一道累次积分小题

青青子衿

\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

回复 1# 青青子衿

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

Proof regarding double integral
math.stackexchange.com/questions/4015137