Integrating $\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx$

Question

When i came across this one i thought of using the substitution $x=\frac{1-t}{1+t}$ that lead to an easy expression. $$\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx\overset{x=\frac{1-t}{1+t}}=\ln \left(2\right)\int _0^1\frac{1}{t^2+1}\:dt+\int _0^1\frac{\ln \left(t\right)}{t^2+1}\:dt-\int _0^1\frac{\ln \left(t+1\right)}{t^2+1}\:dt$$ $$=\frac{\pi }{4}\ln \left(2\right)-G-\frac{\pi }{8}\ln \left(2\right)=\boxed{\frac{\pi }{8}\ln \left(2\right)-G}$$ But then i tried to solve it using feynman's trick with the following parameter: $$I=\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx$$ $$I\left(a\right)=\int _0^1\frac{\ln \left(1-ax\right)}{x^2+1}\:dx$$ $$I'\left(a\right)=-\int _0^1\frac{x}{\left(x^2+1\right)\left(1-ax\right)}\:dx=\frac{1}{a^2+1}\int _0^1\frac{a-x}{\left(x^2+1\right)}-\frac{a}{1-ax}\:dx$$ $$=\frac{1}{a^2+1}\left(\frac{\pi a}{4}-\frac{\ln \left(2\right)}{2}+\ln \left(1-a\right)\right)\:$$ $$\int _0^1I'\left(a\right)\:da=\frac{\pi }{4}\int _0^1\frac{a}{a^2+1}\:da-\frac{\ln \left(2\right)}{2}\int _0^1\frac{1}{a^2+1}\:da+\int _0^1\frac{\ln \left(1-a\right)}{a^2+1}\:da\:$$ $$I=\frac{\pi }{8}\ln \left(2\right)-\frac{\pi }{8}\ln \left(2\right)+I$$ And everything cancels. ¿Am i doing something wrong?, ¿Should i be using another parameter or other upper and lower bounds?

Answer 1

You are not allowed to switch the integral and derivative. The conditions of the theorem about switching those are not met. Denote $\frac{\ln(1-ax)}{x^2+1}$ by $f(x,a)$, one of the conditions is that there must exist a function $g:[0,1] \rightarrow \mathbb{C} $ such that $\int_0^1 |g(x)|dx < \infty$ and $| \frac{\partial f}{\partial a}(x,a) |\leq g(x)$ for all $a$.