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$x-a=(b-a)sin^2 heta$,则有
egin{align*}
int_a^b{fleft( x ight) ext{d}x}&=int_c^d{frac{ ext{d}x}{sqrt{left( x-a ight) left( b-x ight) left( c-x ight) left( d-x ight)}}}\
&=2int_0^{frac{pi}{2}}{frac{ ext{d} heta}{sqrt{left( c-bsin ^2 heta -acos ^2 heta ight) left( d-bsin ^2 heta -acos ^2 heta ight)}}}\
&=2int_0^{frac{pi}{2}}{frac{ ext{d}left( an heta ight)}{sqrt{left( csec ^2 heta -b an ^2 heta -a ight) left( dsec ^2 heta -b an ^2 heta -a ight)}}}\
&=2int_0^{infty}{frac{ ext{d}t}{sqrt{left( cleft( 1+t^2 ight) -bt^2-a ight) left( dleft( 1+t^2 ight) -bt^2-a ight)}}}\
&=2int_0^{infty}{frac{ ext{d}t}{sqrt{left( left( c-b ight) t^2+left( c-a ight) ight) left( left( d-b ight) t^2+d-a ight)}}}\
&=frac{2}{sqrt{left( c-b ight) left( d-b ight)}}int_0^{infty}{frac{ ext{d}t}{sqrt{left( t^2+frac{c-a}{c-b} ight) left( t^2+frac{d-a}{d-b} ight)}}}\
&=frac{2}{sqrt{left( c-b ight) left( d-b ight)}}int_0^{frac{pi}{2}}{frac{ ext{d}varphi}{sqrt{frac{c-a}{c-b}sin ^2varphi +frac{d-a}{d-b}cos ^2varphi}}}~~~left( t=sqrt{frac{c-a}{c-b}} an varphi ight)\
&=2int_0^{frac{pi}{2}}{frac{ ext{d}varphi}{sqrt{left( c-a ight) left( d-b ight) sin ^2varphi +left( c-b ight) left( d-a ight) cos ^2varphi}}}.
end{align*}

对于右边式子,令$x-c=(d-c)sin^2 heta$,则有
egin{align*}
int_c^d{fleft( x ight) ext{d}x}&=int_c^d{frac{ ext{d}x}{sqrt{left( x-a ight) left( x-b ight) left( x-c ight) left( d-x ight)}}}\
&=2int_0^{frac{pi}{2}}{frac{ ext{d} heta}{sqrt{left( dsin ^2 heta +ccos ^2 heta -a ight) left( dsin ^2 heta +ccos ^2 heta -b ight)}}}\
&=2int_0^{frac{pi}{2}}{frac{ ext{d}left( an heta ight)}{sqrt{left( d an ^2 heta +c-asec ^2 heta ight) left( d an ^2 heta +c-bsec ^2 heta ight)}}}\
&=2int_0^{infty}{frac{ ext{d}t}{sqrt{left( dt^2+c-aleft( 1+t^2 ight) ight) left( dt^2+c-bleft( 1+t^2 ight) ight)}}}\
&=2int_0^{infty}{frac{ ext{d}t}{sqrt{left( left( d-a ight) t^2+left( c-a ight) ight) left( left( d-b ight) t^2+left( c-b ight) ight)}}}\
&=frac{2}{sqrt{left( d-a ight) left( d-b ight)}}int_0^{infty}{frac{ ext{d}t}{sqrt{left( t^2+frac{c-a}{d-a} ight) left( t^2+frac{c-b}{d-b} ight)}}}\
&=frac{2}{sqrt{left( d-a ight) left( d-b ight)}}int_0^{frac{pi}{2}}{frac{ ext{d}varphi}{sqrt{frac{c-a}{d-a}sin ^2varphi +frac{c-b}{d-b}cos ^2varphi}}}~~~left(t=sqrt{frac{c-a}{d-a}} an varphi ight)\
&=2int_0^{frac{pi}{2}}{frac{ ext{d}varphi}{sqrt{left( c-a ight) left( d-b ight) sin ^2varphi +left( c-b ight) left( d-a ight) cos ^2varphi}}}=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:
egin{enumerate}
item $displaystyleint_{a}^{b}dfrac{1}{sqrt{-f(x)}}dx=displaystyleint_{c}^{d}dfrac{1}{sqrt{-f(x)}}dx $;

item $displaystyleint_{a}^{b}dfrac{1}{sqrt{-f(x)}}dx=-dfrac{2}{sqrt{(a-c)(b-d)}}Fleft(dfrac{pi}{2},sqrt{dfrac{(a-b)(c-d)}{(a-c)(b-d)}} ight)$;

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