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战巡
发表于 2019-4-26 00:52
回复 1# 青青子衿
首先，探查一下那个积分是怎么回事，令$y=\frac{x}{n}$
\[\int_0^n\frac{\arctan(\frac{x}{n})}{(1+x)(1+x^2)}dx=\int_0^1\frac{n\arctan(y)}{(1+ny)(1+n^2y^2)}dy\]
\[\lim_{n\to\infty}\int_0^1\frac{n\arctan(y)}{(1+ny)(1+n^2y^2)}dy=0\]
因此原极限就是个未定式，那上洛必达定理
\[\lim_{n\to\infty}n\int_0^n\frac{\arctan(\frac{x}{n})}{(1+x)(1+x^2)}dx=\lim_{n\to\infty}\frac{\frac{\arctan(1)}{(1+n)(1+n^2)}-\int_0^n\frac{x}{n^2}\frac{\frac{1}{1+\frac{x^2}{n^2}}}{(1+x)(1+x^2)}dx}{-\frac{1}{n^2}}\]
\[=\lim_{n\to\infty}n^2[\frac{2(n^2+1)\arctan(x)-4n\arctan(\frac{x}{n})-2(n^2-1)\ln(x+1)+\ln(x^2+1)+n^2\ln(x^2+1)-2\ln(x^2+n^2)}{4(n^4-1)}|_0^n-\frac{\arctan(1)}{(1+n)(1+n^2)}]\]
\[=\lim_{n\to\infty}\frac{n^2}{4(n^4-1)}[(2+n^2-2n)\pi-\ln(4)-2(n^2-1)\ln(1+n)+(n^2+1)\ln(1+n^2)]=\frac{\pi}{4}\] |
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