How to show that the integral of$ \frac{1}{x-x\log x}\mathrm{d}x $ from $ 0$ to $1$ diverges?

Question

I have a problem regarding

$$\int_{0^+}^1 \frac{1}{x-x\log x}\mathrm{d}x.$$

I think the integral does not converge. How can I show this?

Answer 1

The integral does not converge. It's easy to prove it, but consider also this: use

$$x \to e^t \qquad \qquad \text{d}x = e^t\ \text{d}t$$

hence we have

$$\int_{0^+}^1 \frac{1}{x - x\ln(x)}\ \text{d}x \rightarrow \int_{-\infty}^0 \frac{1}{e^t - t e^t} e^t\ \text{d}t = \int_{-\infty}^0 \frac{\text{d}t}{1-t} \to \text{does not converge}$$

Even if we tried with Cauchy principal value, we would obtain $\infty$.

Consider indeed

$$\lim_{b\to -\infty} \lim_{a\to 0} \int_b^a \frac{\text{d}t}{1-t} = -\ln(1-t)\bigg|_b^a = \ln\left(\frac{1}{1-a} - \frac{b}{1-a}\right) \to \ln(1-b) \to +\infty$$