Answer: The Integral That Absorbs a Parameter

Key Idea / Intuition

The trick is to notice that $\frac{x^a - x^b}{\ln x}$ looks like it wants to be an integral itself: differentiate $x^t = e^{t \ln x}$ with respect to $t$ and you get $x^t \ln x$, so dividing by $\ln x$ "undoes" that differentiation. This is classic Feynman differentiation under the integral sign: introduce a parameter $t$, differentiate in $t$ to kill the $\ln x$ in the denominator, evaluate the resulting elementary integral, then integrate back in $t$.

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Formal Proof / Solution

Step 1: Introduce a parameter.

Write $x^a - x^b = \int_b^a x^t \ln x\,dt$ (since $\frac{d}{dt}x^t = x^t \ln x$). Therefore:

$I = \int_0^1 \frac{1}{\ln x}\int_b^a x^t \ln x\,dt\,dx = \int_0^1 \int_b^a x^t\,dt\,dx.$

The $\ln x$ factors cancel perfectly.

Step 2: Switch the order of integration.

By Fubini (the integrand $x^t$ is non-negative and integrable for $t > -1$, $x \in (0,1)$):

$I = \int_b^a \left(\int_0^1 x^t\,dx\right)dt.$

Step 3: Inner integral.

$\int_0^1 x^t\,dx = \frac{1}{t+1}, \qquad t > -1.$

Step 4: Outer integral.

$I = \int_b^a \frac{1}{t+1}\,dt = \Big[\ln(t+1)\Big]_b^a = \ln\!\left(\frac{a+1}{b+1}\right).$

Result:

$\boxed{I = \ln\!\left(\frac{a+1}{b+1}\right).}$

Sanity check: When $a = b$, the integrand is identically $0$, and indeed $\ln\!\left(\frac{a+1}{a+1}\right) = 0$. ✓

When $a = 1, b = 0$:

$\int_0^1 \frac{x-1}{\ln x}\,dx = \ln 2,$

a classical result. ✓