โจทย์ข้อ 13. สวยมากครับ สวยจนผมอดใจไม่อยู่จริงๆ
เนื่องจาก $$ \int_0^{\frac12} \{ \frac1x \}\{ \frac{1}{1-x} \} \, dx = \int_{\frac12}^1 \{ \frac{1}{1-y} \}\{ \frac1y \} \, dy \quad ,y=1-x $$ ดังนั้น $$ \int_0^1 \{ \frac1x \}\{ \frac{1}{1-x} \} \, dx = 2 \int_0^{ \frac12 } \{ \frac{1}{x} \}\{ \frac{1}{1-x} \} \, dx $$ $$ =2 \int_1^{\infty} \{u+1\}\{ \frac1u +1\} \frac{du}{(u+1)^2} \quad , u= \frac1x -1 $$ $$ =2 \int_1^{\infty} \{u\}\{ \frac1u \} \frac{du}{(u+1)^2} $$ $$= 2 \int_1^{\infty} \frac{u- \lfloor u \rfloor }{ u(u+1)^2 } \, du $$ $$ =2 \lim_{n \to \infty} \sum_{k=1}^n \left( \int_k^{k+1} \frac{u-k}{ u(u+1)^2 } \, du \right) $$ $$ =2 \lim_{n \to \infty} \sum_{k=1}^n \left( \frac{1}{k+2} + k\ln k + k\ln(k+2) - 2k\ln(k+1) \right) $$ $$ =2 \lim_{n \to \infty} \left( \left( \frac13 + \frac14 + \cdots + \frac{1}{n+2} - \ln(n+1) \right) + n\ln \left( \frac{n+2}{n+1} \right) \right) $$ $$ =2 \left( \left( \gamma -1 -\frac12 \right) +1 \right) = 2\gamma -1 $$
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