I = int_0^1 (x^b-x^a)/(ln(x))"d"x
Make the substitution u=-ln(x). Then ("d"u)/("d"x)=-1/x, "d"x=-x"d"u. As x -> 0, u -> + \infty. And at x=1, u=0.
Then the integral is transformed to,
I = int_(\infty)^(0) (x^b-x^a)/(ln(x)) (-x) "d"u.
As u=-ln(x), x=e^(-u).
Then,
I= int_(\infty)^0 (e^(-bu)-e^(-au))/(-u) (-e^(-u)) "d"u,
I = int_(\infty)^(0) (e^(-(b+1)u)-e^(-(a+1)u))/(u) "d"u.
Integrand can be negated and the limits of integration reversed.
I = int_(0)^(\infty) (e^(-(a+1)u)-e^(-(b+1)u))/(u) "d"u.
This is now, surprisingly, in a standard form. It is called Frullani's integral. It states that
int_(0)^(infty) (f(ax)-f(bx))/(x) "d"x = (f(infty)-f(0))ln(a/b)
if f is such that f(\infty) and f(0) exists. If we choose f(x)=e^(-x), this is transformed into I, with f(\infty)=0 and f(0)=1.
We conclude,
int_(0)^(\infty) (e^(-(a+1)u)-e^(-(b+1)u))/(u) "d"u = -ln((a+1)/(b+1)),
int_(0)^(\infty) (e^(-(a+1)u)-e^(-(b+1)u))/(u) "d"u = ln((b+1)/(a+1)),
I = ln((b+1)/(a+1)).
