Suppose $aSuppose $a<b<c<d$ and $p(x)=(x-a)(x-b)(x-c)(x-d)$. Show that 
$$\int_a^b \frac{dx}{\sqrt{|p(x)|}} = \int_c^d \frac{dx}{\sqrt{|p(x)|}}.$$
My attempt: I perform linear substitution $u=x-a+c$ to change the limits but I'm unable to change it to the required limits. I will end up with the upper limit $b-a+c$, which requires $b+c=a+d$, which is not always true (I think). Can anyone guide me?
 A: 1. Complex Analysis Technique. Consider the function
$$q(z) = i\sqrt{z-a\vphantom{d}}\sqrt{z-b\vphantom{d}}\sqrt{z-c\vphantom{d}}\sqrt{z-d\vphantom{d}}$$
for $z \in \Bbb{C}\setminus([a,b]\cup[c,d])$, where the square root $\sqrt{z} = \exp(\frac{1}{2}\log z)$ means the principal square root. By noting that 
$$ \lim_{\epsilon \downarrow 0} \sqrt{-x+i\epsilon} = i\sqrt{x} \quad \text{for } x > 0, $$
we easily check that
$$ q(x+0^+i) := \lim_{\epsilon \downarrow 0} q(x+i\epsilon) = \begin{cases}
i \sqrt{|p(x)|}, & x \in (-\infty, a) \cup (d, \infty) \\
\sqrt{|p(x)|}, & x \in (a, b) \\
-i \sqrt{|p(x)|}, & x \in (b, c) \\
- \sqrt{|p(x)|}, & x \in (c, d).
\end{cases} $$
Consider an upper semicircular contour $C_{\epsilon, R}$ or radius $R$ with the center $(0, \epsilon)$. Then the Cauchy integration formula gives
$$ \oint_{C_{\epsilon, R}} \frac{dz}{q(z)} = 0. $$
By taking $R \to \infty$ and $\epsilon \downarrow 0$, it is not hard to check that
$$ 0 = \lim_{(\epsilon, R) \to (0, \infty)} \oint_{C_{\epsilon, R}} \frac{dz}{q(z)} = \int_{-\infty}^{\infty} \frac{dx}{q(x+0^+i)}. $$
By extracting the real part, we get the conclusion.
Addendum. This is a special case of the Schwarz-Christoffel mapping which sends the upper-half plane conformally into a polygonal region. In particular, the integral
$$ z \mapsto \int_{z_0}^{z} \frac{d\zeta}{q(\zeta)}, \quad \Im z, \Im z_0 > 0 $$
maps the upper-half plane into a rectangle, and the proposed identity corresponds to the fact that two opposite sides of a rectangle has the same length. This is also explained in Igor Rivin's answer in this answer.

2. Real Analysis Technique(?). Introduce the following Möbius transformation
$$\mu(t) = \frac{(ad-bc)t + (bcd-acd-abd+abc)}{(a-b-c+d)t - (ad-bc)}. $$
Then we can check (by brutal force or by any means that you would like) that


*

*$\mu(a) = d$, $\mu(b) = c$, $\mu(c) = b$, $\mu(d) = a$, and

*With the substitution $x = \mu(t)$,
$$ \frac{dx}{\sqrt{|p(x)|}} = -\frac{dt}{\sqrt{|p(t)|}}. $$


Therefore we have
$$ \int_{a}^{b} \frac{dx}{\sqrt{|p(x)|}} = -\int_{d}^{c} \frac{dt}{\sqrt{|p(t)|}} = \int_{c}^{d} \frac{dt}{\sqrt{|p(t)|}}. $$
A: If $a,b,c,d$ has a center of symmetry $\frac{a+d}{2}= \frac{b+c}{2}$ it's clear. Otherwise, we could consider a "generalized symmetry", an involution $x\leftrightarrow x'$ given by a symmetric equation 
$$u\, x x' + v\, (x+x') + w =0$$
Let's consider this particular case $a,b,c,d= 1,2,4,7$.  We want $1\leftrightarrow 7 $ and $2\leftrightarrow 4$ which gets us $(u:v:w) = (2:1:22)$.
The equality $u \,x x' + v \,(x+x') + w =0$ solved for $x'$ gets us 
$$x' = \iota(x)= \frac{- v\, x  - w}{u\, x + v}$$
which in our case gives $\iota(x)= \frac{-x + 22}{2 x +1}$. We now check that all fits perfectly: the form $\frac{(d x)^2 }{p(x)}$ is invariant under the substitution $x \mapsto \iota(x)$. Indeed, we have
\begin{eqnarray}
\frac{d \iota(x)}{dx}&=& - \frac{45}{(1+2 x)^2}\\
 (\iota(x)-1)(\iota(x)-2)(\iota(x)-4)(\iota(x)-7) &=& \frac{2025}{(2 x+1)^4} \cdot (x-1)(x-2)(x-4)(x-7)
\end{eqnarray}
both following from the  identity $\frac{\iota(x)-\iota(y)}{x-y} = -\frac{45}{(2x+1)(2y+1)}$
Hence $\frac{|dx|}{\sqrt{|p(x)|}}$ is invariant under $\iota$ and so
$$\int_1^2 \frac{dx}{\sqrt{|p(x)|}} = \int_1^2 \frac{|d\iota(x)|}{\sqrt{|p(\iota(x))|}}= \int_4^7 \frac{dx}{\sqrt{|p(x)|}} $$
$\bf{Added:}$
Some numerics: $\int_1^2\frac{dx}{\sqrt{-(x-1)(x-2)(x-4)(x-7)}}=0.8570..=\int_4^7\frac{dx}{\sqrt{-(x-1)(x-2)(x-4)(x-7)}} $.
What about $\int_2^4\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}}$ ? It's $1.1656..$. But note that $\int_{-\infty}^1\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}}=0.72316..$ and 
$\int_7^{\infty}\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}}=0.44245..$ so 
$$\int_2^4\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}} = \int_{-\infty}^1\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}}+ \int_7^{\infty}\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}}$$
the last sum should be thought of as an integral from $1$ to $7$ going around through $\infty$. The proof uses a similar involution. 
