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My question might seem similar to this question -"https://math.stackexchange.com/questions/1965777/substitution-makes-the-integral-bounds-equal" but I'll expand on why it doesn't resolve my query.

First off, let's take a look at this integral - $$ I = \int_0^\infty \frac{x^2-1}{x^4+1} dx$$

Dividing both the numerator and denominator by $x^2$, we get

$$ I = \int_0^\infty \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}} dx = \int_0^\infty \frac{1-\frac{1}{x^2}}{({x+\frac{1}{x}})^2-2} dx$$

Now, if we take $x+\frac{1}{x}=u$, then

$$I= \int_\infty^\infty \frac{du}{u^2-2}$$

Now, the aforementioned question I linked suggested that substitutions must be injective and if not then I should break them in intervals where they are injective, so I did as suggested -

$$I= \int_\infty^2 \frac{du}{u^2-2} + \int_2^\infty \frac{du}{u^2-2} = \int_\infty^2 \frac{du}{u^2-2}-\int_\infty^2 \frac{du}{u^2-2}=0$$

Now, the result $I=0$ is fortunately true in this case but how do I know that the two infinities in the two integrals I compared in the last step are equal? What if the two infinties aren't equal in some other question? Or am I completely off in my handling of this integral? Please help!

Swan
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    The substitution doesn't have to be injective in case of substitution for a single variable. It has to be for multiple variables. You can refer to Integration by substitution. – mathcounterexamples.net Aug 20 '23 at 14:15
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    Also, dividing by $x^2$ is not possible at $x=0$. – mathcounterexamples.net Aug 20 '23 at 14:16
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    When you do an explicit $u$-substitution in the form $u = g(x)$ like what the Integration by Substitution Theorem says, you're never supposed to care about injectivity (at least for one variable). You should only care about injectivity when you make an implicit substitution in the form $g_1 (u)=g_2 (x)$ to get $u = g^{-1}_1 (g_2 (x))$, which requires $g_1$ to have an inverse. – Accelerator Aug 23 '23 at 08:46
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    Not sure if this answers the question.... but if you don't want to split up the integral into two parts and want to find an antiderivative $F$ such that $F' = f$, then you can write $\int_{0}^{\infty}\frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-2}dx = \lim_{b \to \infty}\lim_{a \to 0^+}\int_{a}^{b}\frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-2}dx$, let $u = x + \frac{1}{x}$, find the antiderivative and plug in your new bounds using FTC 2, then take the limits at the end. Whatever your $F(b)$ and $F(a)$ are, be it real or dare I say complex, I guarantee you'll get $0$. – Accelerator Aug 23 '23 at 09:42

2 Answers2

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The expression$$\int_\infty^\infty\frac{\mathrm du}{u^2-2}$$makes no sense.

You have$$\int_0^\infty\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2-2}\,\mathrm dx=\int_0^1\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2-2}\,\mathrm dx+\int_1^\infty\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2-2}\,\mathrm dx.\label{a}\tag1$$Now, if you apply the substitution $u=x+\frac1x$ to the first integral of the RHS of \eqref{a}, then you get$$\int_\infty^2\frac{\mathrm du}{u^2-2}\label{b}\tag2$$and if you apply it to the second integral of the RHS of \eqref{a}, then you get$$\int_2^\infty\frac{\mathrm du}{u^2-2}.\label{c}\tag3$$It is clear that the sum of \eqref{b} with \eqref{c} is equal to $0$.

Or you can prove that the RHS of \eqref{a} is $0$ by applying the substitution $u=\frac1x$ to its first integral. You shall get\begin{align}\int_0^1\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2-2}\,\mathrm dx&=\int_\infty^1\frac{1-u^2}{\left(\frac1u+u\right)^2-2}\times\left(-\frac1{u^2}\right)\,\mathrm du\\&=-\int_1^\infty\frac{1-\frac1{u^2}}{\left(u+\frac1u\right)^2-2}\,\mathrm du,\end{align}which is the symmetric of the second integral of the RHS of \eqref{a}. Therefore, the LHS of \eqref{a} is equal to $0$.

José Carlos Santos
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it works out because the integral is absolutely convergent(for x going to infinity the absolute value of the integrand behaves like x^2/x^4=1/x^2 which is integrable-look up cauchy principal value for a rigorous proof) the problem arises basically because of 1/x in the substitution to avoid it you can factor the denominator(x+sqrt(2)+1)(x-sqrt(2)+1) and then partial fraction decomposition and then the integrals are computable

Spyros Mavrokefalidis
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  • Just because the integral converges absolutely does not imply OP's work is correct. There is nothing wrong with having the $\frac{1}{x}$ in the substitution because $\int_{0}^{\infty}\frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-2}dx = \lim_{b \to \infty}\lim_{a \to \infty}\int_{a}^{b}\frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-2}dx$, and the substitution $u = x + \frac{1}{x}$ on $[a,b]$ works in OP's favor by the Substitution Theorem for integrals. If anything, the only thing I would argue that was "wrong" was writing $\int_{\infty}^{\infty}$ which makes no sense. – Accelerator Aug 23 '23 at 09:15