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一道积分收敛证明征解题

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Czhang271828 Posted at 2021-5-2 11:30:57 |Read mode
设$f$为$[0,\infty)$上的连续可微函数,同时$f(0) >0$、$f'(x) >0$。若
\[ \int_0^\infty\dfrac{\mathrm dx}{f'+f}<\infty \]
求证
\[ \int_0^\infty\dfrac{\mathrm dx}{f}<\infty \]
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abababa Posted at 2021-5-2 13:34:20
回复 1# Czhang271828
发网友的解答:
$f'(x)>0$说明$f(x)$是增函数,所以$f(x) \ge f(0) > 0$,即在积分区间上$f(x)$是正的,还有$\frac{1}{f(x)} \le \frac{1}{f(0)}$,当$x \to \infty$时也有$\frac{1}{f(x)} \le \frac{1}{f(0)}$有界。
\[\int_{0}^{\infty}\frac{1}{f}dx-\int_{0}^{\infty}\frac{1}{f'+f}dx = \int_{0}^{\infty}\frac{f'}{f^2+ff'}dx \le \int_{0}^{\infty}\frac{f'}{f^2}dx = [-\frac{1}{f}]_{0}^{\infty} < \frac{1}{f(0)}\]
移项就知道可积了。

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 Author| Czhang271828 Posted at 2021-5-2 14:13:40
技巧性很棒,谢谢!

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青青子衿 Posted at 2021-5-5 13:45:27
Last edited by 青青子衿 at 2021-5-6 09:08:00回复 1# Czhang271828
Mark一下
PKU数学分析II第五题
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2025-4-21 14:20 GMT+8

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