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\begin{document}

{\bf Question}

Prove that $\ds
\int_0^1\frac{x^p}{1-x}\log\left(\frac{1}{x}\right)dx=
\sum_{n=0}^\infty\frac{1}{(p+n+1)^2}$


\vspace{0.25in}

{\bf Answer}

$\ds \int_0^1\frac{x^p}{1-x}\log\left(\frac{1}{x}\right)= \int_0^1
x^p\log\frac{1}{x}\sum_{n=0}^\infty x^n=\sum_{n=0}^\infty\int_0^1
x^{p+n}\log\frac{1}{x}$

$\ds = \sum_{n=0}^\infty\frac{1}{(p+n+1)^2},\,\,$ since

$\ds \int_0^1 x^\alpha\log\frac{1}{x}=
\frac{1}{\alpha+1}\left[x^{\alpha+1}\log\frac{1}{x}\right]_0^1
+\frac{1}{\alpha+1}\int_0^1 x^{\alpha+1}\frac{x}{x^2}=
\frac{1}{(\alpha+1)^2}$

\end{document}