高数复习（四）
判断敛散性： \[ \int_a^{b} \frac{\mathrm{d} x}{\sqrt{(x-a)(b-x)}} \]
\[ \int_a^{\frac{a+b}{2}} \frac{\mathrm{d} x}{\sqrt{(x-a)(b-x)}}=\lim _{c \rightarrow 0^{+}} \int_{a+c}^{\frac{a b b}{2}} \frac{\mathrm{d} x}{\sqrt{\left(\frac{b-a}{2}\right)^2-\left(x-\frac{b+a}{2}\right)^2}} \]
\[ \text { 令 } x-\frac{b+a}{2}=\frac{b-a}{2} \sin \theta,~于是 \lim _{\varepsilon \rightarrow 0^{+}} \int_{\arcsin \frac{\sigma-(b-a) / 2}{(b-a) / 2}}^0 \mathrm{~d} \theta=\lim _{\delta \rightarrow 0^{+}}\left[-\arcsin \frac{\varepsilon-(b-a) / 2}{(b-a) / 2}\right]=\frac{\pi}{2} \]
又 \[ \lim_{\delta\rightarrow0^+}\int_{\frac{a+b}{2}}^{b-\delta} \frac{\mathrm{d} x}{\sqrt{(x-a)(b-x)}}=\lim _{\delta \rightarrow 0^{+}} \arcsin \frac{(b-a) / 2-\delta}{(b-a) / 2}=\frac{\pi}{2}, \]
故原积分收敛,且 \[ \begin{aligned} \int_a^b \frac{\mathrm{d} x}{\sqrt{(a-x)(b-x)}}= & \lim _{\varepsilon \rightarrow 0^{+}} \int_{a +\varepsilon}^{\frac{a+b}{2}} \frac{\mathrm{d} x}{\sqrt{(a-x)(b-x)}} \\ & +\lim _{\delta \rightarrow 0^{+}} \int_{\frac{a+b}{b}}^{b-\delta} \frac{\mathrm{d} x}{\sqrt{(a-x)(b-x)}}=\pi. \end{aligned} \]
广义积分判断准则： \[ 若0\leq f(x)\leq g(x),~当\int_{a}^{+\infty}g(x)\text{d}x收敛，则\int_{a}^{+\infty}f(x)\text{d}x收敛 \]
\[ 若0\leq f(x)\leq g(x),~当\int_{a}^{+\infty}f(x)\text{d}x发散，则\int_{a}^{+\infty}g(x)\text{d}x发散 \]
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