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(Context) This is a long question, so please bear with me. Pondering about the integral

$$\int_{0}^{\infty}\frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-1}dx$$

that was asked in this question, one of the questions was if letting $u = x+\frac{1}{x}$ was valid. I answered,

"No, because on the interval $(0,\infty)$, the function $x + \frac{1}{x}$ decreases then increases, making it not injective over that interval. You need an interval of $u$'s such that each $x \in (0,\infty)$ matches with one and only one $u$."

Then I basically said they can get away with a piecewise-monotonic situation such that they can break up the integral into $\int_0^1$ and $\int_1^{\infty}$.

However, after clicking on this post, I read,

"However, integration by substitution does not generally/inherently require substitution functions to be invertible/bijective or monotonic—or even injective—on the interval of integration."

That was somewhat shocking to me. I originally thought a substitution required to be injective over the interval of integration, but now I'm wondering if my answer is incorrect.

(Attempt) Trying to see if I messed up, I read in my textbook, "An Introduction to Analysis (3rd ed.)" by William R. Wade, this theorem (page 130):

Let $\phi$ be continuously differentiable on a closed, nondegenerate interval $[a,b]$. If $f$ is continuous on $\phi([a,b])$, or if $\phi$ is strictly increasing on $[a,b]$ and $f$ is integrable on $[\phi (a), \phi (b)]$, then

$$\int_{\phi (a)}^{\phi (b)}f(t)dt = \int_{a}^{b}f(\phi (x))\phi'(x)dx.$$

With that theorem in mind, I experimented with a simple integral:

$$\int_{0}^{3\pi /2}\sin\left(\cos\left(x\right)\right)\left(-\sin\left(x\right)\right)dx.$$

Intuitively, you can let $\phi=\cos{(x)}$ without any problem. Thinking about the theorem above, it doesn't satisfy the second condition because it's not strictly increasing everywhere on $[0,3\pi/2]$. Then I thought, "but I think it satisfies the first condition "$f$ is continuous on $\phi([a,b])$." So I concluded I can do the substitution with no problem.

(Question) For the improper integral in question, can I let $u = x+\frac{1}{x}$ where $x \in (0,\infty)$? It's definitely not strictly increasing everywhere on that interval, so the second condition in the theorem gets thrown out the window. But does the first condition hold? If so, why exactly? Originally, I thought it doesn't hold because $x+\frac{1}{x}$ is not defined at $x=0$, but I can replace $0$ with $\epsilon \to 0$ and replace $\infty$ with $t \to \infty$, which throws me off even more.

P.S. I'm somewhat of a beginner at real/complex analysis, so if you could, please add as many details as possible.

Accelerator
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  • An intuitive explanation is that as $x$ runs from 0 to 1, $u$ runs from $\infty$ to 2, so the limits are the wrong way round, but the $\dfrac{dx}{du}$ factor is negative - so you get the correct positive contribution to the integral; then as $x$ runs from 1 to $\infty$, $u$ runs from 2 to $\infty$, so the substitution works as usual. – mcd Oct 08 '22 at 06:50
  • Note that, in its simplest setting, $u$-substitution is simply an application of the chain rule of differentiation with the fundamental theorem of calculus. In your quoted theorem, the criterion that $f$ be continuous on $\phi([a, b])$ is simply a case where FTC can readily be applied, since continuous functions have antiderivatives. – Dustan Levenstein Oct 08 '22 at 06:55
  • Another way of thinking about it is that integration by substitution is really only "reverse chain rule" formalised into a method: because $\int \dfrac{1}{x^2-1} dx=\frac{1}{2}\ln|\dfrac{1-x}{1+x}|$, the derivative of that, with $x$ replaced by $x+\dfrac{1}{x}$, is the integrand of your integral, so your are just integrating a derivative. – mcd Oct 08 '22 at 06:56
  • I agree with that, and I'm assuming you thought of breaking up the integral like $\int_0^1$ and $\int_1^{\infty}$, right? I just don't know if you can apply that $u$-substitution from the get-go without breaking up the integral. @mcd – Accelerator Oct 08 '22 at 06:56
  • See my comments here. The reason why people state the theorem with injectivity is because it is then easier to apply in calculations. Otherwise, one has to very carefully identify what the functions $f$ and $g$ (in my link’s notation) are, and misidentifying these is what causes all the common errors (especially with trigonometric substitutions). Anyway, here, $f(t)=\frac{1}{t^2-1}$ and $\phi(x)=x+\frac{1}{x}$ makes it mostly work. If anything, you should be more concerned with the limits of $0$ and $\infty$, and take appropriate limits. – peek-a-boo Oct 08 '22 at 07:19
  • So basically, you're saying $\int_{0}^{\infty}\frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-1}dx = \lim_{a,b\to\infty} \int_{a}^{b}\frac{1}{t^{2}-1}dt$, meaning it's safe to apply $\phi(x) = x+\frac{1}{x}$ when $x\in (0,\infty)$, right? @peek-a-boo – Accelerator Oct 08 '22 at 07:34
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    @Accelerator yup, and the final limit is zero by a direct calculation – peek-a-boo Oct 08 '22 at 07:55
  • I think I understand now. Thank you very much for your effort. @peek-a-boo As for everyone else, you can still post an answer if you want and I'll accept it. – Accelerator Oct 08 '22 at 08:05
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    When you look for an antiderivative of a function $f$ you do not care about injectivity. Moreover the steps on the way do not need to be carefully justified. The aim is to obtain a function $F$ such that $F'=f.$ Once you get such function you calculate its increment from $0$ to $\infty.$ – Ryszard Szwarc Oct 08 '22 at 09:02

1 Answers1

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Let's first consider the function $u$ defined on $(0,\infty)$ by $$\tag{1} u=u(x)=x+\frac{1}{x}. $$ Since
$$ u'(x)=\frac{x^2-1}{x^2} $$ it follows that $u$ is decreasing on $(0,1)$, increasing on $(1,\infty)$ and has an absolute minimum at $x=1$. In particular, $u$ isn't invertible on $(0,\infty)$.

We'll therefore split the integral as

$$ \int_0^{\infty}\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2+1}dx= \int_0^1\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2+1}dx +\int_1^{\infty}\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2+1}dx $$ Now, let's use the change of variable (1) in each of the above integrals. We get

\begin{eqnarray} \int_0^{\infty}\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2+1}dx &=&\int_0^1\frac{u\(x)}{[u(x)]^2+1}+\int_1^{\infty}\frac{u'(x)}{[u(x)]^2+1}dx\\ &=&\int_{\infty}^2\frac{du}{u^2+1}+\int_2^{\infty}\frac{du}{u^2+1}\\ &=& -\tan^{-1}(u)\Big|_2^{\infty}+\tan^{-1}(u)\Big|_2^{\infty}\\ &=&-\left[\tan^{-1}(2)-\frac{\pi}{2}\right]+\left[\tan^{-1}(2)-\frac{\pi}{2}\right]\\ &=&0. \end{eqnarray}

HorizonsMaths
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  • It's generally correct, but are you saying $u$ has to be invertible on the interval we're integrating on? Because I learned from the comments that bijectivity isn't a requirement for $u$-substitution. – Accelerator Oct 12 '22 at 03:43