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I'm revisiting a little bit of calc but I can't remember some details. I intentionally came up with the following abomination to try and figure out what goes wrong: $$2=\underbrace{\int_{-1}^1 x+1\,dx\\}_{I_1}\overset*=\underbrace{\int_1^1\frac{\sqrt{u}+1}{2\sqrt{u}}\,du}_{I_2}=0$$ where $$x=\sqrt{u}$$ $$dx=\frac{1}{2\sqrt{u}}du$$ was used.

I learned calc form Spivak which mainly focuses on indefinite integrals and general theory (which I'm both quite comfortable with), my problem arises when it's time to calculate a definite integral. I have the following considerations:

  1. The indefinite integral of $I_2$, once $u$ is replaced by $x^2$, equals the indefinite integral of $I_1$.
  2. For every $a\geq 0$, $\displaystyle\int_0^ax+1\quad dx=\int_0^{a^2}\frac{(\sqrt{u}+1)}{(2\sqrt{u})}du=\frac{a^2}{2}+a$.

My question is the following: Why was the substitution wrong? If a substitution's correctness depends on the domain of integration, when is a substitution correct on the case of the indefinite integral?

I looking for an answer that allows me to know precisely when a substitution may not work. Something of the form: For $x=g(u)$ to be a valid substitution, $g$ must have as co-domain a superset of the whole domain of integration, etc (what about the indefinite integral?) (this sort of rules are not made explicit on Spivak's, in fact, he mentions that this sort of thing works [in the case of the indefinite integral] when $g$ is one-to-one 'for all $x$ under consideration', which I know is not the weakest condition).

insipidintegrator
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Sebastián P. Pincheira
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    $u = x^2$ is not monotonic in $x$. If so, you need to split the integration range so that the transformation is monotonic in each range. – user619894 Sep 15 '22 at 06:28
  • My answer here explains (giving various examples) that the integration by substitution theorem doesn't inherently require injectivity, and that whenever injectivity is called for, the culprit is never the theorem's conditions per se. (This is not to say that mandating injectivity is a bad idea.) Implicit substitutions (x=g(t) instead of u=g(x)) are required to be invertible, and this is an implicit consequence of the theorem. For reference, my full statement of the theorem. – ryang Feb 08 '23 at 16:42

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The thing is, that the lower limit cannot be expressed as a real number. If $x=\sqrt u$, then $x=-1$ does not give any solution(s) for $u$ because the square root of $u$ is always positive. So the substitution won’t work. Note that $u$ is NOT $1$ for the lower limit, as $\sqrt1=1$ ONLY.

  • Here is a link explaining what exactly happens in substitution.

  • Here is a post with a similar problem (vanishing integral upon substitution).

  • Here is a post which reveals that u-substitutions should be preferably (not necessarily) bijective.

  • This post states that substitutions must be injective.

Mikasa
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insipidintegrator
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