与椭圆积分有关的等式证明

(XPS001, 2017年拉丁美洲大学生数学竞赛第7题, Iberoamerican College Math Olympiad 2017, Problem 7)令$a<b<c<d$为实数,记
\[f(x)=\frac1{\sqrt{|x-a|\cdot|x-b|\cdot|x-c|\cdot|x-d|}},\]
求证:
\[\int_{a}^b f(x)dx=\int_c^df(x)dx.\]

 

(向禹)对于左边式子,令$x-a=(b-a)\sin^2\theta$,则有
\begin{align*}
\int_a^b{f\left( x \right) \text{d}x}&=\int_c^d{\frac{\text{d}x}{\sqrt{\left( x-a \right) \left( b-x \right) \left( c-x \right) \left( d-x \right)}}}\\
&=2\int_0^{\frac{\pi}{2}}{\frac{\text{d}\theta}{\sqrt{\left( c-b\sin ^2\theta -a\cos ^2\theta \right) \left( d-b\sin ^2\theta -a\cos ^2\theta \right)}}}\\
&=2\int_0^{\frac{\pi}{2}}{\frac{\text{d}\left( \tan \theta \right)}{\sqrt{\left( c\sec ^2\theta -b\tan ^2\theta -a \right) \left( d\sec ^2\theta -b\tan ^2\theta -a \right)}}}\\
&=2\int_0^{\infty}{\frac{\text{d}t}{\sqrt{\left( c\left( 1+t^2 \right) -bt^2-a \right) \left( d\left( 1+t^2 \right) -bt^2-a \right)}}}\\
&=2\int_0^{\infty}{\frac{\text{d}t}{\sqrt{\left( \left( c-b \right) t^2+\left( c-a \right) \right) \left( \left( d-b \right) t^2+d-a \right)}}}\\
&=\frac{2}{\sqrt{\left( c-b \right) \left( d-b \right)}}\int_0^{\infty}{\frac{\text{d}t}{\sqrt{\left( t^2+\frac{c-a}{c-b} \right) \left( t^2+\frac{d-a}{d-b} \right)}}}\\
&=\frac{2}{\sqrt{\left( c-b \right) \left( d-b \right)}}\int_0^{\frac{\pi}{2}}{\frac{\text{d}\varphi}{\sqrt{\frac{c-a}{c-b}\sin ^2\varphi +\frac{d-a}{d-b}\cos ^2\varphi}}}~~~\left( t=\sqrt{\frac{c-a}{c-b}}\tan \varphi \right)\\
&=2\int_0^{\frac{\pi}{2}}{\frac{\text{d}\varphi}{\sqrt{\left( c-a \right) \left( d-b \right) \sin ^2\varphi +\left( c-b \right) \left( d-a \right) \cos ^2\varphi}}}.
\end{align*}

对于右边式子,令$x-c=(d-c)\sin^2\theta$,则有
\begin{align*}
\int_c^d{f\left( x \right) \text{d}x}&=\int_c^d{\frac{\text{d}x}{\sqrt{\left( x-a \right) \left( x-b \right) \left( x-c \right) \left( d-x \right)}}}\\
&=2\int_0^{\frac{\pi}{2}}{\frac{\text{d}\theta}{\sqrt{\left( d\sin ^2\theta +c\cos ^2\theta -a \right) \left( d\sin ^2\theta +c\cos ^2\theta -b \right)}}}\\
&=2\int_0^{\frac{\pi}{2}}{\frac{\text{d}\left( \tan \theta \right)}{\sqrt{\left( d\tan ^2\theta +c-a\sec ^2\theta \right) \left( d\tan ^2\theta +c-b\sec ^2\theta \right)}}}\\
&=2\int_0^{\infty}{\frac{\text{d}t}{\sqrt{\left( dt^2+c-a\left( 1+t^2 \right) \right) \left( dt^2+c-b\left( 1+t^2 \right) \right)}}}\\
&=2\int_0^{\infty}{\frac{\text{d}t}{\sqrt{\left( \left( d-a \right) t^2+\left( c-a \right) \right) \left( \left( d-b \right) t^2+\left( c-b \right) \right)}}}\\
&=\frac{2}{\sqrt{\left( d-a \right) \left( d-b \right)}}\int_0^{\infty}{\frac{\text{d}t}{\sqrt{\left( t^2+\frac{c-a}{d-a} \right) \left( t^2+\frac{c-b}{d-b} \right)}}}\\
&=\frac{2}{\sqrt{\left( d-a \right) \left( d-b \right)}}\int_0^{\frac{\pi}{2}}{\frac{\text{d}\varphi}{\sqrt{\frac{c-a}{d-a}\sin ^2\varphi +\frac{c-b}{d-b}\cos ^2\varphi}}}~~~\left(t=\sqrt{\frac{c-a}{d-a}}\tan \varphi\right)\\
&=2\int_0^{\frac{\pi}{2}}{\frac{\text{d}\varphi}{\sqrt{\left( c-a \right) \left( d-b \right) \sin ^2\varphi +\left( c-b \right) \left( d-a \right) \cos ^2\varphi}}}=\int_a^bf(x)\mathrm dx.
\end{align*}

 

(XPS002)设$ f(x)=(x-a)(x-b)(x-c)(x-d),a>b>c>d>0 $.证明:
\begin{enumerate}
\item $\displaystyle\int_{a}^{b}\dfrac{1}{\sqrt{-f(x)}}dx=\displaystyle\int_{c}^{d}\dfrac{1}{\sqrt{-f(x)}}dx $;

\item $\displaystyle\int_{a}^{b}\dfrac{1}{\sqrt{-f(x)}}dx=-\dfrac{2}{\sqrt{(a-c)(b-d)}}F\left(\dfrac{\pi}{2},\sqrt{\dfrac{(a-b)(c-d)}{(a-c)(b-d)}}\right)$;

\item $\displaystyle\int_{u}^{b}\dfrac{1}{\sqrt{-f(x)}}dx=-\dfrac{2}{\sqrt{(a-c)(b-d)}}F\left(\arcsin{\sqrt{\dfrac{(a-c)(u-b)}{(a-b)(u-c)}}},\sqrt{\dfrac{(a-b)(c-d)}{(a-c)(b-d)}}\right) $,
\end{enumerate}
其中$F(\varphi,k)$为第一类不完全椭圆积分(incomplete elliptic integral of the first kind),定义成
\[F\left( \varphi ,k \right) =\int_0^{\varphi}{\frac{d\theta}{\sqrt{1-k^2\sin ^2\theta}}}.\]
参考: \url{http://en.wikipedia.org/wiki/Elliptic_integral}