I have:
I(a)= the integral from 0 to 1 of (x^(a)-x^(b))/(ln(x)) where a>b>-1.
How would I show that I'(a)=1/(a+1)?
We have:
$\displaystyle
I(a)= \int_0^1 \frac{x^a-x^b}{\ln(x)}dx= \int_0^1 \frac{x^a}{\ln(x)}dx- \int_0^1 \frac{x^b}{\ln(x)}dx
$
so:
$\displaystyle
\frac{d}{da}I(a)= \frac{d}{da}\int_0^1 \frac{x^a}{\ln(x)}dx= \frac{d}{da}\int_0^1 \frac{e^{a \ln (x)}}{\ln(x)}dx
$
Now differentiating under the integral gives:
$\displaystyle
\frac{d}{da}I(a)= \int_0^1 \ln(x) \frac{e^{a \ln (x)}}{\ln(x)}dx=\int_0^1 x^a dx = \left[ \frac{x^{a+1}}{a+1} \right]_0^1=\frac{1}{a+1}
$
RonL