Value of $\int_{0}^{1}\dfrac{\log x}{1-x}dx$. What is my wrong step?

Question

I would like to evaluate the value of this integral: $$I=\int_{0}^{1}\dfrac{\log x}{1-x}dx.$$

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On one hand, I proceed using integration by parts as follows: $$I=\int_{0}^{1}f(x)g'(x)dx,$$ where $f(x)=\log x$ and $g'(x)=\dfrac{1}{1-x}$. From this, I can write: $f'(x)=\dfrac{1}{x}$ and $g(x)=-\log(1-x)$.

Therefore: $$\begin{equation}\begin{split}I=\int_{0}^{1}f(x)g'(x)dx&=\left[f(x)g(x)\right]_{0}^{1}-\int_{0}^{1}f'(x)g(x)dx,\\&=\left[-\log(1-x)\log x\right]_{0}^{1}+\int_{0}^{1}\dfrac{\log(1-x)}{x}dx,\\&=\int_{0}^{1}\dfrac{\log(1-x)}{x}dx\\&=-\int_{0}^{1}\dfrac{\log(t)}{1-t}dt\\&=-I.\end{split}\end{equation}$$

Hence, $I=0.$

On the other hand, Wolframalpha gives $I=-\dfrac{\pi^2}{6}.$

Answer 1

After integrating by parts, use the power series for $\log(1-x)$:

$$\int_{0}^{1}\frac{\log(1-x)}{x}dx = -\int_{0}^{1}\sum_{k=1}^{\infty}\frac{x^{k-1}}{k}dx\\=-\sum_{k=1}^{\infty}\int_{0}^{1}\frac{x^{k-1}}{k}dx=-\sum_{k=1}^{\infty}\frac{1}{k^2}=-\frac{\pi^2}{6}$$

Answer 2

Don't use integration by parts straight away. Instead, expand the denominator: $\sum_{k=0}^{\infty} x^k$ because the bounds on $x$ are strictly between $0$ and $1$. After this, interchange integration and summation (due to uniform convergence and you'll get integrals of the form $\int_{0}^{1} x^k \log x dx$. Now use integration by parts.