Calculate the integral $\int_0^1 \frac{x-1}{\log(x)} \mathrm dx$

Question

Calculate the integral $$\int_0^1 \frac{x-1}{\log(x)} \mathrm dx$$ using $$\int_0^1 \frac{x^t-1}{\log(x)} \mathrm dx$$.

Answer 1

Using Mattos hint:

$$I(t) = \int_{0}^{1} \frac{x^{t} - 1}{\log(x)} dx$$

Then the derivative with respect to $t$ is

$$\frac{\partial}{\partial t}I(t)= \int_0^1 \frac{x^t\log x }{\log(x)} dx = \int_0^1x^t dx ... = \frac1{1+t}$$ Hope I did not made a mistake there. The inital value $$I(0) = \int_{0}^{1} \frac{x^{0} - 1}{\log(x)} dx=0$$

Then you can integrate to get $I(1)$ which is the requested integral:

$$I(1)=I(0)+ \int_0^1 \frac{\partial}{\partial t}I(t) dt= \int_0^1\frac1{1+t}dt=\log 2$$