Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Isn't it an improper integral? –  B. S. May 21 '13 at 2:05
7  
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." mathworld.wolfram.com/ImproperIntegral.html –  Liu Jin Tsai May 21 '13 at 2:10
add comment

1 Answer 1

up vote 18 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 $$\frac{1}{4i}\cdot2\pi\cdot \ln\frac{A_+}{A_-}=\frac{\pi^2}{6}.$$

share|improve this answer
    
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
1  
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. –  O.L. May 21 '13 at 16:55
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.