How to integrate logarithm and power function? I am trying to solve the following integral
$$\int_{0}^{1}(\ln(1+x))^2 x^{a-1}\,dx  ; a>0.$$
I tried using partial functions but that didn't lead to anything.
Any suggestion?
 A: The integral can be written as
$$ I =\int_{0}^{1}\!dx\,\ln^2(1+x) x^{a-1}= \frac{\partial^2}{\partial b^2} \int_0^1 \!dx\,(1+x)^b x^{a-1} \Bigr|_{b=0} .$$
The remaining integral is an Euler type hypergeometric integral and can be solve by expanding $(1+x)^b$ in a Maclaurin series
$$ (1+x)^b = \sum_{j=0}^\infty \binom{b}{j} x^j.$$
We obtain the result
$$ I =\frac{\partial^2}{\partial b^2} \sum_{j=0}^\infty \binom{b}{j} \frac{1}{a+j} \Bigr|_{b=0}.$$
We have that $$
\binom{b}{j}  = \frac{\Gamma(b+1)}{\Gamma(j+1) \Gamma(1+b-j)}$$
and $(d/dx)\ln \Gamma(x) =\psi(x)$ with $\psi(x)$ the digamma function. We obtain
$$\frac{\partial}{\partial b} \ln\binom{b}{j} = \psi(b+1) - \psi(1+b -j) ; $$
and thus
$$ I= \sum_{j=0}^\infty \binom{b}{j} \frac{ [\psi(b+1) - \psi(1+b -j)]^2
+ \psi'(b+1) - \psi'(1+b -j) }{a+j}  \Bigr|_{b=0}. \tag{1}$$
In a next step, we perform the limit $b\to 0$. For $j\in\mathbb{N}$, we have that $\binom{b}{j} = (-1)^{j+1}b/j +O(b^2)$. So  we need to find, the contribution $\propto b^{-1}$ in the second factor of (1). We know that at close to integer values $j\in\mathbb{N}$, $$\psi(b-j) = -\frac{1}{b} + (H_{j} -\gamma) + O(b)$$ (this, can be derived from $\psi(z+1) = \psi(z) +1/z$)
and $$\psi'(b-j) = \frac{1}{b^2} +O(1).$$
With that we obtain the limit ($j\in\mathbb{N}$)
$$ \lim_{b\to0} \binom{b}{j}\biggl\{ [\psi(b+1) - \psi(1+b -j)]^2
+ \psi'(b+1) - \psi'(1+b -j)\biggr\} = \frac{2 (-1)^j H_{j-1}}{j} $$
with the harmonic numbers $H_n  =\sum_{k=1}^n k^{-1}$.
In conclusion, we obtain
$$I = \sum_{j=1}^\infty \frac{2 (-1)^j H_{j-1}}{j(a+j)}.$$
A: $$\int\log^2(1+x) x^{a-1}\,dx$$
$u=\log^2(1+x),\;\;\;du=\frac{2\log(1+x)}{1+x}dx,\;\;\;v=\frac{x^a}{a},\;\;\;dv=x^{a-1}dx$
$$=\frac{x^a \log^2(1+x)}{a}-\frac{2}{a}\int\frac{x^a \log(1+x)}{1+x} dx$$
Looking at the antiderivative
$$\int\frac{x^a \log(1+x)}{1+x} dx$$
$u=1+x,\;\;\;du=dx$
$$=\int\frac{(u-1)^a \log(u)}{u} du$$
Even assuming that both $a$ and $u$ are positive ($u$ is clearly positive for all positive $x$) doesn't get rid of the hypergeometrics (although it allows one to be represented as a form of the Beta function). Unless I am missing something critical in that something cancels out when evaluating at the limits of the integral, I would guess that an elementary form of the antiderivative does not exist, and thus the integral you provide does not have a representation in terms of elementary functions.
