$\int _{ 0 }^{ 1 }{ \frac { { x }^{ t }-1 }{ \ln { x } } dx } $

Question

How do I solve the following integral:

$$\int _{ 0 }^{ 1 }{ \frac { { x }^{ t }-1}{ \ln { x } } dx } $$

Answer 1

Hint The integral satisfies

$$\frac{d}{dt} \int_0^1 \frac{x^t - 1}{\log x} dx = \int_0^1 \frac{d}{dt} \left(\frac{x^t - 1}{\log x}\right) dx = \int_0^1 x^t \,dx$$

Answer 2

By replacing $x$ with $e^{-u}$, we have:

$$ I(t)=\int_{0}^{1}\frac{x^t-1}{\log x}\,dx = \int_{0}^{+\infty}\frac{e^{-u}-e^{-u(t+1)}}{u}\,du = \color{red}{\log(t+1)}\tag{1}$$ by Frullani's theorem. Another chance is given by differentiation under the integral sign: $$ \frac{d}{dt}\,I(t) = \int_{0}^{1}x^t\,dx =\frac{1}{t+1}.\tag{2}$$