#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))#.
