# Question #23877

Aug 20, 2017

$I = {\int}_{0}^{1} \frac{{x}^{b} - {x}^{a}}{\ln \left(x\right)} \text{d} x = \ln \left(\frac{b + 1}{a + 1}\right)$.

#### Explanation:

$I = {\int}_{0}^{1} \frac{{x}^{b} - {x}^{a}}{\ln \left(x\right)} \text{d} x$

Make the substitution $u = - \ln \left(x\right)$. Then $\left(\text{d"u)/("d} x\right) = - \frac{1}{x}$, $\text{d"x=-x"d} u$. As $x \to 0$, $u \to + \setminus \infty$. And at $x = 1$, $u = 0$.

Then the integral is transformed to,

$I = {\int}_{\setminus \infty}^{0} \frac{{x}^{b} - {x}^{a}}{\ln \left(x\right)} \left(- x\right) \text{d} u$.

As $u = - \ln \left(x\right)$, $x = {e}^{- u}$.

Then,

$I = {\int}_{\setminus \infty}^{0} \frac{{e}^{- b u} - {e}^{- a u}}{- u} \left(- {e}^{- u}\right) \text{d} u$,
$I = {\int}_{\setminus \infty}^{0} \frac{{e}^{- \left(b + 1\right) u} - {e}^{- \left(a + 1\right) u}}{u} \text{d} u$.

Integrand can be negated and the limits of integration reversed.

$I = {\int}_{0}^{\setminus \infty} \frac{{e}^{- \left(a + 1\right) u} - {e}^{- \left(b + 1\right) u}}{u} \text{d} u$.

This is now, surprisingly, in a standard form. It is called Frullani's integral. It states that

${\int}_{0}^{\infty} \frac{f \left(a x\right) - f \left(b x\right)}{x} \text{d} x = \left(f \left(\infty\right) - f \left(0\right)\right) \ln \left(\frac{a}{b}\right)$

if $f$ is such that $f \left(\setminus \infty\right)$ and $f \left(0\right)$ exists. If we choose $f \left(x\right) = {e}^{- x}$, this is transformed into $I$, with $f \left(\setminus \infty\right) = 0$ and $f \left(0\right) = 1$.

We conclude,

${\int}_{0}^{\setminus \infty} \frac{{e}^{- \left(a + 1\right) u} - {e}^{- \left(b + 1\right) u}}{u} \text{d} u = - \ln \left(\frac{a + 1}{b + 1}\right)$,
${\int}_{0}^{\setminus \infty} \frac{{e}^{- \left(a + 1\right) u} - {e}^{- \left(b + 1\right) u}}{u} \text{d} u = \ln \left(\frac{b + 1}{a + 1}\right)$,
$I = \ln \left(\frac{b + 1}{a + 1}\right)$.