# Closed form for $\int_{-1}^1\frac{\ln\left(2+x\,\sqrt3\right)}{\sqrt{1-x^2}\,\left(2+x\,\sqrt3\right)^n}dx$

I'm trying to find a closed form for the following integral: $$\mathcal{J}(n)=\int_{-1}^1\frac{\ln\left(2+x\,\sqrt3\right)}{\sqrt{1-x^2}\,\left(2+x\,\sqrt3\right)^n}dx\tag1$$ I have conjectured values of $\mathcal{J}(n)$ (supported by numerical inegration) for some integer values of $n$: \begin{align}&\mathcal{J}(1)\stackrel?=-\pi\ln\left(\frac32\right),\\&\mathcal{J}(2)\stackrel?=-\pi\left(1+2\ln\left(\frac32\right)\right),\\&\mathcal{J}(3)\stackrel?=-\pi\left(\frac{15}4+\frac{11}2\ln\left(\frac32\right)\right),\\&\mathcal{J}(4)\stackrel?=-\pi\left(\frac{77}6+17\ln\left(\frac32\right)\right).\end{align}\tag2 These values suggest that a general form for $n\in\mathbb{N}$ is $$\mathcal{J}(n)\stackrel?=-\pi\left(a_n+b_n\ln\left(\frac32\right)\right),\tag3$$ where $a_n, b_n$ are some rational coefficients. Moreover, I conjecture that $$b_n\stackrel?={_2F_1}\left(1-n,n;\,1;\,-\frac12\right).\tag4$$

Are my conjectures true? Can we find a formula or recurrence relation for $a_n$? Can we find a general formula for $\mathcal{J}(z)$ for non-integer values of $z$?

• $$I(a)=\int_{-1}^1\Big(2+x\sqrt3\Big)^ad\big(\arcsin x\big)\quad\iff\quad\mathcal{I}(n)=I'(-n)$$ Commented Feb 27, 2014 at 9:08
• It appears that $\mathcal{J}(1/2)=0$. Commented Feb 27, 2014 at 20:26

Define: $$J_{{n}} \left( x,y,z \right) =\int _{-1}^{1}\!{\frac {\ln \left( xz+y \right) }{\sqrt {1-{x}^{2}} \left( xz+y \right) ^{n}}}{dx}\quad: n\in \mathbb{Z},\, 0\le z\le y \in \mathbb{R} \tag{1}$$ and note that:

\begin{aligned} {\frac {\partial }{\partial y}}J_{{n}} \left( x,y,z \right) &=\int _{-1 }^{1}\!{\frac {1}{\sqrt {1-{x}^{2}} \left( xz+y \right) ^{1+n}}}{dx}- nJ_{{1+n}} \left( x,y,z \right)\\ &=\frac{\left( -1 \right) ^{n}}{n!}{\frac {\partial ^{n}}{\partial {y}^{n }}}\int _{-1}^{1}\!{\frac {1}{\sqrt {-{x}^{2}+1} \left( xz+y \right) } }{dx}-nJ_{{1+n}} \left( x,y,z \right) \end{aligned} \tag{2} Now consider the integral in $(2)$: \begin{aligned} \int _{-1}^{1}\!{\frac {1}{\sqrt {-{x}^{2}+1} \left( xz+y \right) }}{d x}&=\frac{1}{2z}\,\int _{-\pi }^{\pi }\! \frac{1}{\left( \sin \left( \theta \right) +{ \frac {y}{z}} \right)}{d\theta} \,:\,x=\sin(\theta)\\ &={\frac {i}{2\sqrt {{y}^{2}-{z}^{2} }}}\oint\! \frac{1}{\left( w-\xi_{{1}} \right)}- \frac{1}{\left( w-\xi_{{2}} \right)}{dw} \end{aligned} \tag{3} with: $$\xi_{{1}}=i{\frac {y+\sqrt {{y}^{2}-{z}^{2}}}{z}},\quad\xi_{{2}}=i{\frac {y-\sqrt {{y}^{2}-{z}^{2}}}{z}},\quad w=e^{i\theta} \tag{4}$$ and so by the conditions in $(1)$ only $\xi_2$ is inside the unit circle contour and so, by the residue theorem: \begin{aligned} \int _{-1}^{1}\!{\frac {1}{\sqrt {-{x}^{2}+1} \left( xz+y \right) }}{d x}&={\frac {i}{2\sqrt {{y}^{2}-{z}^{2} }}}(- 2i\pi)={\frac {\pi}{\sqrt {{y}^{2}-{z}^{2} }}} \end{aligned} \tag{5} and thus $(2)$ becomes: $$J_{{1+n}} \left( x,y,z \right) =\frac{\pi}{n}\frac { \left( -1 \right) ^{n} \, }{n!}\frac {\partial ^{n}}{\partial {y}^{n}} \left( {\frac {1}{\sqrt {{y}^ {2}-{z}^{2}}}} \right)-\frac{1}{n}{\frac {\partial }{\partial y}} J_{{n}} \left( x,y,z \right) \tag{6}$$ Successive differentiation then yields: $$J_{{n}} \left( x,y,z \right) = \frac{\left( -1 \right) ^{n-1}}{(n-1)!}{\frac { \partial ^{n-1}}{\partial {y}^{n-1}}} \left( {\frac { \pi\left( \Psi \left( n \right) +\gamma \right) }{\sqrt {{y}^{2}-{z}^{2}}}}+J_{{ 1}} \left( x,y,z \right) \right) \tag{7}$$ where: $$\Psi\left( n \right) +\gamma=\sum_{m=1}^{n-1}\frac{1}{m} \tag{8}$$ It remains to begin the recursion, to do so we note: \begin{aligned} J_{{1}} \left( x,y,z \right) &=\int _{-1}^{1}\!{\frac {\ln \left( xz+y \right) }{\sqrt {-{x}^{2}+1} \left( xz+y \right) }}{dx}\\ &=\frac{1}{2}{\frac {\partial }{\partial y}}\int _{-1}^{1}\!{\frac { \ln^2 \left( xz+y \right) }{\sqrt {-{x}^{2}+1}}}{dx}\\ &=\frac{1}{4}{\frac {\partial }{\partial y}}\int _{-\pi }^{\pi }\! \ln^2 \left( \sin \left( \theta \right) z+y \right){d\theta} \end{aligned} \tag{9} we then use: $$\ln \left( \sin \left( \theta \right) z+y \right) =\ln \left( \frac{q z}{2} \right) -\sum _{l=1}^{\infty }\frac{1}{l} \left( {{\rm e}^{i\theta\,l}}+ \left( -1 \right) ^{l}{{\rm e}^{-i\theta\,l}} \right) \left( {\frac {i}{q}} \right) ^{l}\\ q=\sqrt {{\frac {{y}^{2}}{{z}^{2}}}-1}+{\frac {y}{z}}\ge1 \tag{10}$$ together with the Kronecker integral: $$\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{(n-m)i\theta}d\theta=\delta_{n,m} \tag{11}$$ to show that: \begin{aligned} \frac{1}{4}\int _{-\pi }^{\pi }\! \ln^{2} \left( \sin \left( \theta \right) z+y \right) {d\theta}=&\frac{\pi}{2}\ln^2 \left( \left( \sqrt {{\frac {{y}^{2}}{{z}^{2}}}-1}+{\frac {y}{z}} \right) \frac{z}{2} \right)\\& +\,\pi \mathrm{Li_2}\left( \left( {\frac {y}{z}}-\sqrt {{\frac {{y}^{2}}{{z}^{2}}}-1}\right)^{\!{2}} \right) \end{aligned} \tag{12} where $\mathrm{Li_2}$ is a polylogarithm, and thus after differentiating $(12)$ w.r.t $y$ and simplifying we get: $$J_{{1}} \left( x,y,z \right) = {\frac {\pi \, \left( \ln \left( 2 \right) -\ln \left( y+\sqrt {{y}^{2}-{z}^{2}} \right) +\ln \left( { y}^{2}-{z}^{2} \right) \right) }{\sqrt {{y}^{2}-{z}^{2}}}} \tag{13}$$ and finally:

$$\int _{-1}^{1}\!{\frac {\ln \left( xz+y \right) }{\sqrt {-{x}^{2}+1} \left( xz+y \right) ^{n+1}}}{dx}=\pi \frac{\left( -1 \right) ^{n}}{n!}{\frac {\partial ^{n}}{\partial {y}^{n}}} \left( {\frac {\Psi \left( n+1 \right) +\gamma+\ln \left( 2 \right) -\ln \left( y+\sqrt {{y}^{2}-{ z}^{2}} \right) +\ln \left( {y}^{2}-{z}^{2} \right) }{\sqrt {{y}^{2}- {z}^{2}}}} \right)$$ $$\tag{14}$$

Equation $(14)$ can be used to find closed form for arbitrary integer $n$ and arbitrary $0<z<y\in \mathbb{R}$. It confirms the ops's conjectured cases and gives the $n=5$ case for example as: $$\int _{-1}^{1}\!{\frac {\ln \left( 2+x\sqrt {3} \right) }{\sqrt {-{x} ^{2}+1} \left( 2+x\sqrt {3} \right) ^{5}}}{dx}=-\pi \, \left( {\frac { 4213}{96}}+{\frac {443}{8}}\,\ln \left( 3/2 \right) \right)$$ As an extension, it may be worth applying the generalized Leibniz formula to obtain a finite sum form; the $\log$ part could be pulled out and treated separately and it's derivatives would be rational functions of $\sqrt{y^2-z^2}$. We also note that $2^2-(\sqrt{3})^2=1$ which would presumably help.

Update

The Legendre P function for $x\in\mathbb{R}\, ,s\in\mathbb{C}$ is defined as: $$P_s\left( y \right) =\frac{1}{\pi}\int _{0}^{1}\!{\frac { \left( y+ \sqrt {{y}^{2}-1} \left( 2\,t-1 \right) \right) ^{s}}{\sqrt {t \left( 1-t \right) }}}{dt} \tag{i}$$ make the substitution $t=x/2+1/2$ and use the symmetry property that stems from the defining differential equation: $$P_{-s}(y)=P_{s-1}(y)\tag{ii}$$ differentiate w.r.t to the parameter $s$ and you have: $${\frac {\partial }{\partial s}}P_s \left( y \right) = -\frac{1}{\pi}\int _{-1}^{1}\!{\frac { \ln \left( y+\sqrt {{y}^{2}-1}x \right) }{\left( y+\sqrt {{y}^{2}-1}x \right)^{s+1}\sqrt {1-{x}^{2}}}}{dx}\tag{iii}$$ then using the result attributed to Bromwich, but which I took from this paper by Szmytkowski (who has written a few papers on differentiation w.r.t to the parameter) we get: $${\frac {\partial }{\partial s}}{P_s} \left( y \right)|_{s=n} ={P_n} \left( y \right) \ln \left( \frac{y+1}{2} \right) +\sum _{k=1}^{n}{\frac { \left( {P_k} \left( y \right) -P_{k-1} \left( y \right) \right) {P_{n-k} } \left( y \right) }{k}}\tag{iv}$$ equate $\mathrm{(iii)}$ and $\mathrm{(iv)}$ and evaluate at $y=2$ and we have:

$$\int _{-1}^{1}\!{\frac {\ln \left( 2+\sqrt {3}x \right) }{ \left( 2+ \sqrt {3}x \right) ^{n+1}\sqrt {1-{x}^{2}}}}{dx}=\\ -\pi\sum _{k=1}^{n}{\frac { \left( {P_k} \left( 2 \right) -P_{k-1} \left( 2 \right) \right) {P_{n-k} } \left( 2 \right) }{k}}-\pi{P_n} \left( 2 \right) \ln \left( \frac{3}{2} \right)\tag{v}$$ $$a_n=\sum _{k=1}^{n}{\frac { \left( {P_k} \left( 2 \right) -P_{k-1} \left( 2 \right) \right) {P_{n-k} } \left( 2 \right) }{k}},\quad b_n={P_n} \left( 2 \right),\quad a_n,b_n\in\mathbb{Q} \tag{vi}$$ and: $${P_{n-1}} \left(2 \right) = {\mbox{_2F_1}(n,-n+1;\,1;\,-1/2)}\tag{vii}$$ which confirms the conjecture.

with Maple's help, I get $$\mathcal{J}(n)= \frac{\pi}{{2}^{n}}\, \left[ {\mbox{_2F_1}\left(\frac{n}{2},\frac{n+1}{2};\,1;\frac{3}{4}\right)\log 2}-{\frac {d}{dn}}\; {\mbox{_2F_1}\left(\frac{n}{2},\frac{n+1}{2};\,1;\frac{3}{4}\right)} \right]$$

That derivative in there makes it somewhat unsatisfactory.

Note $${}_2F_1\left(\frac{n}{2},\frac{n+1}{2};1;\frac{3}{4}\right) = 2^n P_{n-1}(2) = 2^n\;{}_2F_1\left(n,-n+1;1;-\frac{1}{2}\right)$$ where $P_{n-1}$ is a Legendre function. This perhaps (?) justifies the expression for $b_n$ in the question. But these alternate expressions also presumably have no convenient expression for the derivative with respect to $n$.

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\cal J}\pars{n} \equiv \int_{-1}^{1} {\ln\pars{2 + x\root{3}} \over \root{1-x^{2}}\pars{2 + x\root{3}}^{n}}\,\dd x}$

Let's consider $\ds{{\cal K}\pars{\mu} = \int_{-1}^{1} {\pars{2 + x\root{3}}^{\mu} \over \root{1-x^{2}}}\,\dd x}$ such that $\ds{{\cal J}\pars{n} = \lim_{\mu\ \to\ -n}\totald{{\cal K}\pars{\mu}}{\mu}}$

\begin{align} {\cal K}\pars{\mu}&=\int_{-\pi/2}^{\pi/2}\bracks{\root{3}\sin\pars{\theta} + 2}^{\mu} \,\dd\theta =2^{\mu} \int_{-\infty}^{\infty}\pars{{\root{3} \over 2}\,{2t \over 1 + t^{2}} + 1}^{\mu}\,{2\,\dd t \over 1 + t^{2}} \\[3mm]&=2^{\mu + 1}\int_{-\infty}^{\infty} {\pars{t^{2} + \root{3}t + 1}^{\mu} \over \pars{1 + t^{2}}^{\mu + 1}}\,\dd t =2^{\mu + 1}\int_{-\infty}^{\infty} {\pars{t - w}^{\mu}\pars{t - w^{*}}^{\mu} \over \pars{t - \ic}^{\mu + 1}\pars{t + \ic}^{\mu + 1}}\,\dd t \end{align} where $\ds{w \equiv -\root{3} + \ic}$. The integral in the upper complex plane has two branch cut at $\ic$ and $w$. I guess it will be helpful.

• Dear @Felix Marin, why you have J(n)=lim…? Could the derivative of $K(\mu)$ product a $\ln\dots$?
– Lion
Commented Mar 2, 2014 at 23:55
• @Lion Yes. Note that $\displaystyle{\large {{\rm d}z^{\mu} \over {\rm d}\mu} = z^{\mu}\ln\left(z\right)}$. Commented Mar 2, 2014 at 23:59
• Is that not $\frac{dz^\mu}{d\mu}=\mu z^{(\mu-1)}$?
– Lion
Commented Mar 3, 2014 at 0:04
• I get it. My mistake. Please forgive my naivete.
– Lion
Commented Mar 3, 2014 at 0:13
• @Lion Any way it was helpful. Thanks. Commented Mar 3, 2014 at 0:14

Following Lucian's comment, Let's write $$I(n)=\int^1_{-1}(2+\sqrt3x)^{-n}\frac{dx}{\sqrt{1-x^2}}$$

so that $J(n)=-I'(n)$.

Mathematica calculates that $$I(n)=2^{-n}\pi~{}_2F_1(\frac{1+n}{2},\frac{n}{2};1;\frac34)\\ =(2/3)^n\pi~{}_2F_1(n,n;1;\frac13)\\ =\pi~{}_2F_1(n,1-n;1;-\frac12).$$ From DLMF 15.8.15 with $a=b=n,z=1/3$ and DLMF 15.8.1.

One should be able to get $a_n$ and $b_n$ from this.

Update 2014.03.08: From the above expression we know that $I(n)=I(1-n)$. Also, Using DLMF15.8.22, we can write $I_n$ in a form where $n$ only appears once in the paremeters. That is, we have $I(n)=\pi~{}_2F_1(n,1-n;1;-\frac12)=(2-\sqrt3)^n\pi~{}_2F_1(n,\frac12;1;4\sqrt3-6).$

Using the Contiguous Functions for 2F1 functions (DLMF 15.5.11), we know that $$n(4\sqrt3-7){}_2F_1(n+1,\frac12;1;4\sqrt3-6)+(2n-1+(\frac12-n)(4\sqrt3-6)){}_2F_1(n,\frac12;1;4\sqrt3-6)+(1-n){}_2F_1(n-1,\frac12;1;4\sqrt3-6)=0.$$ That is, $$-nI(n+1)+(4n-2)I(n)-(n-1)I(n-1)=0.$$ Taking derivative with respect to $n$, we have $$-I(n+1)+4I(n)-I(n-1)+(-nI'(n+1)+(4n-2)I'(n)-(n-1)I'(n-1))=0.$$ If we can somehow get a closed form for $-I(n+1)+4I(n)-I(n-1)$, we will have a recurrance relation for $I'(n)$. I think (and should be able to prove) that $-I(n+1)+4I(n)-I(n-1)=\pi~{}_2F_1(n,1-n;2;-\frac12)$, which will always be a rational multiple of $\pi$. That gives a possible recurrence relation concerning $b_n$: $$-nb_{n+1}+(4n-2)b_n-(n-1)b_{n-1}=0.$$

OP's original guess is that $b_n=I(n)$. We see here that $b_n$ and $I_n$ obey the same recurrence relation, so that's one step towards proving OP's claim.