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青青子衿 发表于 2020-12-20 15:14

一道累次积分小题

[i=s] 本帖最后由 青青子衿 于 2020-12-20 20:21 编辑 [/i]

\begin{align*}
\int_0^u\dfrac{{\mathrm{d}}v}{\sqrt{u-v}}\int_0^v\dfrac{f'(w)}{\sqrt{v-w}}{\mathrm{d}}w=\pi\Big[f(u)-f(0)\Big]
\end{align*}

战巡 发表于 2020-12-21 12:00

[b]回复 [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=37802&ptid=7564]1#[/url] [i]青青子衿[/i] [/b]

换个积分次序就完事了啊...
\[原式=\int_0^uf'(w)dw\int_w^u\frac{dv}{\sqrt{(u-v)(v-w)}}\]
\[=\int_0^uf'(w)dw\cdot\left[2\arctan(\frac{\sqrt{v-w}}{\sqrt{u-v}})|_w^u\right]\]
\[=\int_0^uf'(w)dw\left[2\lim_{v\to u^-}\arctan(\frac{\sqrt{v-w}}{\sqrt{u-v}})\right]\]
\[=\pi\int_0^uf'(w)dw=\pi[f(u)-f(0)]\]

青青子衿 发表于 2021-4-4 22:21

Proof regarding double integral
[url]https://math.stackexchange.com/questions/4015137[/url]

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