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战巡
Posted at 2020-9-15 18:59:22
回复 1# 血狼王
\[\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{x^2\cos^2(\theta)+\sin^2(\theta)}}=\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-(1-x^2)\cos^2(\theta)}}\]
按照定义，这就是第一类椭圆积分，有
\[\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-(1-x^2)\cos^2(\theta)}}=\int_{\frac{\pi}{2}}^0\frac{d(\frac{\pi}{2}-\theta)}{\sqrt{1-(1-x^2)\cos^2(\frac{\pi}{2}-\theta)}}=\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-(1-x^2)\sin^2(\theta)}}=F(\frac{\pi}{2}|1-x^2)\]
注意
\[\lim_{x\to0^+}F(\frac{\pi}{2}|1-x^2)=\int_0^{\frac{\pi}{2}}\frac{d\theta}{\cos(\theta)}=+\infty\]
这里就可以直接洛必达了，有
\[\lim_{x\to0^+}\frac{F(\frac{\pi}{2}|1-x^2)}{\ln(x)}=\lim_{x\to0^+}x\cdot(-2x)\left[\frac{E(\frac{\pi}{2}|1-x^2)}{2(1-x^2)x^2}-\frac{F(\frac{\pi}{2}|1-x^2)}{2(1-x^2)}\right]\]
其中$E(\frac{\pi}{2}|1-x^2)$为第二类椭圆积分，有
\[E(\frac{\pi}{2}|1-x^2)=\int_0^{\frac{\pi}{2}}\sqrt{1-(1-x^2)\sin^2(\theta)}d\theta\]
然后原式变成
\[=\lim_{x\to0^+}x^2F(\frac{\pi}{2}|1-x^2)-E(\frac{\pi}{2}|1)\]
这其中
\[\lim_{x\to0^+}x^2F(\frac{\pi}{2}|1-x^2)=\lim_{x\to0^+}\int_0^{\frac{\pi}{2}}\frac{x^2d\theta}{\sqrt{1-(1-x^2)\sin^2(\theta)}}\]
而
\[E(\frac{\pi}{2}|1)=\int_0^{\frac{\pi}{2}}\cos(\theta)d\theta=1\]
这里
\[0<\lim_{x\to0^+}\frac{x^2}{\sqrt{1-(1-x^2)\sin^2(\theta)}}<\lim_{x\to0^+}\frac{x^2}{\sqrt{1-(1-x^2)}}=0\]
故此
\[\lim_{x\to0^+}x^2F(\frac{\pi}{2}|1-x^2)=\lim_{x\to0^+}\int_0^{\frac{\pi}{2}}\frac{x^2d\theta}{\sqrt{1-(1-x^2)\sin^2(\theta)}}=0\]
于是
\[原式=\lim_{x\to0^+}x^2F(\frac{\pi}{2}|1-x^2)-E(\frac{\pi}{2}|1)=0-1=-1\] |
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