Evaluation of $\int^1_0\frac{x^b-x^a}{\ln x} d x=\ln{{\frac{1+b}{1+a}}}$ $$\int^1_0\frac{x^b-x^a}{\ln x} d x =\ln{{\frac{1+b}{1+a}}}$$ 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.
 A: 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:
$$
   \int_0^\infty \int_{b+1}^{a+1} \mathrm{e}^{-x t} \mathrm{d}x \mathrm{d}t = \int_{b+1}^{a+1}  \int_0^\infty  \mathrm{e}^{-x t}  \mathrm{d}t \mathrm{d}x = \int_{b+1}^{a+1}   \frac{\mathrm{d}x}{x} = \ln \frac{1+b}{1+a}
$$
A: Define $$f(z) = \int^1_0 \frac{ x^{bz} - x^{az} }{\log x} dx.$$ Differentiating gives $$f'(z) = \int^1_0 \left(bx^{bz} - ax^{az} \right) dx = \frac{b}{bz+1} - \frac{a}{az+1}$$
and integrating this gives $$f(z) = \log(bz+1) - \log(az+1)+C$$ and setting $z=0$ indicates the constant is $0.$ So 
$$\int^1_0 \frac{x^b-x^a}{\log x}dx=f(1) = \log \frac{b+1}{a+1}.$$

The integrand is not defined over the interval of integration, so I assume you wanted $$\lim_{p\to 0} \int^1_p \frac{x^b-x^a}{\log x} dx.$$
Define $$ f(z) = \int^1_p \frac{x^{bz}-x^{az}}{\log x} dx.$$ Then $$f'(z) = \frac{b(1-p^{bz+1})}{bz+1} - \frac{a (1-p^{az+1})}{az+1} $$ so $$f(z) = \log \frac{bz+1}{az+1} + \frac{1}{a} \text{Ei}( (az+1)\log p ) - \frac{1}{b} \text{Ei}( (bz+1)\log p) + \frac{a-b}{ab} \text{Ei}(\log p)$$
where $\text{Ei}$ is the Exponential Integral and the constant was found by letting $z=0.$ Since $\lim_{x\to -\infty} \text{Ei}(x)=0$ the result follows. 
