Complex result or solution of an improper integral

Question

The indefinite integral

$\displaystyle \int \frac{\ln (x)}{x^2-2 x+1} \, dx$

has the solution

$\ln (1-x)+\frac{x \ln (x)}{1-x} +C$

Taking the limit as $x$ approaches infinity

$\displaystyle \lim_{x\to \infty } \, \left(\ln (1-x)+\frac{x \ln (x)}{1-x}\right)= i \pi$

Also

$\displaystyle \lim_{x\to 0 } \, \left(\ln (1-x)+\frac{x \ln (x)}{1-x}\right)= 0$

This means that

$\displaystyle \int_0^{\infty } \frac{\ln (x)}{x^2-2 x+1} \, dx = i \pi$

However there is the argument that the integration is over the real plane, the limit does not exist as $x$ approaches zero, and a complex solution was found, so there might be a contradiction.

How can this be explained or resolved?

Answer 1

Since$$\ln (1-x)+\frac{x \ln (x)}{1-x}$$is defined only when $x\in(0,1)$, it is not true that$$\lim_{x\to+\infty}\ln (1-x)+\frac{x \ln (x)}{1-x}=i\pi.$$

Answer 2

Complex analysis methods to compute integrals don't prove that the integral exists, they just give you a candidate answer. You can blindly apply many rules to get nonsensical answers. An example in a similar vein: an antiderivative of $x^{-2}$ by power rule is $-x^{-1}$ and hence $$\int_{-1}^1 \frac{dx}{x^2} = -2.$$ But observe $1/x^2 > 0$, so surely we should also have $-2>0$?