show this integral $$I=\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}=\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}$$ where $0<a,b<1$

my idea: let \begin{align*}&(-2ax+a^2+1)(-2bx+b^2+1)=4abx^2-2(a+b+a^2b+b^2a)x+(a^2+1)(b^2+1)\\ &=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a^2+1)(b^2+1)-\dfrac{4ab(a+b+a^2b+b^2a)^2}{4ab}\\ &=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a^2+1)(b^2+1)-(a+b+a^2b+b^2a)^2\\ &=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a+b)^2+(ab-1)^2+(a+b)^2(1+ab)^2 \end{align*} so I think this idea is not good, maybe this have good methods,because this reslut is nice

  • $\begingroup$ what kind of numbers are $a$ and $b$? $\endgroup$ – Dr. Sonnhard Graubner Dec 20 '14 at 15:21
  • $\begingroup$ @Dr.SonnhardGraubner $0<a,b<1$ on the 3rd line of the question... $\endgroup$ – user2345215 Dec 20 '14 at 15:28
  • 1
    $\begingroup$ Just an idea but, denoting the integral by $I(a,b)$ then, if you could just show that $I(a,b)=I(1,ab)$ this would imply $I(a,b)=I(\sqrt{ab},\sqrt{ab})$ and then the square root vanishes. $\endgroup$ – Myself Dec 20 '14 at 15:32

Notice for $|t| < 1$, $\frac{1}{\sqrt{1 - 2xt + t^2}}$ is the generating function for the Legendre polynomials:

$$\frac{1}{\sqrt{1 - 2xt + t^2}} = \sum_{n=0}^\infty P_n(x) t^n\tag{*1}$$

It is known that for $x \in [-1,1]$, $|P_n(x)| \le P_n(1) = 1$. This means as long as $t$ is fixed and $|t| < 1$, the absolute values of the $n^{th}$ term is bounded by a geometric series $|t|^n$. As a result, the convergence in $(*1)$ is absolute and uniform for $x$ over $[-1,1]$.

Recall the orthogonality relation for Legendre polynomials:

$$\int_{-1}^1 P_n(x) P_m(x) dx = \begin{cases} \frac{2}{2n+1}, & n = m\\ 0, & n \ne m\end{cases} $$ We find $$\begin{align} &\int_{-1}^1 \frac{dx}{\sqrt{1-2ax+a^2}\sqrt{1-2bx+b^2}}\\ =& \int_{-1}^1 \left(\sum_{n=0}^\infty P_n(x) a^n\right)\left(\sum_{m=0}^\infty P_m(x) b^m\right) dx\\ =& \sum_{n=0}^\infty \sum_{m=0}^\infty a^n b^m \int_{-1}^1 P_n(x) P_m(x) dx\\ =& \sum_{n=0}^\infty \frac{2}{2n+1} (ab)^n\\ =& \sum_{k=0}^\infty \frac{1}{(k+1)}\left( (\sqrt{ab})^k + (-\sqrt{ab})^k \right)\\ =& - \frac{ \log(1 - \sqrt{ab})}{\sqrt{ab}} + \frac{ \log(1 + \sqrt{ab})}{\sqrt{ab}}\\ =& \frac{1}{\sqrt{ab}}\log\left(\frac{1+\sqrt{ab}}{1-\sqrt{ab}}\right) \end{align} $$

  • $\begingroup$ 哇,It's very very nice!!+1+1+1 $\endgroup$ – china math Dec 20 '14 at 16:21
  • $\begingroup$ @china math You are Chinese?I am a Chinese student. Actually,the indefinite integral of this can be calculated.Certainly,achille hui's method is very skillful. $\endgroup$ – gcy-rolle Dec 21 '14 at 12:16

$$$$ Let us start to calculate it. \begin{eqnarray} I&=&\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}\\ &=&\frac{1}{2\sqrt{ab}}\int_{-1}^{1}\dfrac{dx}{\sqrt{\frac{a^2+1}{2a}-x}\sqrt{\frac{b^2+1}{2b}-x}}\\ &=&\frac{1}{2\sqrt{ab}}\int_{-1}^{1}\dfrac{dx}{\sqrt{m-x}\sqrt{n-x}}\\ &=&\frac{1}{\sqrt{ab}}\int_{-1}^{1}\dfrac{-d(\sqrt{n-x})}{\sqrt{(\sqrt{n-x})^2+m-n}}\\ &=&-\frac{1}{\sqrt{ab}}\ln(\sqrt{n-x}+\sqrt{m-x}\sqrt{n-x})|_{-1}^{1}\\ &=&\frac{1}{\sqrt{ab}}\frac{ln(\sqrt{n+1}+\sqrt{m+1}\sqrt{n+1})}{ln(\sqrt{n-1}+\sqrt{m-1}\sqrt{n-1})}\\ &=&\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}} \end{eqnarray} Where $m=\frac{a^2+1}{2a} \\ n=\frac{b^2+1}{2b}$

So we can get it.


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