Integration technique of writing $\int_0^{\infty}$ as $\int_0^1 + \int_1^{\infty}$ and using the substitution $\frac{1}{x} \leftrightarrow u$

Question

I've recently seen this idea applied a few times to evaluate otherwise difficult integrals. Suppose you have an integral $\int_0^{\infty} f(x)\,\mathrm{d}x$. You can break up the domain of integration and make the substitution $\frac{1}{x}\leftrightarrow u$ to arrive at an integral which may be easier to evaluate.

$$\begin{align} \int\limits_0^{\infty} f(x)\,\mathrm{d}x \;\;&=\;\; \int\limits_0^1 f(x)\,\mathrm{d}x + \int\limits_1^{\infty} f(u)\,\mathrm{d}u \\\;\;&=\;\;\int\limits_0^1 f(x)\,\mathrm{d}x + \int\limits_0^1 \frac{f\!\left(\frac{1}{x}\right)}{x^2}\,\mathrm{d}x \;\;=\;\; \int\limits_0^1 \frac{x^2f(x)+f\!\left(\frac{1}{x}\right)}{x^2}\,\mathrm{d}x \end{align}$$

Does this technique have a name? Are there broad families of functions for which this trick is essential for integration, or does this only help in very specific cases? Can this idea be applied generally enough to be worth showing students?

It is probably good to have this link here, as well as the link to the Glasser master theorem — Giuseppe Negro, Dec 15 '16 at 19:44
That trick was crucial in evaluating this wonderful integral: http://math.stackexchange.com/questions/562694/integral-int-11-frac1x-sqrt-frac1x1-x-ln-left-frac2-x22-x1/565626#565626 — Ron Gordon, Dec 15 '16 at 20:14
Just a thought, but if one couldn't evaluate the integral (and it converged upon coming to +- infinity) couldn't you use this to approximate the integral? For example, using this technique and calculating bounds from 0 to 1000 as opposed to infinity (without calculating the remaining improper integral)? — Gil Keidar, Jan 19 '18 at 17:42

Mike Pierce · Accepted Answer · 2019-07-27T21:17:55.437

Since no one has responded that there is a name, I think we can conclude that, no, this technique doesn't have a notable name. Sometimes, though, this technique is useful in the wild:

Show that $\displaystyle \int\limits_{-\infty}^{+\infty} f(x) \, \mathrm{d}x = \int\limits_{-\infty}^{+\infty} f\left( x - \frac{1}{x} \right) \,\mathrm{d}x$
Evaluate $\displaystyle \int\limits_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\,\mathrm{d}x$
Show that $\displaystyle \int\limits_{0}^{\infty} \frac {\ln x}{1+x^2} \,\mathrm{d}x = 0$
Show that $\displaystyle \int\limits_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)} = \int\limits_0^1 \frac{\mathrm{d}x}{1+x^2} = \frac{\pi}{4}\,$ for all values of $a \neq 0$
Improper integral of natural log over a quadratic

A neat generalisation of the first is $\int_{\mathbb{R}} f(x),\mathrm{d}x=\int_{\mathbb{R}} f\left(x-\sum_{i=1}^n \frac{\alpha_i}{x+\beta_i}\right),\mathrm{d}x$ for any $\alpha_i \in\mathbb{R}^+$ and $\beta_i\in\mathbb{R}$ — ocg, Jul 28 '19 at 00:25

Integration technique of writing $\int_0^{\infty}$ as $\int_0^1 + \int_1^{\infty}$ and using the substitution $\frac{1}{x} \leftrightarrow u$

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