TITLE: Evaluation of $\int^1_0\frac{x^b-x^a}{\ln x} d x=\ln{{\frac{1+b}{1+a}}}$
QUESTION [10 upvotes]: $$\int^1_0\frac{x^b-x^a}{\ln x} d x =\ln{{\frac{1+b}{1+a}}}$$ The  part inside $\ln$ is absolute value. A solution including integration under the integral can be found here. Which used the identity $$\int^1_0x^{\alpha} dx =\frac{1}{1+\alpha}$$. Multiplies both side by $d{\alpha}$ and integrates with respect to $\alpha$. Are there other methods to prove this identity?
Thank You.

REPLY [3 votes]: Let's change variables $x = \exp(-t)$:
$$
   \int_0^1 \frac{x^b-x^a}{\ln x} \mathrm{d}x = \int_0^\infty \frac{\mathrm{e}^{-(b+1)t}-\mathrm{e}^{-(a+1)t}}{t} \mathrm{d}t = \int_0^\infty \int_{b+1}^{a+1} \mathrm{e}^{-x t} \mathrm{d}x \mathrm{d}t
$$
Interchange the order of integration by Tonelli theorem:
$$
   \int_0^\infty \int_{b+1}^{a+1} \mathrm{e}^{-x t} \mathrm{d}x \mathrm{d}t = \int_{b+1}^{a+1}  \int_0^\infty  \mathrm{e}^{-x t}  \mathrm{d}t \mathrm{d}x = \int_{b+1}^{a+1}   \frac{\mathrm{d}x}{x} = \ln \frac{1+b}{1+a}
$$