Integrating : $\int_0^1 {\frac {x^a-x^b} {\ln x} dx}$ [duplicate]

We are given parameters $a > 0, b > 0$.
Task is to integrate that: $\displaystyle \int_0^1 {\frac {x^a-x^b} {\ln x} dx}$.
I have tried approaching problem from different angles with no luck. I tried integration by parts(tried all combinations of possible $v$ and $u$), u-substitution with no luck.
Also I tried to integrate this two similar terms separately.

Tried to get some idea of how to go from answer, got nice answer from MATLAB: $\displaystyle \ln{\frac{a+1}{b+1}}$, but no idea how to reach it.

I would appreciate some suggestions.

Solution 1:

First notice that: $$I = \int_{0}^{1} \frac{x^b-x^a}{\ln x} dx = \int_{0}^{1} \Big[\int_{a}^{b} x^y dy\Big]dx $$ The function $f(x,y)=x^y$ is continous in the set $[0,1]\times[a,b]$, therefore:

$$ I= \int_{0}^{1} \Big[\int_{a}^{b} x^y dy\Big]dx = \int_{a}^{b} \Big[\int_{0}^{1} x^y dx\Big]dy = \int_{a}^{b} \frac{1}{y+1}dy = \ln \Big(\frac{b+1}{a+1} \Big)$$

Solution 2:

If we let $t=-\ln x$ then, $dx =-e^{-t}dt$ and the result follows frullani's theorem $$ \int_{0}^{1} \frac{x^b-x^a}{\ln x} dx =- \int_{0}^{\infty} \frac{e^{-tb}-e^{-ta}}{t} e^tdt=-\int_{0}^{\infty} \frac{e^{-t(b+1)}-e^{-t(a+1)}}{t} dt= \ln\left(\frac{b+1}{a+1}\right)$$

Here is the Frullani's theorem :Proof of Frullani's theorem

Also see this https://math.stackexchange.com/q/2521520