So I feel a bit strange asking a Calculus question, but this came up today while teaching.

One can check that if you start with some integral, which can be see as an "obvious u-substitution problem", that you can instead use integration by parts, and wind up with the scenario where you have have the original integral on both sides of your equation so you solve for the integral.

Example: Given $I=\int g^n(x)g'(x)dx$ we can clearly use u-substitution, but if we use integration by parts we get the equation $I=-nI+g^{n+1}$. This is nothing exciting or surprising, but it yields the observation that u-sub leads to one of these int by parts equations.

Question Is the opposite true?

What I mean to say is, if you do integration by parts and you wind up with an equation of this type, does it mean that you could of used some very clever u-substitution?

I feel like I should know this, but I have thought about it today, and asked a friend or two, and we don't see an immediate proof of this.


  • $\begingroup$ I have to admit my ignorance... what exactly is a "u substitution"? I'm perfectly aware of integration by parts, but your other technique looks new to me. $\endgroup$ Oct 27 '10 at 3:33
  • $\begingroup$ @J.M.: It's just another name for integration by substitution. You substitute $u = g(x)$, $du = g'(x) dx$ in the integrand to get, in this case, $I = \int u^n du$. $\endgroup$
    – user856
    Oct 27 '10 at 3:45
  • $\begingroup$ @Rahul: I see. I know the technique, but only now have I seen it called that way. $\endgroup$ Oct 27 '10 at 3:48
  • 3
    $\begingroup$ @J.M.: Well, $\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$, so substituting you get $\frac{1}{2i}\int(e^{(1+i)x}-e^{(1-i)x})dx$ for which the substitutions are fairly obvious... $\endgroup$ Oct 27 '10 at 3:57
  • 2
    $\begingroup$ @J.M. Use the substitution $y = \exp(x)(\sin(x) - \cos(x)) / 2$! (Generalizing this trivial comment shows there is always some "very clever ... substitution" for any integral you know how to evaluate.) $\endgroup$
    – whuber
    Oct 27 '10 at 4:18

Added: I interpreted the question as follows: suppose that we use the integration by parts formula, $$\int u(x)v'(x)\,dx = u(x)v(x) - \int v(x)u'(x)\,dx$$ and $\int v(x)u'(x)\,dx = \alpha \int u(x)v'(x)\,dx$ for some constant $\alpha$, so that we can "solve" for the original integral as $$\int u(x)v'(x)\,dx = \frac{1}{1+\alpha}u(x)v(x) + C.$$ Is it the case that we can solve the original integral by doing a direct substitution instead of integration by parts?

I think this works:

Suppose that $v(x)u'(x) = \alpha u(x)v'(x)$ for some $\alpha\neq -1$. (I think this is essentially what you have, since the integral on the left hand side is $\int u(x)v'(x)\,dx$, and the integral on the right is $\int v(x)u'(x)\,dx$). I restrict to $\alpha\neq -1$, because if $\alpha=-1$, then you cannot "solve" for the original integral. But in fact, from the work below we will get that $u=\frac{B}{|v|}$, so that integration by parts will simply result in the true (but useless) $\int u\,dv = B + \int u\,dv$.

If $u$ or $v$ are zero, then the original integral was the integral of $0$, so we may discard that case. So we get that $\frac{u'}{u} = \alpha \frac{v'}{v}$, and integrating we get $\ln|u|=\alpha\ln|v|+C$, or $|u| = A|v|^{\alpha}$ for some $A\gt 0$; hence $u=B|v|^{\alpha}$ for some $B\neq 0$. So the original integral was $$\int u(x)v'(x)\,dx = \int B|v(x)|^{\alpha}v'(x)\,dx$$ which suggests the substitution $g=v(x)$ to get $B\int |g|^{\alpha}\,dg$.

Added: Note that if $\alpha = -1$, we still get $u=B|v|^{-1}$, and so again the substitution $g=|v(x)|$ will do the trick.

  • $\begingroup$ Should be "integral on RHS is.....and the integral on the LHS is" $\endgroup$ Oct 27 '10 at 4:31
  • $\begingroup$ @Trevor, huh? @Arturo, I think this works! I don't see any issues here. $\endgroup$
    – BBischof
    Oct 27 '10 at 4:36
  • $\begingroup$ @Bbischof integral on LHS is $\int v(x)u'(x)$ and integral on RHS is $\alpha \int u(x)v'(x)$. $\endgroup$ Oct 27 '10 at 4:41
  • $\begingroup$ Also doesn't $I = -\alpha I+ g^{\alpha}$? So its not just $v(x)u'(x) = \alpha u(x)v'(x)$? $\endgroup$ Oct 27 '10 at 4:50
  • $\begingroup$ @Trevor: The usual way to state integration by parts (at least in the US) is $\int u\,dv = uv - \int v\,du$; so $uv'$ on the left, $vu'$ on the right. And I thought the example of a power was just one case, that BBischof was interested in the general case when the integral on the RHS equals that on the LHS, so that we can "solve" for the integral, regardless of what the $uv$ part of the integration by parts formula might be. I'll edit the answer to make it clear. $\endgroup$ Oct 27 '10 at 15:57

This expands on a comment that might not have been perfectly clear (perhaps because of its obviousness, for which I apologize).

Suppose you have in hand some magic method of integration (such as the one in the question). By this I mean any procedure that takes a functional expression $f(x)$ and returns some expression $F(x)$ for its indefinite integral. (I also assume you accept $F(x)$ as a valid solution to your integration; e.g., some people might not accept an elliptic function, or imaginary values are verboten, or they might not like power series, etc. I leave the criterion of acceptability to you as a matter of taste. All I require is that you know how to differentiate $F$ and are able to compare that result to $f$ in order to check the validity of your magic method.) A trivial application of the Fundamental Theorem of Calculus asserts that the "very clever" substitution

$$u = F(x)$$

will enable you to perform the integral. This happens, of course, because you can calculate that $du = F'(x)dx = f(x)dx$, whence the substitution converts the original integral into $\int{du}$ with general solution $u + C = F(x) + C$ by back-substitution. In other words, the question as posed merely asks whether the FTC is true in a special case.

  • $\begingroup$ So perhaps one should say "nontrivial substitution" (one might also substitute $u=x$ and get an integral that is doable, since we are assuming we can do the integral in the first place!). $\endgroup$ Oct 27 '10 at 20:59
  • $\begingroup$ @Arturo Point taken. But my comment was intended in a different spirit: given that the integral can be "performed," there is always a nontrivial substitution that renders the integrand immediately identifiable as a derivative. That is different than your sense of "doable", which refers to a mathematical operation, not a symbolic one. $\endgroup$
    – whuber
    Oct 27 '10 at 21:11
  • $\begingroup$ I agree with Arturo, but I also understand what you are saying here and in your comment. This is a fair point, that should be made. I am going to upvote this, and accept Arturo's. Arturo's is what I was looking for, while this is a pedants response. That being said, in mathematics, precision is essential, and thus I appreciate your response. $\endgroup$
    – BBischof
    Oct 28 '10 at 2:36
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    $\begingroup$ @BBischof Agreed (although it would be nice to maintain a distinction between a "pedant's response" and a "pedantic response," preferring the latter characterization as being more gracious than the ad hominem one, which wasn't really called for). I offered it with some diffidence--remember, it started out as an offhand comment--but was moved to promote it to a full response because you indicated that the comment wasn't completely clear and also I think it reveals something important about substitutions and what we really mean by "integration." $\endgroup$
    – whuber
    Oct 28 '10 at 3:03
  • $\begingroup$ wow you are totally right, the phrasing I used was wrong and inflammatory, it is not what I meant. I am very sorry if you were offended. $\endgroup$
    – BBischof
    Oct 28 '10 at 3:29

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