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Citing from Wikipedia and one of its references, MathWorld, following hold:

Glasser's master Theorem For $f$ integrable, $\Phi(x) = |a|x - \sum_{i=1}^N \frac{|\alpha_i|}{x-\beta_i}$ and $a$, $\alpha_i$, $\beta_i$ arbitrary real constants the identity \begin{equation} \mathrm{PV}\int_{-\infty}^\infty f(\Phi(x)) dx = \mathrm{PV} \int_{-\infty}^\infty f(x) dx \label{Glasser} \tag{1} \end{equation} holds.

Now consider \begin{align*} \Phi_1(x) &= |a|x - \sum_{i=1}^N \frac{|\alpha_i|}{x-\beta_i} \\ \Phi_2(x) &= x - \sum_{i=1}^N \frac{|a\alpha_i|}{x-|a|\beta_i} \end{align*} Then, by Glasser's theorem \ref{Glasser} $$\mathrm{PV}\int_{-\infty}^\infty f(\Phi_1(x)) dx = \mathrm{PV} \int_{-\infty}^\infty f(x) dx = \mathrm{PV}\int_{-\infty}^\infty f(\Phi_2(x)).$$

However, under the change of variables $y = |a| x$ \begin{equation} \mathrm{PV}\int_{-\infty}^\infty f(\Phi_1(x)) dx = \frac{1}{|a|}\mathrm{PV} \int_{-\infty}^\infty f(\Phi_2(y)) dy. \label{my Idea} \tag{2} \end{equation}

Thus, I assume Glasser's theorem only holds for $|a| = 1$; a quick numerical check seems to support Eq. \ref{my Idea}. Is Wikipedia and MathWorld wrong about this?

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Thank you for noticing this. Your argument is correct. (An alternative way to see the error is simply to set every $α_i$ to 0.) I fixed the Wikipedia page (it now uses $x-a$ in place of $|a|x$) and submitted a correction to MathWorld.

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