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I asked WolframAlpha to solve a certain differential equation and it gave it in this form: $f(x)=(x-1)(\ln(x-1)-i\pi-1)$. Now I am only interested in this function when $x$ is in the interval $(0,1)$. In this case, I think $f(x)$ will always be a real number.

My question is, is there a way to simplify this expression in the interval $(0,1)$, so that it makes no reference to complex logarithms or imaginary numbers? I just want to express it in terms of real functions.

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  • $\begingroup$ Probably not; even if $x$ is real you'll still have that $i\pi$. $\endgroup$ Mar 15 '19 at 2:09
  • $\begingroup$ @ParclyTaxel But if $x$ is between 0 and 1, then $f(x)$ is a real number. So you should be able to express one real number in terms of another real number without referring to imaginary numbers. $\endgroup$ Mar 15 '19 at 2:10
  • $\begingroup$ What is the differential equation in question, out of curiosity? $\endgroup$ Mar 15 '19 at 2:12
  • $\begingroup$ @EeveeTrainer This one. $\endgroup$ Mar 15 '19 at 2:13
  • $\begingroup$ @EeveeTrainer FYI, here is the question that prompted this question: math.stackexchange.com/q/3148842/71829 $\endgroup$ Mar 15 '19 at 3:26
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We assume principal values for the multivalued functions we invoke here. Since $x\in(0,1)$, $x-1$ is negative and $$\ln(x-1)=\ln(1-x)+i\pi$$ Thus $$(x-1)(\ln(x-1)-i\pi-1)=(x-1)(\ln(1-x)+i\pi-i\pi-1)=(x-1)(\ln(1-x)-1)$$

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