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不清楚下面的写法是否正确,还望指教
首先有
\[n\int_0^1{\frac{x^n}{1+x^n}\text{d}x}=\int_0^1{\frac{nx^{n-1}\cdot x}{1+x^n}\text{d}x}
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=\int_0^1{\frac{x}{1+x^n}\text{d}\left( 1+x^n \right)}=\int_0^1{x\text{d}\ln \left( 1+x^n \right)}
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=x\ln \left( 1+x^n \right) \mid_{0}^{1}-\int_0^1{\ln \left( 1+x^n \right) \text{d}x}
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=\ln 2-\int_0^1{\ln \left( 1+x^n \right) \text{d}x}\]
对\(\forall c\in \left( 0,1 \right) \),有
\[0\leqslant \int_0^1{\ln \left( 1+x^n \right) \text{d}x}=\int_0^c{\ln \left( 1+x^n \right) \text{d}x}+\int_c^1{\ln \left( 1+x^n \right) \text{d}x}
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\leqslant \int_0^c{\ln \left( 1+c^n \right) \text{d}x}+\int_c^1{\ln \left( 1+1 \right) \text{d}x}
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=c\ln \left( 1+c^n \right) +\left( 1-c \right) \ln 2\rightarrow \left( 1-c \right) \ln 2,\left( n\rightarrow \infty \right) \]
为保证上式对\(\forall c\in \left( 0,1 \right) \)恒成立,则应该有
\[\int_0^1{\ln \left( 1+x^n \right) \text{d}x}\leqslant \left[ \left( 1-c \right) \ln 2 \right] _{\min}=0\]
由夹逼准则可知
\[\lim_{n\rightarrow \infty} \int_0^1{\ln \left( 1+x^n \right) \text{d}x}=0\]
所以有
\[\lim_{n\rightarrow \infty} n\int_0^1{\frac{x^n}{1+x^n}\text{d}x}=\ln 2-\lim_{n\rightarrow \infty} \int_0^1{\ln \left( 1+x^n \right) \text{d}x}=\ln 2\] |
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