Evaluating the sum of the series $sumlimits_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} $
$begingroup$
Wolfram MathWorld states that $$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Bigg[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg]- frac{4}{3} zeta(3) , , $$
where $psi_{1}(x)$ is the trigamma function.
I can't seem to get the answer in that form.
Using the Taylor expansion $ displaystyle arcsin^{2}(x) = frac{1}{2} sum_{n=1}^{infty} frac{1}{n^{2} binom{2n}{n}} (2x)^{2n}$, I get
$$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx. $$
Then I integrating by parts, I get
$$ begin{align} &4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx \ &= frac{pi^{2}}{9} ln left(frac{1}{2} right) - 8 int_{0}^{frac{1}{2}} frac{arcsin (x) ln (x)}{sqrt{1-x^{2}}} , dx \ &= - frac{pi^{2}}{9} ln 2 - 8 int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 ln 2 int_{0}^{frac{pi}{6}} u , du - 8int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) , du \ &= -8 text{Re} int_{0}^{frac{pi}{6}} u ln (1-e^{2iu}) , du \ &= 8 text{Re} int_{0}^{frac{pi}{6}} u sum_{n=1}^{infty} frac{e^{2in u}}{n} , du \ &= 8 sum_{n=1}^{infty} frac{1}{n} int_{0}^{frac{pi}{6}} u cos (2nu) , du \ &= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3) \ &= frac{2 pi}{3} Bigg( frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{2}} + frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{2}} Bigg) \ &+ 2 Bigg( frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{3}} - sum_{n=0}^{infty} frac{1}{(6n+3)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{3}} \ &+ frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{3}} + sum_{n=1}^{infty} frac{1}{(6n)^{3}} Bigg) - 2 zeta(3) \ &= frac{pi sqrt{3}}{108} Bigg( psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) Bigg) + frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &-28 zeta(3) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) + 4 zeta(3)Bigg) - 2 zeta (3) .end{align}$$
From here I've been going in circles trying to get the answer in the form given above.
EDIT:
Using the duplication formula for the trigamma function (i.e., $ displaystyle 4 psi_{1}(2x) = psi_{1}(x) + psi_{1} left(x + frac{1}{2} right) $),
$$ begin{align} &psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) \ &= 4 psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) - 4 psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) \ &= 6 psi_{1} left(frac{1}{3} right) - 6 psi_{1} left(frac{2}{3} right). end{align}$$
Therefore,
$$ begin{align} sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} &= frac{sqrt{3} pi}{18} Bigg( psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg) +
frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &+ psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) - 2 zeta(3)Bigg) - 2 zeta (3). end{align}$$
So it comes down to somehow showing that $$-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) = 312 zeta(3) .$$
SECOND EDIT:
Using the duplication formula for $psi_{2}(x)$, we get
$$ begin{align} &-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) \ &= -8 psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) - 8 psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) \ &= -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) . end{align}$$
And using the more general multiplication formula, we get
$$ psi_{2} left(frac{1}{3} right) + psi_{2} left( frac{2}{3} right) + psi_{2}(1) = 27 psi_{2} (1) .$$
Therefore,
$$ -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) = -156 psi_{2} (1) = 312 zeta(3) .$$
sequences-and-series complex-analysis special-functions binomial-coefficients gamma-function
edited Dec 25 '18 at 10:30
Did
249k23227466
asked Sep 20 '13 at 21:20
Random VariableRandom Variable
25.6k173139
$endgroup$
add a comment |
$begingroup$
Wolfram MathWorld states that $$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Bigg[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg]- frac{4}{3} zeta(3) , , $$
where $psi_{1}(x)$ is the trigamma function.
I can't seem to get the answer in that form.
Using the Taylor expansion $ displaystyle arcsin^{2}(x) = frac{1}{2} sum_{n=1}^{infty} frac{1}{n^{2} binom{2n}{n}} (2x)^{2n}$, I get
$$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx. $$
Then I integrating by parts, I get
$$ begin{align} &4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx \ &= frac{pi^{2}}{9} ln left(frac{1}{2} right) - 8 int_{0}^{frac{1}{2}} frac{arcsin (x) ln (x)}{sqrt{1-x^{2}}} , dx \ &= - frac{pi^{2}}{9} ln 2 - 8 int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 ln 2 int_{0}^{frac{pi}{6}} u , du - 8int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) , du \ &= -8 text{Re} int_{0}^{frac{pi}{6}} u ln (1-e^{2iu}) , du \ &= 8 text{Re} int_{0}^{frac{pi}{6}} u sum_{n=1}^{infty} frac{e^{2in u}}{n} , du \ &= 8 sum_{n=1}^{infty} frac{1}{n} int_{0}^{frac{pi}{6}} u cos (2nu) , du \ &= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3) \ &= frac{2 pi}{3} Bigg( frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{2}} + frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{2}} Bigg) \ &+ 2 Bigg( frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{3}} - sum_{n=0}^{infty} frac{1}{(6n+3)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{3}} \ &+ frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{3}} + sum_{n=1}^{infty} frac{1}{(6n)^{3}} Bigg) - 2 zeta(3) \ &= frac{pi sqrt{3}}{108} Bigg( psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) Bigg) + frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &-28 zeta(3) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) + 4 zeta(3)Bigg) - 2 zeta (3) .end{align}$$
From here I've been going in circles trying to get the answer in the form given above.
EDIT:
Using the duplication formula for the trigamma function (i.e., $ displaystyle 4 psi_{1}(2x) = psi_{1}(x) + psi_{1} left(x + frac{1}{2} right) $),
$$ begin{align} &psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) \ &= 4 psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) - 4 psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) \ &= 6 psi_{1} left(frac{1}{3} right) - 6 psi_{1} left(frac{2}{3} right). end{align}$$
Therefore,
$$ begin{align} sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} &= frac{sqrt{3} pi}{18} Bigg( psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg) +
frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &+ psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) - 2 zeta(3)Bigg) - 2 zeta (3). end{align}$$
So it comes down to somehow showing that $$-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) = 312 zeta(3) .$$
SECOND EDIT:
Using the duplication formula for $psi_{2}(x)$, we get
$$ begin{align} &-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) \ &= -8 psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) - 8 psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) \ &= -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) . end{align}$$
And using the more general multiplication formula, we get
$$ psi_{2} left(frac{1}{3} right) + psi_{2} left( frac{2}{3} right) + psi_{2}(1) = 27 psi_{2} (1) .$$
Therefore,
$$ -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) = -156 psi_{2} (1) = 312 zeta(3) .$$
sequences-and-series complex-analysis special-functions binomial-coefficients gamma-function
edited Dec 25 '18 at 10:30
Did
249k23227466
asked Sep 20 '13 at 21:20
Random VariableRandom Variable
25.6k173139
$endgroup$
$begingroup$
Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:15
$begingroup$
That sum is the alternating version of this one.
$endgroup$
– Random Variable
Sep 21 '13 at 1:31
$begingroup$
And the techniques that apply to the alternating version don't work for the one you ask about?
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:41
$begingroup$
They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
$endgroup$
– Random Variable
Sep 21 '13 at 1:56
$begingroup$
use Clausen function
$endgroup$
– alexjo
Nov 13 '13 at 21:08
add a comment |
$begingroup$
Wolfram MathWorld states that $$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Bigg[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg]- frac{4}{3} zeta(3) , , $$
where $psi_{1}(x)$ is the trigamma function.
I can't seem to get the answer in that form.
Using the Taylor expansion $ displaystyle arcsin^{2}(x) = frac{1}{2} sum_{n=1}^{infty} frac{1}{n^{2} binom{2n}{n}} (2x)^{2n}$, I get
$$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx. $$
Then I integrating by parts, I get
$$ begin{align} &4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx \ &= frac{pi^{2}}{9} ln left(frac{1}{2} right) - 8 int_{0}^{frac{1}{2}} frac{arcsin (x) ln (x)}{sqrt{1-x^{2}}} , dx \ &= - frac{pi^{2}}{9} ln 2 - 8 int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 ln 2 int_{0}^{frac{pi}{6}} u , du - 8int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) , du \ &= -8 text{Re} int_{0}^{frac{pi}{6}} u ln (1-e^{2iu}) , du \ &= 8 text{Re} int_{0}^{frac{pi}{6}} u sum_{n=1}^{infty} frac{e^{2in u}}{n} , du \ &= 8 sum_{n=1}^{infty} frac{1}{n} int_{0}^{frac{pi}{6}} u cos (2nu) , du \ &= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3) \ &= frac{2 pi}{3} Bigg( frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{2}} + frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{2}} Bigg) \ &+ 2 Bigg( frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{3}} - sum_{n=0}^{infty} frac{1}{(6n+3)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{3}} \ &+ frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{3}} + sum_{n=1}^{infty} frac{1}{(6n)^{3}} Bigg) - 2 zeta(3) \ &= frac{pi sqrt{3}}{108} Bigg( psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) Bigg) + frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &-28 zeta(3) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) + 4 zeta(3)Bigg) - 2 zeta (3) .end{align}$$
From here I've been going in circles trying to get the answer in the form given above.
EDIT:
Using the duplication formula for the trigamma function (i.e., $ displaystyle 4 psi_{1}(2x) = psi_{1}(x) + psi_{1} left(x + frac{1}{2} right) $),
$$ begin{align} &psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) \ &= 4 psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) - 4 psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) \ &= 6 psi_{1} left(frac{1}{3} right) - 6 psi_{1} left(frac{2}{3} right). end{align}$$
Therefore,
$$ begin{align} sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} &= frac{sqrt{3} pi}{18} Bigg( psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg) +
frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &+ psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) - 2 zeta(3)Bigg) - 2 zeta (3). end{align}$$
So it comes down to somehow showing that $$-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) = 312 zeta(3) .$$
SECOND EDIT:
Using the duplication formula for $psi_{2}(x)$, we get
$$ begin{align} &-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) \ &= -8 psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) - 8 psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) \ &= -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) . end{align}$$
And using the more general multiplication formula, we get
$$ psi_{2} left(frac{1}{3} right) + psi_{2} left( frac{2}{3} right) + psi_{2}(1) = 27 psi_{2} (1) .$$
Therefore,
$$ -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) = -156 psi_{2} (1) = 312 zeta(3) .$$
sequences-and-series complex-analysis special-functions binomial-coefficients gamma-function
edited Dec 25 '18 at 10:30
Did
249k23227466
asked Sep 20 '13 at 21:20
Random VariableRandom Variable
25.6k173139
$endgroup$
Wolfram MathWorld states that $$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Bigg[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg]- frac{4}{3} zeta(3) , , $$
where $psi_{1}(x)$ is the trigamma function.
I can't seem to get the answer in that form.
Using the Taylor expansion $ displaystyle arcsin^{2}(x) = frac{1}{2} sum_{n=1}^{infty} frac{1}{n^{2} binom{2n}{n}} (2x)^{2n}$, I get
$$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx. $$
Then I integrating by parts, I get
$$ begin{align} &4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx \ &= frac{pi^{2}}{9} ln left(frac{1}{2} right) - 8 int_{0}^{frac{1}{2}} frac{arcsin (x) ln (x)}{sqrt{1-x^{2}}} , dx \ &= - frac{pi^{2}}{9} ln 2 - 8 int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 ln 2 int_{0}^{frac{pi}{6}} u , du - 8int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) , du \ &= -8 text{Re} int_{0}^{frac{pi}{6}} u ln (1-e^{2iu}) , du \ &= 8 text{Re} int_{0}^{frac{pi}{6}} u sum_{n=1}^{infty} frac{e^{2in u}}{n} , du \ &= 8 sum_{n=1}^{infty} frac{1}{n} int_{0}^{frac{pi}{6}} u cos (2nu) , du \ &= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3) \ &= frac{2 pi}{3} Bigg( frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{2}} + frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{2}} Bigg) \ &+ 2 Bigg( frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{3}} - sum_{n=0}^{infty} frac{1}{(6n+3)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{3}} \ &+ frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{3}} + sum_{n=1}^{infty} frac{1}{(6n)^{3}} Bigg) - 2 zeta(3) \ &= frac{pi sqrt{3}}{108} Bigg( psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) Bigg) + frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &-28 zeta(3) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) + 4 zeta(3)Bigg) - 2 zeta (3) .end{align}$$
From here I've been going in circles trying to get the answer in the form given above.
EDIT:
Using the duplication formula for the trigamma function (i.e., $ displaystyle 4 psi_{1}(2x) = psi_{1}(x) + psi_{1} left(x + frac{1}{2} right) $),
$$ begin{align} &psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) \ &= 4 psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) - 4 psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) \ &= 6 psi_{1} left(frac{1}{3} right) - 6 psi_{1} left(frac{2}{3} right). end{align}$$
Therefore,
$$ begin{align} sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} &= frac{sqrt{3} pi}{18} Bigg( psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg) +
frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &+ psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) - 2 zeta(3)Bigg) - 2 zeta (3). end{align}$$
So it comes down to somehow showing that $$-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) = 312 zeta(3) .$$
SECOND EDIT:
Using the duplication formula for $psi_{2}(x)$, we get
$$ begin{align} &-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) \ &= -8 psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) - 8 psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) \ &= -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) . end{align}$$
And using the more general multiplication formula, we get
$$ psi_{2} left(frac{1}{3} right) + psi_{2} left( frac{2}{3} right) + psi_{2}(1) = 27 psi_{2} (1) .$$
Therefore,
$$ -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) = -156 psi_{2} (1) = 312 zeta(3) .$$
sequences-and-series complex-analysis special-functions binomial-coefficients gamma-function
sequences-and-series complex-analysis special-functions binomial-coefficients gamma-function
edited Dec 25 '18 at 10:30
Did
249k23227466
asked Sep 20 '13 at 21:20
Random VariableRandom Variable
25.6k173139
edited Dec 25 '18 at 10:30
Did
249k23227466
asked Sep 20 '13 at 21:20
Random VariableRandom Variable
25.6k173139
edited Dec 25 '18 at 10:30
Did
249k23227466
edited Dec 25 '18 at 10:30
Did
249k23227466
edited Dec 25 '18 at 10:30
Did
249k23227466
249k23227466
asked Sep 20 '13 at 21:20
Random VariableRandom Variable
25.6k173139
asked Sep 20 '13 at 21:20
Random VariableRandom Variable
25.6k173139
asked Sep 20 '13 at 21:20
Random VariableRandom Variable
25.6k173139
25.6k173139
$begingroup$
Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:15
$begingroup$
That sum is the alternating version of this one.
$endgroup$
– Random Variable
Sep 21 '13 at 1:31
$begingroup$
And the techniques that apply to the alternating version don't work for the one you ask about?
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:41
$begingroup$
They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
$endgroup$
– Random Variable
Sep 21 '13 at 1:56
$begingroup$
use Clausen function
$endgroup$
– alexjo
Nov 13 '13 at 21:08
add a comment |
$begingroup$
Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:15
$begingroup$
That sum is the alternating version of this one.
$endgroup$
– Random Variable
Sep 21 '13 at 1:31
$begingroup$
And the techniques that apply to the alternating version don't work for the one you ask about?
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:41
$begingroup$
They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
$endgroup$
– Random Variable
Sep 21 '13 at 1:56
$begingroup$
use Clausen function
$endgroup$
– alexjo
Nov 13 '13 at 21:08
$begingroup$
Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:15
$begingroup$
Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:15
$begingroup$
That sum is the alternating version of this one.
$endgroup$
– Random Variable
Sep 21 '13 at 1:31
$begingroup$
That sum is the alternating version of this one.
$endgroup$
– Random Variable
Sep 21 '13 at 1:31
$begingroup$
And the techniques that apply to the alternating version don't work for the one you ask about?
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:41
$begingroup$
And the techniques that apply to the alternating version don't work for the one you ask about?
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:41
$begingroup$
They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
$endgroup$
– Random Variable
Sep 21 '13 at 1:56
$begingroup$
They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
$endgroup$
– Random Variable
Sep 21 '13 at 1:56
$begingroup$
use Clausen function
$endgroup$
– alexjo
Nov 13 '13 at 21:08
$begingroup$
use Clausen function
$endgroup$
– alexjo
Nov 13 '13 at 21:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using the Taylor expansion for $arcsin^2(x)$ and integrating, it's easy to show (as you made) that
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&= 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} operatorname{d}x = - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) operatorname{d}u \
&= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3)\
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)
end{align}
$$
recalling that the Clausen function are defined as
$$
operatorname{Cl}_{m}(theta)=
begin{cases}displaystyle
sum_{n=1}^{infty} frac{sin (ntheta)}{n^{m}}& text{for } mtext{ even}\
displaystylesum_{n=1}^{infty} frac{cos(ntheta)}{n^{m}}& text{for } mtext{ odd}
end{cases}
$$
The value of the Clausen function $operatorname{Cl}_3$ at $frac{pi}{3}$ is
$$
operatorname{Cl}_3left(frac{pi}{3}right)=frac{1}{2}left(1-2^{-2}right)left(1-3^{-2}right)zeta(3)=frac{1}{3}zeta(3)
$$
From the duplication formula
$$
operatorname{Cl}_{2m}(2theta)=2^{2m-1}left[operatorname{Cl}_{2m}(theta)-operatorname{Cl}_{2m}(pi-theta)right]
$$
we find
$$
operatorname{Cl}_{2}left(frac{2pi}{3}right)=frac{2}{3}operatorname{Cl}_{2}left(frac{pi}{3}right).
$$
From the identities for the trigamma function at $1/3$ and $2/3$
$$
begin{align}
psi_1left(frac{1}{3}right) &=frac{2pi^2}{3}+3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)\
psi_1left(frac{2}{3}right) &=frac{2pi^2}{3}-3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
end{align}
$$
one has
$$
psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)=6sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
$$
and then
$$
operatorname{Cl}_{2}left(frac{pi}{3}right)=frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]
$$
Finally, putting all together, we have
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
&=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
&=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
end{align}
$$
answered Nov 13 '13 at 3:09
alexjoalexjo
12.5k1430
$endgroup$
add a comment |
$begingroup$
$newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
$ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n{2n choose n}}
=sum_{n = 1}^{infty}x^{n},
{Gammapars{n}Gammapars{n + 1} over Gammapars{2n + 1}}
=sum_{n = 1}^{infty}x^{n}int_{0}^{1}t^{n - 1}pars{1 - t}^{n},dd t
\[3mm] & =
int_{0}^{1}sum_{n = 1}^{infty}bracks{txpars{1 - t}}^{n}
,{dd t over t}
=int_{0}^{1}bracks{{1 over 1 - txpars{1 - t}} - 1},{dd t over t}
\[3mm] & =
int_{0}^{1}{xpars{1 - t} over 1 - txpars{1 - t}},dd t
end{align}
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n^{2}{2n choose n}}
=int_{0}^{x}dd yint_{0}^{1}{1 - t over 1 - typars{1 - t}},dd t
=-int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
end{align}
begin{align}
&color{#66f}{largesum_{n = 1}^{infty}{1 over n^{3}{2n choose n}}}
=-int_{0}^{1}{dd x over x}
int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
\[3mm] = &
int_{0}^{1}
{{rm Li}_{2}pars{tbracks{1 - t}} over t},dd t
\[3mm] = &
color{#66f}{{pi over 36root{3}}, left[psi ^{(1)}left(frac{1}{3}right)-psi ^{(1)}left(frac{2}{3}right)+psi ^{(1)}left(frac{1}{6}right)-psi ^{(1)}left(frac{5}{6}right)right]-frac{4 zeta (3)}{3}}
\[3mm] approx & {tt 0.5229}
end{align}
answered Sep 17 '14 at 4:17
Felix MarinFelix Marin
68.9k7110147
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f499926%2fevaluating-the-sum-of-the-series-sum-limits-n-1-infty-frac1n3-bin%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using the Taylor expansion for $arcsin^2(x)$ and integrating, it's easy to show (as you made) that
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&= 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} operatorname{d}x = - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) operatorname{d}u \
&= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3)\
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)
end{align}
$$
recalling that the Clausen function are defined as
$$
operatorname{Cl}_{m}(theta)=
begin{cases}displaystyle
sum_{n=1}^{infty} frac{sin (ntheta)}{n^{m}}& text{for } mtext{ even}\
displaystylesum_{n=1}^{infty} frac{cos(ntheta)}{n^{m}}& text{for } mtext{ odd}
end{cases}
$$
The value of the Clausen function $operatorname{Cl}_3$ at $frac{pi}{3}$ is
$$
operatorname{Cl}_3left(frac{pi}{3}right)=frac{1}{2}left(1-2^{-2}right)left(1-3^{-2}right)zeta(3)=frac{1}{3}zeta(3)
$$
From the duplication formula
$$
operatorname{Cl}_{2m}(2theta)=2^{2m-1}left[operatorname{Cl}_{2m}(theta)-operatorname{Cl}_{2m}(pi-theta)right]
$$
we find
$$
operatorname{Cl}_{2}left(frac{2pi}{3}right)=frac{2}{3}operatorname{Cl}_{2}left(frac{pi}{3}right).
$$
From the identities for the trigamma function at $1/3$ and $2/3$
$$
begin{align}
psi_1left(frac{1}{3}right) &=frac{2pi^2}{3}+3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)\
psi_1left(frac{2}{3}right) &=frac{2pi^2}{3}-3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
end{align}
$$
one has
$$
psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)=6sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
$$
and then
$$
operatorname{Cl}_{2}left(frac{pi}{3}right)=frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]
$$
Finally, putting all together, we have
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
&=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
&=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
end{align}
$$
answered Nov 13 '13 at 3:09
alexjoalexjo
12.5k1430
$endgroup$
add a comment |
$begingroup$
Using the Taylor expansion for $arcsin^2(x)$ and integrating, it's easy to show (as you made) that
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&= 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} operatorname{d}x = - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) operatorname{d}u \
&= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3)\
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)
end{align}
$$
recalling that the Clausen function are defined as
$$
operatorname{Cl}_{m}(theta)=
begin{cases}displaystyle
sum_{n=1}^{infty} frac{sin (ntheta)}{n^{m}}& text{for } mtext{ even}\
displaystylesum_{n=1}^{infty} frac{cos(ntheta)}{n^{m}}& text{for } mtext{ odd}
end{cases}
$$
The value of the Clausen function $operatorname{Cl}_3$ at $frac{pi}{3}$ is
$$
operatorname{Cl}_3left(frac{pi}{3}right)=frac{1}{2}left(1-2^{-2}right)left(1-3^{-2}right)zeta(3)=frac{1}{3}zeta(3)
$$
From the duplication formula
$$
operatorname{Cl}_{2m}(2theta)=2^{2m-1}left[operatorname{Cl}_{2m}(theta)-operatorname{Cl}_{2m}(pi-theta)right]
$$
we find
$$
operatorname{Cl}_{2}left(frac{2pi}{3}right)=frac{2}{3}operatorname{Cl}_{2}left(frac{pi}{3}right).
$$
From the identities for the trigamma function at $1/3$ and $2/3$
$$
begin{align}
psi_1left(frac{1}{3}right) &=frac{2pi^2}{3}+3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)\
psi_1left(frac{2}{3}right) &=frac{2pi^2}{3}-3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
end{align}
$$
one has
$$
psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)=6sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
$$
and then
$$
operatorname{Cl}_{2}left(frac{pi}{3}right)=frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]
$$
Finally, putting all together, we have
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
&=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
&=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
end{align}
$$
answered Nov 13 '13 at 3:09
alexjoalexjo
12.5k1430
$endgroup$
add a comment |
$begingroup$
Using the Taylor expansion for $arcsin^2(x)$ and integrating, it's easy to show (as you made) that
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&= 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} operatorname{d}x = - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) operatorname{d}u \
&= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3)\
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)
end{align}
$$
recalling that the Clausen function are defined as
$$
operatorname{Cl}_{m}(theta)=
begin{cases}displaystyle
sum_{n=1}^{infty} frac{sin (ntheta)}{n^{m}}& text{for } mtext{ even}\
displaystylesum_{n=1}^{infty} frac{cos(ntheta)}{n^{m}}& text{for } mtext{ odd}
end{cases}
$$
The value of the Clausen function $operatorname{Cl}_3$ at $frac{pi}{3}$ is
$$
operatorname{Cl}_3left(frac{pi}{3}right)=frac{1}{2}left(1-2^{-2}right)left(1-3^{-2}right)zeta(3)=frac{1}{3}zeta(3)
$$
From the duplication formula
$$
operatorname{Cl}_{2m}(2theta)=2^{2m-1}left[operatorname{Cl}_{2m}(theta)-operatorname{Cl}_{2m}(pi-theta)right]
$$
we find
$$
operatorname{Cl}_{2}left(frac{2pi}{3}right)=frac{2}{3}operatorname{Cl}_{2}left(frac{pi}{3}right).
$$
From the identities for the trigamma function at $1/3$ and $2/3$
$$
begin{align}
psi_1left(frac{1}{3}right) &=frac{2pi^2}{3}+3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)\
psi_1left(frac{2}{3}right) &=frac{2pi^2}{3}-3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
end{align}
$$
one has
$$
psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)=6sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
$$
and then
$$
operatorname{Cl}_{2}left(frac{pi}{3}right)=frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]
$$
Finally, putting all together, we have
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
&=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
&=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
end{align}
$$
answered Nov 13 '13 at 3:09
alexjoalexjo
12.5k1430
$endgroup$
Using the Taylor expansion for $arcsin^2(x)$ and integrating, it's easy to show (as you made) that
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&= 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} operatorname{d}x = - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) operatorname{d}u \
&= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3)\
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)
end{align}
$$
recalling that the Clausen function are defined as
$$
operatorname{Cl}_{m}(theta)=
begin{cases}displaystyle
sum_{n=1}^{infty} frac{sin (ntheta)}{n^{m}}& text{for } mtext{ even}\
displaystylesum_{n=1}^{infty} frac{cos(ntheta)}{n^{m}}& text{for } mtext{ odd}
end{cases}
$$
The value of the Clausen function $operatorname{Cl}_3$ at $frac{pi}{3}$ is
$$
operatorname{Cl}_3left(frac{pi}{3}right)=frac{1}{2}left(1-2^{-2}right)left(1-3^{-2}right)zeta(3)=frac{1}{3}zeta(3)
$$
From the duplication formula
$$
operatorname{Cl}_{2m}(2theta)=2^{2m-1}left[operatorname{Cl}_{2m}(theta)-operatorname{Cl}_{2m}(pi-theta)right]
$$
we find
$$
operatorname{Cl}_{2}left(frac{2pi}{3}right)=frac{2}{3}operatorname{Cl}_{2}left(frac{pi}{3}right).
$$
From the identities for the trigamma function at $1/3$ and $2/3$
$$
begin{align}
psi_1left(frac{1}{3}right) &=frac{2pi^2}{3}+3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)\
psi_1left(frac{2}{3}right) &=frac{2pi^2}{3}-3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
end{align}
$$
one has
$$
psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)=6sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
$$
and then
$$
operatorname{Cl}_{2}left(frac{pi}{3}right)=frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]
$$
Finally, putting all together, we have
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
&=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
&=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
end{align}
$$
answered Nov 13 '13 at 3:09
alexjoalexjo
12.5k1430
answered Nov 13 '13 at 3:09
alexjoalexjo
12.5k1430
answered Nov 13 '13 at 3:09
alexjoalexjo
12.5k1430
answered Nov 13 '13 at 3:09
alexjoalexjo
12.5k1430
12.5k1430
add a comment |
add a comment |
$begingroup$
$newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
$ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n{2n choose n}}
=sum_{n = 1}^{infty}x^{n},
{Gammapars{n}Gammapars{n + 1} over Gammapars{2n + 1}}
=sum_{n = 1}^{infty}x^{n}int_{0}^{1}t^{n - 1}pars{1 - t}^{n},dd t
\[3mm] & =
int_{0}^{1}sum_{n = 1}^{infty}bracks{txpars{1 - t}}^{n}
,{dd t over t}
=int_{0}^{1}bracks{{1 over 1 - txpars{1 - t}} - 1},{dd t over t}
\[3mm] & =
int_{0}^{1}{xpars{1 - t} over 1 - txpars{1 - t}},dd t
end{align}
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n^{2}{2n choose n}}
=int_{0}^{x}dd yint_{0}^{1}{1 - t over 1 - typars{1 - t}},dd t
=-int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
end{align}
begin{align}
&color{#66f}{largesum_{n = 1}^{infty}{1 over n^{3}{2n choose n}}}
=-int_{0}^{1}{dd x over x}
int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
\[3mm] = &
int_{0}^{1}
{{rm Li}_{2}pars{tbracks{1 - t}} over t},dd t
\[3mm] = &
color{#66f}{{pi over 36root{3}}, left[psi ^{(1)}left(frac{1}{3}right)-psi ^{(1)}left(frac{2}{3}right)+psi ^{(1)}left(frac{1}{6}right)-psi ^{(1)}left(frac{5}{6}right)right]-frac{4 zeta (3)}{3}}
\[3mm] approx & {tt 0.5229}
end{align}
answered Sep 17 '14 at 4:17
Felix MarinFelix Marin
68.9k7110147
$endgroup$
add a comment |
$begingroup$
$newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
$ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n{2n choose n}}
=sum_{n = 1}^{infty}x^{n},
{Gammapars{n}Gammapars{n + 1} over Gammapars{2n + 1}}
=sum_{n = 1}^{infty}x^{n}int_{0}^{1}t^{n - 1}pars{1 - t}^{n},dd t
\[3mm] & =
int_{0}^{1}sum_{n = 1}^{infty}bracks{txpars{1 - t}}^{n}
,{dd t over t}
=int_{0}^{1}bracks{{1 over 1 - txpars{1 - t}} - 1},{dd t over t}
\[3mm] & =
int_{0}^{1}{xpars{1 - t} over 1 - txpars{1 - t}},dd t
end{align}
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n^{2}{2n choose n}}
=int_{0}^{x}dd yint_{0}^{1}{1 - t over 1 - typars{1 - t}},dd t
=-int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
end{align}
begin{align}
&color{#66f}{largesum_{n = 1}^{infty}{1 over n^{3}{2n choose n}}}
=-int_{0}^{1}{dd x over x}
int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
\[3mm] = &
int_{0}^{1}
{{rm Li}_{2}pars{tbracks{1 - t}} over t},dd t
\[3mm] = &
color{#66f}{{pi over 36root{3}}, left[psi ^{(1)}left(frac{1}{3}right)-psi ^{(1)}left(frac{2}{3}right)+psi ^{(1)}left(frac{1}{6}right)-psi ^{(1)}left(frac{5}{6}right)right]-frac{4 zeta (3)}{3}}
\[3mm] approx & {tt 0.5229}
end{align}
answered Sep 17 '14 at 4:17
Felix MarinFelix Marin
68.9k7110147
$endgroup$
add a comment |
$begingroup$
$newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
$ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n{2n choose n}}
=sum_{n = 1}^{infty}x^{n},
{Gammapars{n}Gammapars{n + 1} over Gammapars{2n + 1}}
=sum_{n = 1}^{infty}x^{n}int_{0}^{1}t^{n - 1}pars{1 - t}^{n},dd t
\[3mm] & =
int_{0}^{1}sum_{n = 1}^{infty}bracks{txpars{1 - t}}^{n}
,{dd t over t}
=int_{0}^{1}bracks{{1 over 1 - txpars{1 - t}} - 1},{dd t over t}
\[3mm] & =
int_{0}^{1}{xpars{1 - t} over 1 - txpars{1 - t}},dd t
end{align}
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n^{2}{2n choose n}}
=int_{0}^{x}dd yint_{0}^{1}{1 - t over 1 - typars{1 - t}},dd t
=-int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
end{align}
begin{align}
&color{#66f}{largesum_{n = 1}^{infty}{1 over n^{3}{2n choose n}}}
=-int_{0}^{1}{dd x over x}
int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
\[3mm] = &
int_{0}^{1}
{{rm Li}_{2}pars{tbracks{1 - t}} over t},dd t
\[3mm] = &
color{#66f}{{pi over 36root{3}}, left[psi ^{(1)}left(frac{1}{3}right)-psi ^{(1)}left(frac{2}{3}right)+psi ^{(1)}left(frac{1}{6}right)-psi ^{(1)}left(frac{5}{6}right)right]-frac{4 zeta (3)}{3}}
\[3mm] approx & {tt 0.5229}
end{align}
answered Sep 17 '14 at 4:17
Felix MarinFelix Marin
68.9k7110147
$endgroup$
$newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
$ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n{2n choose n}}
=sum_{n = 1}^{infty}x^{n},
{Gammapars{n}Gammapars{n + 1} over Gammapars{2n + 1}}
=sum_{n = 1}^{infty}x^{n}int_{0}^{1}t^{n - 1}pars{1 - t}^{n},dd t
\[3mm] & =
int_{0}^{1}sum_{n = 1}^{infty}bracks{txpars{1 - t}}^{n}
,{dd t over t}
=int_{0}^{1}bracks{{1 over 1 - txpars{1 - t}} - 1},{dd t over t}
\[3mm] & =
int_{0}^{1}{xpars{1 - t} over 1 - txpars{1 - t}},dd t
end{align}
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n^{2}{2n choose n}}
=int_{0}^{x}dd yint_{0}^{1}{1 - t over 1 - typars{1 - t}},dd t
=-int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
end{align}
begin{align}
&color{#66f}{largesum_{n = 1}^{infty}{1 over n^{3}{2n choose n}}}
=-int_{0}^{1}{dd x over x}
int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
\[3mm] = &
int_{0}^{1}
{{rm Li}_{2}pars{tbracks{1 - t}} over t},dd t
\[3mm] = &
color{#66f}{{pi over 36root{3}}, left[psi ^{(1)}left(frac{1}{3}right)-psi ^{(1)}left(frac{2}{3}right)+psi ^{(1)}left(frac{1}{6}right)-psi ^{(1)}left(frac{5}{6}right)right]-frac{4 zeta (3)}{3}}
\[3mm] approx & {tt 0.5229}
end{align}
answered Sep 17 '14 at 4:17
Felix MarinFelix Marin
68.9k7110147
edited Dec 27 '18 at 16:55
answered Sep 17 '14 at 4:17
Felix MarinFelix Marin
68.9k7110147
answered Sep 17 '14 at 4:17
Felix MarinFelix Marin
68.9k7110147
answered Sep 17 '14 at 4:17
Felix MarinFelix Marin
68.9k7110147
68.9k7110147
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f499926%2fevaluating-the-sum-of-the-series-sum-limits-n-1-infty-frac1n3-bin%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:15
$begingroup$
That sum is the alternating version of this one.
$endgroup$
– Random Variable
Sep 21 '13 at 1:31
$begingroup$
And the techniques that apply to the alternating version don't work for the one you ask about?
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:41
$begingroup$
They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
$endgroup$
– Random Variable
Sep 21 '13 at 1:56
$begingroup$
use Clausen function
$endgroup$
– alexjo
Nov 13 '13 at 21:08