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I need to solve the to following integral: $$\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\arctan\frac{11-6\,x}{4\,\sqrt{21}}\mathrm dx.$$

I tried this integral in Mathematica, but it was not able to solve it. An approximate numeric integration gives $1.6449340668482264364724151666460251892189...$ that is close to $\frac{\pi^2}6$. But when I tried to increase the precision above 60 decimal digits, I began to see a tiny difference, which could be interpreted either as a numerical algorithm glitch, or as $\frac{\pi^2}6$ being just an accidentally close value and not the exact answer. Indeed, $\frac{\pi^2}6$ would be a suspiciously nice result for this integral. Anyway, I need your help with this.

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Isn't it an improper integral? – Babak S. May 21 '13 at 2:05
Yes, it is. "An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration." – Liu Jin Tsai May 21 '13 at 2:10
up vote 21 down vote accepted

I think the most ecological approach to the problem is as follows:

  1. Denote $x=\sin\varphi$ and recall that $\arctan x=\frac{1}{2i}\ln\frac{1+ix}{1-ix}$. One then obtains the integral $$\frac{1}{2i}\int_{-\pi/2}^{\pi/2}\ln\frac{4\sqrt{21}+i(11-6\sin\varphi)}{4\sqrt{21}-i(11-6\sin\varphi)}d\varphi=\frac{1}{4i}\int_{0}^{2\pi}\ln\frac{4\sqrt{21}+i(11-6\cos\varphi)}{4\sqrt{21}-i(11-6\cos\varphi)}d\varphi,\tag{1}$$ where at the last step we first used the symmetry of sine function to extend the integration to interval $[-\pi,\pi]$, and then made use of periodicity to shift the integration interval and to replace $\sin$ by $\cos$.

  2. There is a well-known integral (see, for example, here) $$\int_{0}^{2\pi}\ln\left(1+r^2-2r\cos\varphi\right)d\varphi=\begin{cases} 0, &\text{for}\; |r|<1,\\ 2\pi\ln r^2, &\text{for}\; |r|>1. \end{cases}\tag{2}$$ Obviously, our integral above is a difference of two integrals of this type. Some care should be taken over multivaluedness of logarithms. This can be handled by saying that the arguments of $4\sqrt{21}\pm i(11-6\cos\varphi)$ belong to $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

  3. Now we have \begin{align} 4\sqrt{21}\pm i(11-6\cos\varphi)=A_{\pm}\left(1+r_{\pm}^2-2r_{\pm}\cos\varphi\right) \end{align} with $$r_{\pm}=\frac{11-4\sqrt7}{3}e^{\pm i\pi/3},\qquad A_{\pm}=(11+4\sqrt7)e^{\pm i\pi /6}.$$ Since $|r_{\pm}|<1$, the integral (1) reduces to $$\frac{1}{4i}\cdot2\pi\cdot \ln\frac{A_+}{A_-}=\frac{\pi^2}{6}.$$

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Nice answer, thanks! So, there is nothing "magical" about $\sqrt{21}$, right? Is it possible to find a general formula with a parameter(s) such that this integral would be its special case? – Liu Jin Tsai May 21 '13 at 16:42
Yes, exactly. One could put into $\arctan$ any fraction $\frac{ax+b}{cx+d}$ and would still be able to obtain the solution. However, keeping parameters arbitrary, it may become more tedious to choose logarithm branches correctly. – Start wearing purple May 21 '13 at 16:55

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