# Closed form $\int_{-1}^{1} \frac{\ln (\sqrt{3} x +2)}{\sqrt{1-x^{2}} (\sqrt{3} x + 2)^{n}}\ dx$

Does the following integral $$\int_{-1}^{1} \frac{\ln (\sqrt{3} x +2)}{\sqrt{1-x^{2}} (\sqrt{3} x + 2)^{n}}\ dx, \; \; n \in \mathbb{N}$$

have a nice closed form? Basically I cannot tackle it in any direction. Symmetry is useless . Applyig parts , well , gets things worse than they actually are. Could complex analysis help us here? That is, integrating around a dog bone contour ... I highly doubt it but it is just a thought.

Any help on this one?

• Do you have reason to believe there is a closed form of answer? May 11, 2015 at 23:18
• Well, I actually do not have a specific reason. That is why I am asking the question. The original question I was trying to solve was: $$\int_{-1}^{1} \frac{\ln (\sqrt{3} x +2)}{\sqrt{1-x^{2}} (\sqrt{3} x + 2)^{2}}\ dx$$ and for this one I still have no progress. The value of this integral is: $$2\pi\left(\log2-\log3-\frac{1}{2} \right)$$ but I do not know how to get it. May 11, 2015 at 23:22
• By the way I also tried the Feymann trick by inserting a parameter $a>0$ in the integral, that is: $$\int_{-1}^{1} \frac{\ln (\sqrt{3} x a +2)}{\sqrt{1-x^{2}} (\sqrt{3} x + 2)^{2}}\ dx$$ but I think things get much more complicated. The derivative appears not have such a clear formula or it is very difficult by hand to get it. W|A cannot seem to find a form for the derivative. Let's see if someone has a clever idea. May 13, 2015 at 11:53
• This never hurts, @Cleo May 13, 2015 at 19:41

Let's integrate by $n$ first: $$K(n) = \int I(n) dn = \int \int_{-1}^1 \frac{\log(\sqrt{3}x+2)}{\sqrt{1-x^2}(\sqrt{3}x+2)^n} dx dn = -\int_{-1}^1 \frac{dx}{\sqrt{1-x^2}(\sqrt{3}x+2)^n} + C$$ Long path. (original) According to Wolfram Mathematica $$K(n) = -\int_{-1}^1 \frac{dx}{\sqrt{1-x^2}(\sqrt{3}x+2)^n} + C = -\frac{\pi}{2^n} \,{_2F_1}\left(\frac{n}{2},\frac{n+1}{2};1;\frac{3}{4}\right) + C.$$ So $$I(n) = K'(n) = \frac{\pi \log 2}{2^n} {_2F_1}\left(\frac{n}{2},\frac{n+1}{2};1;\frac{3}{4}\right) -\\ -\frac{\pi}{2^{n+1}}\left( \frac{\partial _2F_1}{\partial a}\left(\frac{n}{2},\frac{n+1}{2};1;\frac{3}{4}\right) + \frac{\partial _2F_1}{\partial b}\left(\frac{n}{2},\frac{n+1}{2};1;\frac{3}{4}\right) \right)$$ Let $$G(n,z) = \frac{1}{2^{n}}{_2F_1}\left(\frac{n}{2},\frac{n+1}{2};1;z\right)$$ Or rewriting in terms of Legendre functions (A&S 15.4.10) $$G(n,z) = \left(\frac{1}{4-4z}\right)^{n/2} P_{n-1}\left(\frac{1}{\sqrt{1-z}}\right)\\ G\left(n,\frac{3}{4}\right) = P_{n-1}\left(2\right)$$ Short path. Note that $$P_\lambda(x) = \frac{1}{2\pi}\int_{-\pi}^\pi (x + \sqrt{x^2 - 1}\cos \theta)^\lambda dx = \frac{1}{\pi}\int_0^\pi (x + \sqrt{x^2 - 1}\cos \theta)^\lambda dx.$$ And $$\int_{-1}^1 \frac{dx}{\sqrt{1-x^2}(\sqrt{3}x+2)^n} = \int_{0}^\pi \frac{d\theta}{(2+\sqrt{3}\cos \theta)^n} = \pi P_{-n}(2) = \pi P_{n-1}(2)$$

Next, following this paper and this paper $$\frac{\partial P_{\nu}(z)}{\partial \nu}\Big|_{\nu = n} = -P_n(z) \log\frac{z+1}{2} + \frac{1}{2^{n-1}n!}\frac{d^n}{dz^n} \left[ (z^2-1)\log \frac{z+1}{2} \right]$$ so $$\frac{\partial}{\partial n} G\left(n, \frac{3}{4}\right) = -P_{n-1}(2) \log\frac{3}{2} + \frac{1}{2^{n-2}(n-1)!}\frac{d^{n-1}}{dz^{n-1}} \left[ (z^2-1)^{n-1}\log \frac{z+1}{2} \right]_{z=2}$$ And finally $$I(n) = \pi P_{n-1}(2) \log\frac{3}{2} - \frac{\pi}{2^{n-2}(n-1)!}\frac{d^{n-1}}{dz^{n-1}} \left[ (z^2-1)^{n-1}\log \frac{z+1}{2} \right]_{z=2} = -\pi A_n \log\frac{3}{2} - \pi B_n$$ Here's a table for some $n$: $$\begin{array}{ccc} n & A_n & B_n \\\hline 0 & -1 & 0\\ 1 & 1 & 0 \\ 2 & 2 & 1 \\ 3 & \frac{11}{2} & \frac{15}{4} \\ 4 & 17 & \frac{77}{6} \\ 5 & \frac{443}{8} & \frac{4213}{96} \\ 6 & \frac{743}{4} & \frac{36353}{240} \\ 7 & \frac{10159}{16} & \frac{168833}{320} \\ 8 & \frac{17593}{8} & \frac{2074197}{1120} \\ 9 & \frac{984467}{128} & \frac{234465461}{35840} \\ 10 & \frac{1734443}{64} & \frac{3746664781}{161280} \\ \end{array}$$

Just a partial answer. I've found closed-form expressions for $n=0,\dots,4$.

$$I(n) := \int_{-1}^{1} \frac{\ln (\sqrt{3} x +2)}{\sqrt{1-x^{2}} (\sqrt{3} x + 2)^{n}}\ dx, \; \; n \in \mathbb{N}.$$

Then these are the following: \begin{align} I(0) &= -\pi\ln\left(\frac{2}{3}\right),\\ I(1) &= \pi\ln\left(\frac{2}{3}\right),\\ I(2) &= 2\pi\ln\left(\frac{2}{3}\right)-\pi,\\ I(3) &= \frac{11\pi}{2}\ln\left(\frac{2}{3}\right)-\frac{15\pi}{4},\\ I(4) &= 17\pi\ln\left(\frac{2}{3}\right)-\frac{77\pi}{6}. \end{align}

I've conjectured that the general form is $a\pi\ln(2/3)+b\pi$ for some rational $a,b$ constans, but with PSLQ algorithm I've found nothing more. After all, I think there's a chance to find closed-form for this integral.

• How have you calculated the integrals? W|A does not present a result, only a numerical approach, neither does Maple. May 15, 2015 at 17:02