I came across this seemingly interesting, but poorly exemplified, method of integrating:

Since integration is the inverse of differentiation, you can think of integration as “dividing” by $d$.

J. P. Ballantine [1] shows that you can formally divide by $d$ and get the correct integral. For example, he arrives at $$\int x^2 \sin x\,dx=(2-x^2)\cos x + 2x\sin x + C$$ using long division!

poor example

[1] J. P. Ballantine. Integration by Long Division. The American Mathematical Monthly, Vol. 58, No. 2 (Feb., 1951), pp. 104-105

Surely the claim that this is rigorous is questionable. However, it would be fascinating to understand how to follow this method and maybe even understand why it works; the example has no explanation, and I can’t put the steps together via reverse engineering.

Could someone shed some light on this technique?

addendum: An additional example would really complete the perfect response $\ddot\smile$

  • $\begingroup$ "Formal" doesn't mean rigorous here, it means based on form; messing around with the symbols without worrying too much about whether they mean anything. Euler famously did this sort of thing a lot. $\endgroup$ Nov 1 '18 at 19:45
  • $\begingroup$ @QiaochuYuan Gotcha! Thank you for specifying that; that distinction makes a lot more sense than what I thought. I’ve made a revision ;) $\endgroup$ Nov 1 '18 at 19:46

I managed to "reverse engineer" the steps and I am going to go through them for you!

First, write the function you wish to integrate on the left

$$x^2 \sin x\ dx$$

Now find a function such that, when you differentiate it, the $x^2 \sin x$ comes up. Ah! When you differentiate $-x^2 \cos x$ you get $x^2 \sin x - 2x \cos x$, so it has the bit you want and the other bit seems "smaller" in the sense of "degree" or something. Just write that $-x^2 \cos x$ on the side, differentiate it and subtract the derivative, getting:

enter image description here

Now it's like we found an anti-derivative for the function we want to integrate, but in doing so we got the extra term $2x \cos x$ that we need to cancel... Think of something that, when differentiated, arrives at $-2x \cos x$... Ah! When you differentiate $2x \sin x$, you get the $2x \cos x$ and an extra $2 \sin x$. Write the $2x \sin x$ on the right, differentiate it, and subtract from whatever you had, getting:

enter image description here

and you keep going. The point being, you are always trying to come up with something (in some sense, smaller) that counters the leftovers you have from finding the anti-derivatives of the other terms!

Was I clear enough?

Let us integrate $x \log x$ just for the fun of it. Can you come up with something that, when differentiated, gives $x \log x + $ something? Of course you can, just notice that $(\frac12x^2\log x)' = x\log x + \frac12 x$

So we now have to deal with the extra $-\frac12 x$.

Can you come up with something that cancels that $-\frac12 x$? Of course you can, because $-\frac14 x^2$ does it! We now get


meaning we are done and

$$\int x \log x = \frac12 x^2 \log x - \frac14 x^2 + C$$

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    $\begingroup$ It’s like only doing half of the product rule! isn’t it? $\endgroup$ Nov 1 '18 at 20:00
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    $\begingroup$ “$x\log x +\text{something}$”: That’s the key point right there. If you don’t mind, I’m going to ruminate on that for a bit, then I’ll probably come back and mark this answer $\color{green}{\checkmark}$—but first I’ll let other people contribute their fair shares too $\ddot\smile$ $\endgroup$ Nov 1 '18 at 20:05
  • $\begingroup$ sure thing @ChaseRyanTaylor . I tried to include a picture of the division process but I couldn't. Do you want me to further develop my answer? $\endgroup$
    – RGS
    Nov 1 '18 at 20:07
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    $\begingroup$ @RGS I followed it just fine. And actually, I think “long division” is a misleading name. The subtraction pattern is a way to keep track of the terms you have left to reconcile, like you do with place values / partial fractions in long division, but whence the author got this “‘dividing’ by $d$” notion is beyond me. . . . $\endgroup$ Nov 1 '18 at 20:10
  • $\begingroup$ @ChaseRyanTaylor sorry nobody chipped in in the meantime $\endgroup$
    – RGS
    Nov 12 '18 at 9:39

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