Limit matters in improper integral

Recently, one of my best friends, Joking, asked me to solve an integral $$\begin{equation}\int_0^1 \frac{x-1}{\log x}dx\,.\end{equation}$$

His intention is to show me a trick as in this video: Let $I(b)\equiv \int_0^1 \frac{x^b-1}{\log x}dx$. By taking derivative to $b$, we obtain an simple differential equation $I'(b)=\frac{1}{b+1}$. With $I(0)=0$, we solve $I(b)$ and thus have $I(1)=\log 2$.

However, my first attempt gave the answer of zero: Let $y=x^2$, $\int_0^1 \frac{x}{\log x}dx=\int_0^1 \frac{dx^2}{\log x^2}=\int_0^1 \frac{dy}{\log y}$. As a result, $\int_0^1 \frac{x-1}{\log x}dx=\int_0^1 \frac{x}{\log x}dx-\int_0^1 \frac{dx}{\log x}=\int_0^1 \frac{dy}{\log y}-\int_0^1 \frac{dx}{\log x}=0$. Clearly, the integral cannot be zero. So what's wrong here?

The reason is that this is an improper integral. Strictly speaking, we should solve $$\begin{equation} \lim_{\epsilon\rightarrow 0} \int_0^{1-\epsilon} \frac{x-1}{\log x}dx\,.\end{equation}$$ The limit here matters. With this limit, we end with $$\begin{equation} \int_0^{1-\epsilon} \frac{x-1}{\log x}dx= \int_0^{(1-\epsilon)^2} \frac{dy}{\log y}-\int_0^{(1-\epsilon)} \frac{dx}{\log x}=\int_{(1-\epsilon)}^{(1-\epsilon)^2}\frac{dx}{\log x}\,,\tag{1}\end{equation}$$ which is not necessarily zero.

To solve (1), we define $f(x)\equiv \int_{(1-\epsilon)}^x \frac{dt}{\log t}$. The question is to solve the value $f((1-\epsilon)^2)$. By Taylor expansion at $1-\epsilon$, we have

$$\begin{equation} f((1-\epsilon)^2)=\sum_{k=0}^{\infty} \frac{1}{k!} f^{(k)}(1-\epsilon)\left[(1-\epsilon)^2-(1-\epsilon)\right]^{k} = \sum_{k=0}^2 \frac{1}{k!} f^{(k)}(1-\epsilon) (-\epsilon)^k\,. \end{equation}$$

Because of the existence of $\epsilon^k$, we only have to compute $f^{(k)}(1-\epsilon)$ to the leading order $1/\epsilon^k$. Keeping this in mind, we can compute $f'(x)=\frac{1}{\log x}$, $f''(x)=-\frac{1}{(\log x)^2 x}$, $f^{(3)}(x)\approx\frac{2}{(\log x)^3 x^2}$, etc. So $f^{(k)}(x)\approx-\frac{(k-1)!}{(-\log x)^k x^{k-1}}$, and thus $f^{(k)}(1-\epsilon)\approx-\frac{(k-1)!}{(-\log(1-\epsilon))^k}=-\frac{(k-1)!}{\epsilon^k}$. Put it back to the Taylor expansion, we finally have $$\begin{equation} f((1-\epsilon)^2)=-\sum_{k=0}^{\infty}\frac{1}{k!} \frac{(k-1)!}{\epsilon^k} (-\epsilon)^k=-\sum_{k=0}^{\infty}\frac{(-1)^k}{k}=\log 2\,,\end{equation}$$ where we used the series $\sum_{k=0}^{\infty}{x^k}/{k}=-\log(1-x)$ in the last equal sign. The final answer is the same as that from the trick. Cheers!