# Breaking $u$-Substitution

While learning $u$-substitution, I noticed a potential glitch in its application to definite integrals. If $u$ is defined as a non-injective function of the original integrand variable, then given only the $u$ definition along with the $u$-substituted definite integral, it would be impossible to recover the bounds of the original definite integral. For example, given $$\int_4^9e^u\ du,\text{ where } u = x^2,\text{ thus } du = 2x\ dx$$ it would be impossible to determine whether the original integral had bounds $-2$ to $-3$, $-2$ to $3$, $2$ to$-3$, or $2$ to $3$. While the u-substitution definite integral algorithm can be completed without the need to inverse-substitute the bounds at the end, and while at least in this problem, the answer turns out the same regardless of which pair of original bounds was present, the geometric picture for each case is different; each definite integral corresponds to the area under a different part of the curve. Seemingly, the reason the signs of the original bounds are unimportant in this case is that the original integrand, ${2xe}^{(x^2)}$, is an odd function, so adding, i.e., the area between $x = -2$ and $x = 2$ to the area between $x = 2$ and $x = 3$ won't change the result.

So a good candidate for a case where the information loss due to a u-substitution would matter might involve

1) A non-injective $u$ definition, and

2) A non-odd original integrand, that doesn't become odd when shifted in any direction by a constant, and

3) An integrand that is locally linear everywhere

This is actually trickier to find than one might expect. The integrand ${2xe}^{(x^2)}$ won't work because it's odd, while the integrand $3x^2e^{(x^3)}$ won't work because $u = x^3$ is injective. $4x^3e^{(x^4)}$ is odd, $u = x^5$ is injective, and so on. The absolute value, ramp, and digital root functions aren't odd, but they aren't locally linear everywhere, so they can't really be integrated. The most promising integrand I found is $\cos(x)e^{\sin(x)}$. Integrating it from $0$ to $-3\pi$ on Desmos yields a different answer than integrating from $0$ to $-2\pi$, but the difference is approximately $10^{-15}$, so it could simply be a rounding anomaly.

Has anyone found an integrand and corresponding $u$ definition that fit this bill? If not, do we know for certain that the information lost in a non-injective u-substitution is never important?

• My questions here and here are in a similar vein, but I'm not sure if these actually lead to an answer for your precise question. – Omnomnomnom Jul 4 '18 at 17:34
• In fact, it seems that the answer to my second question shows that, as you've put it, "the information lost in a non-injective u-substitution is never important". – Omnomnomnom Jul 4 '18 at 18:11