\( \int_{0}^{1}\int_{0}^{1}\frac{dx\,dy}{1-xy} \)
- Tolaso J Kos
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\( \int_{0}^{1}\int_{0}^{1}\frac{dx\,dy}{1-xy} \)
Prove that: \( \displaystyle \int_{0}^{1}\int_{0}^{1}\frac{dx\,dy}{1-xy}=\frac{\pi^2}{6} \).
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- Grigorios Kostakos
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Re: \( \int_{0}^{1}\int_{0}^{1}\frac{dx\,dy}{1-xy} \)
\begin{align*}
\displaystyle \int_{0}^{1}{\int_{0}^{1}{\frac{1}{1-xy}\,dy}\,dx}&= -\int_{0}^{1}{\frac{\log(1-x)}{x}\,dx}\\
&=-\int_{0}^{1}{\frac{1}{x}\biggl({-\mathop{\sum}\limits_{n=1}^{\infty}{\frac{x^n}{n}}}\biggr)\,dx}\\
&=\mathop{\sum}\limits_{n=1}^{\infty}\biggl({\frac{1}{n}\,\int_{0}^{1}{x^{n-1}\,dx}}\biggr)\\
&=\mathop{\sum}\limits_{n=1}^{\infty}{\frac{1}{n}\,\frac{1}{n}}\\
&=\mathop{\sum}\limits_{n=1}^{\infty}{\frac{1}{n^2}}\\
&=\zeta(2)\\
&=\frac{\pi^2}{6}\,.
\end{align*}
\displaystyle \int_{0}^{1}{\int_{0}^{1}{\frac{1}{1-xy}\,dy}\,dx}&= -\int_{0}^{1}{\frac{\log(1-x)}{x}\,dx}\\
&=-\int_{0}^{1}{\frac{1}{x}\biggl({-\mathop{\sum}\limits_{n=1}^{\infty}{\frac{x^n}{n}}}\biggr)\,dx}\\
&=\mathop{\sum}\limits_{n=1}^{\infty}\biggl({\frac{1}{n}\,\int_{0}^{1}{x^{n-1}\,dx}}\biggr)\\
&=\mathop{\sum}\limits_{n=1}^{\infty}{\frac{1}{n}\,\frac{1}{n}}\\
&=\mathop{\sum}\limits_{n=1}^{\infty}{\frac{1}{n^2}}\\
&=\zeta(2)\\
&=\frac{\pi^2}{6}\,.
\end{align*}
Grigorios Kostakos
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