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I am tackling some minor Analysis topic, which has nonetheless to do with rational functions and how they are integrated. My textbook says the following:

If a rational function $R(u, v)$ is unaltered by the change of sign of some one of its arguments, e. g. $R(u, v) = R(-u, v)$ (i.e. the function is even in $u$), then it can be brought to the form $R(u, v) = R_1(u^2, v)$, containing only even powers of $u$.

I sort of know how to prove this for polynomials, but the general case of rational functions eludes me completely. How would one go about this? Any help would be much appreciated.

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    $\begingroup$ if $f(u) = p(u)/q(u)$ is an even rational function then $f(u) = p(u) q(-u) / (q(u) q(-u))$ and both numerator and denominator are even polynomials, as you'll easily verify $\endgroup$ Commented May 8 at 12:13
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    $\begingroup$ My holy God this actually makes sense. Thanks! $\endgroup$ Commented May 8 at 12:19
  • $\begingroup$ @user8268 The assertion in your comment is incorrect as stated. The numerator and denominator could have a common factor like $\frac{x^3+x^2}{2x+2}$. Or the numerator and denominator could be both odd instead of both even. If one eliminates all common factors, it should be possible to prove that both numerator and denominator are in fact even, but that requires proof. $\endgroup$ Commented May 8 at 15:36
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    $\begingroup$ @GregMartin That's not a counterexample. Both the numerator $(x^3+x^2)(-2x+2)$ and the denominator $(2x+2)(-2x+2)$ are even polynomials. $\endgroup$ Commented May 8 at 16:00
  • $\begingroup$ You're right, my reading comprehension was bad. In that case, the comment should be made an answer. $\endgroup$ Commented May 8 at 19:43

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