TITLE: How prove this integral equation $\int_{a}^{b}\frac{1}{\sqrt{|f(x)|}}dx=\int_{c}^{d}\frac{1}{\sqrt{|f(x)|}}dx$
QUESTION [13 upvotes]: let $a>b>c>d$,and
$$f(x)=(x-a)(x-b)(x-c)(x-d)$$
show that
$$\int_{a}^{b}\dfrac{1}{\sqrt{|f(x)|}}dx=\int_{c}^{d}\dfrac{1}{\sqrt{|f(x)|}}dx$$
my try: maybe let
$$u=x+( )$$
such when $x=a,b$ then $u=c,d$?
if $$a+d=b+c$$
then we take
$$y=a+d-x$$
then we have $$\int_{a}^{b}\dfrac{1}{\sqrt{|f(x)|}}dx=\int_{c}^{d}\dfrac{1}{\sqrt{|f(x)|}}dx$$
other case,
 I can't,Thank you

REPLY [8 votes]: You need a fractional linear (a.k.a. projective linear) change of variable
$$
x' = \frac{rx+s}{tx+u},
$$
with coefficients $r,s,t,u$ chosen so that the map $x \mapsto x'$
is an involution taking $a,b,c,d$ to $c,d,a,b$.  Explicitly you can take
$$
r = ac-bd, \quad
s = bcd - acd + abd - abc, \quad
t = a - b + c - d, \quad
u = bd-ac
$$
(note that $u = -r$, which is necessary and sufficient for a
fractional linear transformation to be its own functional inverse).
This transformation takes each of the integrals
$\int_a^b dx \, / \sqrt{|f(x)|}$ and $\int_c^d dx \, / \sqrt{|f(x)|}$ 
to the other, proving that they are equal.