TITLE: Evaluating $\int_{0}^{1}\left(\frac{\ln{(1+x)}}{1+x}\right)^n dx $
QUESTION [6 upvotes]: I wonder if this integral $$\int_{0}^{1}\left(\frac{\ln{(1+x)}}{1+x}\right)^n dx \quad n=1,2,3,...$$ admits a general formula for integers $n$. I've found $$\int_{0}^{1}\frac{\ln{(1+x)}}{1+x} dx = \left[ \frac{1}{2}\left(\ln(1+x)\right)^2 \right]_0^1=\frac{1}{2}\left(\ln2\right)^2. $$ Using integration by parts twice,
$$\begin{align}
\int_{0}^{1}\left(\frac{\ln{(1+x)}}{1+x}\right)^2 dx &= \left[ \frac{-1}{(1+x)}\left(\ln(1+x)\right)^2 \right]_0^1-\int_{0}^{1}\frac{\ln{(1+x)}}{\left(1+x\right)^2} dx \\
&=-\frac{1}{2}\left(\ln2\right)^2+ \left[ \frac{1}{(1+x)}\ln(1+x) \right]_0^1-\int_{0}^{1}\frac{1}{\left(1+x\right)^2} dx\\
&=-\frac{1}{2}\left(\ln2\right)^2+\frac{1}{2}\ln2-\frac{1}{2}.
\end{align}$$
Could you find a general formula? Any help is welcome! Thank you.

REPLY [2 votes]: \begin{align}\int^{1}_{0}\Big(\frac{\ln(1+x)}{1+x}\Big)^n\,dx&=\int^{\ln(2)}_{0}\Big(\frac{u}{e^u}\Big)^ne^u\,du\\&=\int^{\ln(2)}_{0}u^ne^{-(n-1)u}\,du=\frac{1}{(n-1)^{n+1}}\int^{(n-1)\ln(2)}_{0}v^ne^{-v}\,dv\\&=\frac{1}{(n-1)^{n+1}}\Big(\int^{\infty}_{0}v^ne^{-v}\,dv-\int^{\infty}_{(n-1)\ln(2)}v^ne^{-v}\,dv\Big)\\&=\frac{1}{(n-1)^{n+1}}\Big(\Gamma(n+1)-\Gamma(n+1,(n-1)\ln(2))\Big)\\&=\frac{n!}{(n-1)^{n+1}}-\frac{\Gamma(n+1,(n-1)\ln(2))}{(n-1)^{n+1}}\end{align}
where $\Gamma(n,x)$ is the upper incomplete gamma function.