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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

3 Answers3

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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.

Noam D. Elkies
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The general integral represents the Schwartz-Christoffel transformation of the upper halfplane to a rectangle, and the two integrals represent the differences between the images of the points $a$ and $b$ and $c$ and $d$ respectively. Since the opposite sides of a rectangle are equal, the two integral is the same (I am not sure why @Braindead, who is clearly aware of this fact, post this as an answer, but someone should :)

Igor Rivin
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  • I didn't want to bring in Complex Analysis into this. I think there is a simpler way to show this. I rescind my previous comment about not using change of variables, because I think OP's observation is right on the money. – Braindead Jan 06 '14 at 19:13
  • @Braindead there is probably a more elementary answer, but it would be inspired by complex analysis (for example, the right change of variables is PROBABLY the one that sends $a$ to $0,$ $b$ to $1,$ $c$ to infinity, and $d$ to where ever it goes, and that seeing what happens, but this is a very complex analytic thing to do). – Igor Rivin Jan 06 '14 at 19:16
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NOTE: I read the inequalities backwards, as $a<b<c<d$. Maybe that's what you meant too?

So here are some "elementary" methods that don't involve (too much) complex analysis. My naive suspicion is that a simple linear transformation will not suffice, and that you actually need to use fractional linear transformation:

  1. Using a fractional linear transformation, you can actually map the interval $(a,b)$ to $(c,d)$ and $(c,d)$ to $(a,b)$ simultaneously with the right orientation. If you imagine that the real number line is joined as a circle with the point at infinity, it's sort of like a 'rotation' (with some distortion, of course.)

  2. You can map $a,b,c,d$ into $a',b',c',d'$ so that $b'-a' = d'-c'$, which is just another form of your equation $a'+d' = b'+c'$.

I personally like approach 2, but it all depends on what level of prior knowledge you have, and what sort of techniques you are willing / allowed to use.

I'll leave the detailed computation to you, but the motivation for using this method and also just "knowing" why the equation holds is essentially complex analysis (see @IgorRivin's answer).

Note2: The first method requires you to verify that the particular linear fractional transformation keeps the integrand invariant. So you probably need to explicitly compute the coefficients.

Note3: The second method has the advantage that you don't actually need to explicitly compute the linear fractional transformation. Both integrals $\int_a^b$ and $\int_c^d$ will transform into $\int_{a'}^{b'}$ and $\int_{c'}^{d'}$ with a common integrand. You still need to verify a few things though. Again, I'll let you think about it.

Braindead
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