thats an excellent suggestion. Because I'm looking for something I can explain to a first-year calculus student, should I go in this direction, I would need an elementary function and a non-rigorous explanation, understandable by such a student, for why its antiderivative is not elementary. Does something like that exist at this level? – robinson Jul 10 '11 at 20:38
@user6312 - I suppose I neglected to say what, exactly, I mean by "good." I would be happy with any expression for $\int fg$ in terms of $\int f, f, f', f'', \ldots$ and $\int g, g, g', g'', \ldots$, since such an expression would turn the computation of antiderivatives of products into simple mechanics. – robinson Jul 10 '11 at 20:40
up vote 9 down vote accepted

Expanding on the idea of @Bill Dubuque, let us hypothesize that we have a differentiable function $p(x_1,x_2,x_3,x_4)$ such that $\int fg=p(f,g,\int f,\int g)$ for all differentiable functions $f$ and $g$. (Assuming $f$ and $g$ are merely continuous is not good enough, since then the right-hand side may not be differentiable, whereas the left-hand side must be.)

More precisely, we assume that whenever $F$ and $G$ are twice differentiable, we have $$ \frac d{dx}p(F'(x),G'(x),F(x),G(x)) = F'(x)G'(x). $$ Let us take $F(x)=ux^2+ax+c$ and $G(x)=vx^2+bx+d$. Using the multi-variable chain rule, we get \begin{align*} ab &= F'(0)G'(0) = \frac d{dx}p(F'(x),G'(x),F(x),G(x))\bigg|_{x=0}\\ &= 2up_1(a,b,c,d) + 2vp_2(a,b,c,d) + ap_3(a,b,c,d) + bp_4(a,b,c,d). \end{align*} (Here, $p_j=\partial p/\partial x_j$.) Since the left-hand side does not depend on $u$ and $v$, we must have $p_1=p_2=0$, for all $a,b,c,d$. But this implies that the function $p$ depends only on $x_3$ and $x_4$. So the above reduces to $$ ab = ap_3(c,d) + bp_4(c,d). $$ Taking $a=1$ and $b=0$ gives $p_3=0$, for all $c,d$. Taking $a=0$ and $b=1$ gives $p_4=0$, for all $c,d$. Finally, then, we have that the function $p$ is a constant function. Since a constant function cannot satisfy our hypothesis, this shows that there can be no such function.

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HINT $\ $ If there were a product rule $\rm\:\int f\:g\ =\ p(f,g,\int f,\int g)\:$ for some polynomial $\rm\:p\:$ then for $\rm\:f = x,\ g = 1/x^2\:$ this would imply that $\rm\: log(x)\:$ is a rational function.

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Thanks, this is exactly the sort of thing I'm looking for. Now suppose the student to whom I'm explaining this were to object that perhaps we should look for a non-polynomial $p$? I do see that this begins to look rather unlikely, but is there a simple argument to shut this possibility down? – robinson Jul 10 '11 at 21:32
I'd ask what they mean by "non-polynomial", because if they mean some rational function, then I believe the same argument works. If they mean any arbitrary continuous function or something, then it will be a bit trickier I'd guess, but I'm not sure I'd consider that a "rule" anymore. – Matt Jul 10 '11 at 22:02
+1 This is a very good answer. Although, I wonder, is there a intuitive explanation for why we shouldn't expect $\log(x)$ to be a rational function? (Besides a formal proof that it is not, that is.) – Zev Chonoles Jul 10 '11 at 23:23
Here's another one: If there were a product rule for antiderivatives then you could integrate $$e^{-x^2/2}=\bigl({1\over x}\bigr)\cdot\bigl(x e^{-x^2/2}\bigr)$$ in elementary terms. – Christian Blatter Jul 11 '11 at 8:07

My answer:

Differentiating is somewhat like taking the difference of two things (that are close together).

Integrating is somewhat like taking the sum of many things (that may be fluctuating).

Suppose $f$ and $g$ are functions that we "know" how to differentiate and integrate. That's somewhat analogous to saying we have expressions for





Wanting to differentiate or integrate the product $fg$ is somewhat analogous to wanting formulas for $f_2g_2-f_1g_1$ or for $f_1g_1+\cdots+f_ng_n$.

With a difference of two things, you can do algebraic "tricks" such as $f_2g_2-f_1g_1=f_2g_2-f_1g_2+f_1g_2-f_1g_1$.

It's not so easy to find an algebraic "trick" for expressing $f_1g_1+\cdots+f_ng_n$ in terms of $f_1+\cdots+f_n$ and $g_1+\cdots+g_n$.

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Yes, this is more or less how I think about it. Another way to phrase the same observation is that differentiation is local, while integration is global. – Qiaochu Yuan Jul 10 '11 at 22:14

Maybe use the notion of "elementary function". There are elementary functions $A,B$ such that both $\int A$ and $\int B$ are elementary, but $\int(AB)$ is not elementary. So there can be no simple "product rule" for integration, since any simple combinations of elementary functions is again elementary.

Example: $A(x) = \cos x$, $B(x) = \frac{1}{x}$. Then $\int \cos x\,dx = \sin x + C$ and $\int \frac{1}{x}\,dx = \ln x +C$ are elementary, but $\int \frac{\cos x}{x}\,dx$ is not.

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Ok, but is there any simple way to convince a first year calculus student that $\int cos(x)/x$ is not elementary? That is the intended audience. – robinson Jul 10 '11 at 21:27

Arguably integration by parts is the anti-derivative form of the product rule; in fact it is derived from the differentiation (i.e. the usual) product rule.

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Sure, but it expresses the antiderivative of one product in terms of the antiderivative of another product. I'm wondering if there is an intuitive reason we can't do even better and get rid of the products entirely. – robinson Jul 10 '11 at 18:49
I'm not sure I have a good explanation beyond "that's just how it works" :) Hopefully someone else will be able to give you a more satisfying, intuitive answer. – Zev Chonoles Jul 10 '11 at 19:01

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