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战巡
发表于 2013-10-13 11:06
本帖最后由 战巡 于 2013-10-13 13:08 编辑 \[\int_{0}^{\pi}ln(1+sin(x))dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}ln(1+cos(x))dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}ln(2cos^2(\frac{x}{2})dx\]
\[=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}ln(2)dx+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}4ln(cos(x))dx={\pi}ln(2)+8\int_{0}^{\frac{\pi}{4}}ln(cos(x))dx\]
令$8\int_{0}^{\frac{\pi}{4}}ln(cos(x))dx=a$，有$a=8\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}ln(sin(x))dx$，则有
\[8\int_{0}^{\frac{\pi}{2}}ln(sin(x))dx-8\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}ln(sin(x))dx=-4{\pi}ln(2)-a=8\int_{0}^{\frac{\pi}{4}}ln(sin(x))dx\]
\[8\int_{0}^{\frac{\pi}{4}}ln(tan(x))dx=8\int_{0}^{\frac{\pi}{4}}[ln(sin(x))-ln(cos(x))]dx=-4{\pi}ln(2)-2a\]
\[8\int_{0}^{\frac{\pi}{4}}ln(tan(x))dx=8xln(tan(x))|_0^{\frac{\pi}{4}}-8\int_0^{\frac{\pi}{4}}xd[ln(tan(x))]=-8\int_0^{\frac{\pi}{4}}\frac{x}{sin(x)cos(x)}dx\]
\[=-8\int_0^{\frac{\pi}{4}}\frac{x}{tan(x)cos^2(x)}dx=-8\int_0^{\frac{\pi}{4}}\frac{x}{tan(x)}d[tan(x)]=-8\int_0^1\frac{arctan(x)}{x}dx\]
对这个泰勒展开得到
\[-8\int_0^1\frac{arctan(x)}{x}dx=-8\int_0^1\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2n+1}dx=-8\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}=-8G\]
其中$G$为卡塔兰常数
因此有
\[a=4G-2{\pi}ln(2)\]
\[\int_0^{\pi}ln(1+sin(x))dx={\pi}ln(2)+a=4G-{\pi}ln(2)\] |
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kuing
发表于 2013-10-18 23:30
