Integration by Parts implies U-substitution?

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.

Thanks!

Solution 1:

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.

Solution 2:

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.