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对于任意 \(a>0\) 有
\[\int_0^a f(x)\,\mathrm{d}x=\int_0^a f(a-x)\,\mathrm{d}x\]
所以
\[\int_0^{\pi/2}\ln{(\sin{x})}\,\mathrm{d}x=\int_0^{\pi/2}\ln{(\cos{x})}\,\mathrm{d}x=-\frac{\pi}{2}\ln{2}\]
然后
\[I=\int_0^{\pi/2}x\ln{(\sin{x})}\,\mathrm{d}x=\int_0^{\pi/2}\left(\frac{\pi}{2}-x\right)\ln{(\cos{x})}\,\mathrm{d}x\]
所以
\[\begin{aligned}
2I
&=\int_0^{\pi/2}x\ln{(\sin{x})}\,\mathrm{d}x
+\int_0^{\pi/2}\left(\frac{\pi}{2}-x\right)\ln{(\cos{x})}\,\mathrm{d}x \\
&=\int_0^{\pi/2}x\ln{(\tan{x})}\,\mathrm{d}x
+\frac{\pi}{2}\int_0^{\pi/2}\ln{(\cos{x})}\,dx \\
&=\int_0^{\pi/2}x\ln{(\tan{x})}\,\mathrm{d}x-\frac{\pi^2}{4}\ln{2}
\end{aligned}\]
因为有
\[\int_0^{\pi/2}x\ln{(\tan{x})}\,\mathrm{d}x=\frac{7}{8}\,\zeta(3)\]
所以
\[I=\int_0^{\pi/2}x\ln{(\sin{x})}\,\mathrm{d}x=\frac{7}{16}\,\zeta(3)-\frac{\pi^2}{8}\ln{2}\]
即可得原结果
\[\frac{7}{8}\zeta(3)=\sum_{k=0}^\infty\frac{1}{(2k+1)^3}=\frac{\pi^2}{2}\ln{2}+2\int_0^{\pi/2}x\ln{(\sin{x})}\,\mathrm{d}x\] |
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