Evaluate $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}$. [duplicate]

Solution 1:

Here is an explicit way to proceed, we are calculating the last integral in the OP, denoted here by $J$: $$ \begin{aligned} J &= \int_0^{1/2}\frac 4u\;\arcsin^2 u\; du\qquad\text{(Substitution: $u=\sin x$)}\\ &= \int_0^a\frac 4{\sin x}\;x^2\; \cos x\; dx\qquad\text{(where $a=\arcsin (1/2)=\pi/6$)}\\ &= 4 \int_0^a\frac x{\tan x}\;x\; dx\\ &= 4 \int_0^a\left(\frac {2ix}{e^{2ix}-1}+ix\right)\;x\; dx\\ &= 4\Re \int_0^a\frac {2ix}{e^{2ix}-1}\;x\; dx\\ &= -\Re \int_0^a\frac {2ix}{e^{2ix}-1}\;2ix\; d(2ix)\\ &= -\Re \int_0^{2ia}\frac {X^2}{e^X-1}\;dX\\ &= -\Re \left[ \ \color{gray}{ -\frac 13X^3} + X^2\log(1-e^X) + 2X\operatorname{Li}_2(e^X) - 2\operatorname{Li}_3(e^X) \ \right]_0^{2ia} \\ &= -\Re \left[ \ -a^2\log(1-e^{2ia}) + 4ia\operatorname{Li}_2(e^{2ia}) - 2\operatorname{Li}_3(e^{2ia}) + 2\operatorname{Li}_3(e^0) \ \right] \\ &= -\Re \left[ \ 0 + 4ia\operatorname{Li}_2(e^{2ia}) - 2\operatorname{Li}_3(e^{2ia}) + 2\zeta(3) \ \right] \\ &= -\Re \left[ \ 4ia\operatorname{Li}_2(b) - 2\operatorname{Li}_3(b) + 2\zeta(3) \ \right] \ . \end{aligned} $$ The logarithmic term disappeared because $$ \Re \log(1-e^{2ia}) = \log |1-e^{2ia}| = \log |1-e^{2i\pi/6}| = \log |e^{-2i\pi/6}| =\log 1 =0\ . $$ The value of $b$ is explicitly $$ b = e^{2ia} =e^{2i\pi/6} =e^{i\pi/3} \ , $$ so $b$ is the above primitive root of unity of order six.

One can show the relation: $$ \Re \operatorname{Li}_3(b) =\frac 13\operatorname{Li}_3(1) =\frac 13\zeta(3)\ . $$ This is seen for instance after a series expansion: $$ \begin{aligned} &2\Re\operatorname{Li}_3(b) = \operatorname{Li}_3(b) + \operatorname{Li}_3(\bar b) \\ &= \frac 1{1^3}\underbrace{(b+\bar b)}_{=1} + \frac 1{2^3}\underbrace{(b^2+\bar b^2)}_{=-1} + \frac 1{3^3}\underbrace{(b^3+\bar b^3)}_{=-2} + \frac 1{4^3}\underbrace{(b^4+\bar b^4)}_{=-1} + \frac 1{5^3}\underbrace{(b^5+\bar b^5)}_{=1} + \frac 1{6^3}\underbrace{(b^6+\bar b^6)}_{=2} + \dots\text{ with $6$-periodic repetitions} \\ &= \frac 1{1^3}(1) + \frac 1{2^3}(1-\color{blue}{2}) + \frac 1{3^3}(1-\color{darkgreen}{3}) + \frac 1{4^3}(1-\color{blue}{2}) + \frac 1{5^3}(1) + \frac 1{6^3}(1-\color{blue}{2}-\color{darkgreen}{3}+\color{red}{6}) + \dots\text{ with $6$-periodic repetitions} \\ &= \zeta(3)\left( 1 - \frac{\color{blue}{2}}{2^3} - \frac{\color{darkgreen}{3}}{3^3} + \frac{\color{red}{6}}{6^3} \right) \\ &= \zeta(3) \left( 1 - \frac{\color{blue}{2}}{2^3} \right) \left( 1- \frac1{\color{darkgreen}{3}}{3^3} \right) =\zeta(3)\cdot\frac 34\cdot\frac 89 =\zeta(3)\cdot\frac 23\ . \end{aligned} $$ In a similar manner we can (try to) look into the dilogarithmic part. $$ \begin{aligned} &2\Re i\operatorname{Li}_2(b) =\Re i( \operatorname{Li}_2(b) - \operatorname{Li}_2(\bar b)) \\ &= \frac i{1^2}\underbrace{(b-\bar b)}_{=i\sqrt 3} + \frac i{2^2}\underbrace{(b^2-\bar b^2)}_{=i\sqrt 3} + \frac i{3^2}\underbrace{(b^3-\bar b^3)}_{=0} + \frac i{4^2}\underbrace{(b^4-\bar b^4)}_{=-i\sqrt 3} + \frac i{5^2}\underbrace{(b^5-\bar b^5)}_{=-i\sqrt 3} + \frac i{6^2}\underbrace{(b^6-\bar b^6)}_{=0} + \dots\text{ with $6$-periodic repetitions} \\ &= \sqrt3\left( \frac1{1^2} + \frac 1{2^2} - \frac1{4^2} - \frac 1{5^2} + \dots \right)\text{ with $6$-periodic repetitions} \\ &= \frac {\sqrt3}{18} \left( \psi'\left(\frac 16\right) + \psi'\left(\frac 13\right) \right) -\frac {4 {\sqrt3}}{27}\pi^2 \ . \end{aligned} $$ We have a thus some formula for the expression involving corresponding special values $L(2,\chi)$ of the Dirichlet $L$-function computed in $2$ w.r.t. some characters $\chi$ with periodicity modulo six, and in general we can expect some $\psi$-values as in:

https://mathworld.wolfram.com/DirichletL-Series.html

We can do slightly better. Using $b^4=-b$ and the dilogarithmic relation $\operatorname{Li}_2(x)+\operatorname{Li}_2(-x)=\frac 12\operatorname{Li}_2(x^2)$, we obtain: $\operatorname{Li}_2(b)+\operatorname{Li}_2(b^4)=\frac 12\operatorname{Li}_2(b^2)$. Here, $b^2$ and $b^4$ are cubic primitive units, so the periodicity is reduced. Writing series for instance, we get $\operatorname{Li}_2(b^4)= -\operatorname{Li}_2(b^2)$. This gives: $$ \begin{aligned} \Im \operatorname{Li}_2(b) &=\frac 32 \Im \operatorname{Li}_2(b^2) \\ &= \frac {3\sqrt 3}4 \left( \frac1{1^2} - \frac 1{2^2} + \frac1{4^2} - \frac 1{5^2} + \dots \right)\text{ with $3$-periodic repetitions} \\ &= \frac {3\sqrt 3}4 \left( \frac 29\psi'\left(\frac 13\right) -\frac4{27}\pi^2 \right)\ . \end{aligned} $$ Here, for instance, the sum of the inverses of $1^2$, $4^2$, $7^2$, ... is $\frac 19\psi'\left(\frac 13 \right)$.

Putting all together: $$ \color{blue}{ \begin{aligned} J &= \frac 23\pi\cdot \Im \operatorname{Li}_2(b) - \frac 43\zeta(3) \\ &= \pi\cdot \Im \operatorname{Li}_2(b^2) - \frac 43\zeta(3) \ . \end{aligned} } $$ (And the dilogarithmic values have expressions in terms of $\psi'$ computed in $1/3$.)

Numerical check using pari/gp:

? \p 50
? J = intnum( u=0, 1/2, 4/u*asin(u)^2);
? b = exp(I*Pi/3);
? J
%158 = 0.52294619213333510849118518352730354016304459174398
? 2/3*Pi*imag(dilog(b)) - 4/3*zeta(3)
%159 = 0.52294619213333510849118518352730354016304459174398
?     Pi*imag(dilog(b^2)) - 4/3*zeta(3)
%160 = 0.52294619213333510849118518352730354016304459174398

Numerical check using sage:

sage: var('x,u,k');
sage: J = integral( 4/u * arcsin(u)^2, u, 0, 1/2, hold=True )
sage: J.n()
0.5229461921333352
sage: b = (1 + i*sqrt(3))/2
sage: ( 2/3 * pi * imag(dilog(b)) - 4/3 * zeta(3) ).n( digits=50 )
0.52294619213333510849118518352730354016304459174398
sage: (       pi * imag(dilog(b^2)) - 4/3 * zeta(3) ).n( digits=50 )
0.52294619213333510849118518352730354016304459174398

sage: imag( dilog(b) ).n( digits=50 )
1.0149416064096536250212025542745202859416893075303

sage: imag( 3/2 * dilog(b^2) ).n( digits=50 )
1.0149416064096536250212025542745202859416893075303

sage: ( sum( 1/(3*k+1)^2 - 1/(3*k+2)^2, k, 0, oo) * 3*sqrt(3)/4 ).n( digits=50 )
1.0149416064096536250212025542745202859416893075303

sage:   sum( 1/(3*k+1)^2 - 1/(3*k+2)^2, k, 0, oo)
-4/27*pi^2 + 2/9*psi(1, 1/3)

sage: var('X');
sage: integral( X^2 / (exp(X) - 1), X)
-1/3*X^3 + X^2*log(-e^X + 1) + 2*X*dilog(e^X) - 2*polylog(3, e^X)

(Please omit this coding section, if this feels misplaced, it was done only for my calm sleep, so that i can easily double check the computations next day with better eyes.)

Solution 2:

Hint: We find in Evaluations of Binomial Series by J. M. Borwein and R. Girgensohn the formula (47) \begin{align*} \sum_{n\geq 1}\frac{1}{n^3\binom{2n}{n}}=\frac{2}{3}\pi\Im\left(L_2\left(e^{i\pi/3}\right)\right) -\frac{4}{3}\zeta(3) \end{align*} with $L_p(z)=\sum_{n>0}\frac{z^n}{n^p}$ the polylogarithms.