Hello.

$\displaystyle \displaystyle \[I(p) = \int\limits_0^{ + \infty } {\frac{{{x^{p - 1}} \cdot \ln (x)}}{{x + 1}}dx} \]$

and

$\displaystyle \displaystyle \[\frac{d}{{dp}}I(p) = \int\limits_0^{ + \infty } {\frac{{{x^{p - 1}} \cdot {{(\ln (x))}^2}}}{{x + 1}}dx} \]$

seems useless to me.

So I tried substituting

$\displaystyle \displaystyle \[\ln (x + 1) = u\]$

which apparently gave me

$\displaystyle \displaystyle \[I(p) = \int\limits_0^{ + \infty } {{e^{(u - 1)(p - 1)}}(u - 1)du} \]$

I am not sure if this is really true. Moreover, I don't seem to be able to solve them (I actually expect to get an Euler's gamma function somehow).