¿Existe una forma cerrada para $\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}?$

Question

encontré

$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=-8\int_0^{\pi/2}x^2\cot x\ln(\cos x)\ dx=I\tag1.$$

Mathematica no pudo encontrar$I$, por lo que no estoy seguro de si existe un formulario cerrado para ello. Solo lo estoy intentando aquí.

La primera idea que se me ocurrió es utilizar la serie de Fourier de $-\ln(\cos x)=\ln(2)+\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}{n}$ y tenemos

$$I=8\ln(2)\underbrace{\int_0^{\pi/2}x^2\cot x\ dx}_{\frac32\ln(2)\zeta(2)-\frac78\zeta(3)}+8\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^{\pi/2}x^2 \cot x\cos(2nx)\ dx.$$

Me quedé atrapado aquí. Cualquier ayuda será muy apreciada.

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Prueba de $(1)$

de aqui tenemos

$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2{2n\choose n}}$$

reemplazar $x$ por $\sqrt{x}$ obtenemos $$\sum_{n=1}^\infty\frac{2^{2n}x^n}{n^2{2n\choose n}}=2\arcsin^2(\sqrt{x})$$

multiplica ambos lados por $-\frac{\ln(1-x)}{x}$ entonces $\int_0^1$ y use $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ obtenemos

$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=2\int_0^1\frac{\arcsin^2(\sqrt{x})\ln(1-x)}{x}dx\overset{\sqrt{x}=\sin\theta}{=}-8\int_0^{\pi/2}x^2\cot x\ln(\cos x)\ dx$$

Answer 1

$$S=-8 \text{Li}_4\left(\frac{1}{2}\right)+\frac{\pi ^4}{90}-\frac{1}{3} \log ^4(2)+\frac{4}{3} \pi ^2 \log ^2(2)$$ Prueba $1$. Hay una publicación sobre MSE que se centra exactamente en la misma integral.

Prueba $2$. Hay un artículo que se centra en numerosos problemas que cubre exactamente la misma suma.

Prueba $3$. Hay otro artículo aplastando todas estas series de un solo golpe .

Pequeña bonificación: $$\small \int_0^{\frac{\pi }{2}} x^3 \cot (x) \log (\cos (x)) \, dx=\frac{3}{2} \pi \text{Li}_4\left(\frac{1}{2}\right)+\frac{9}{16} \pi \zeta (3) \log (2)-\frac{\pi ^5}{120}+\frac{1}{16} \pi \log ^4(2)-\frac{1}{8} \pi ^3 \log ^2(2)$$

QED

Answer 2

Gracias a @ user97357329 por su pista en los comentarios.

En el libro, Integrales casi imposibles, sumas y series , página$247$ Eq $(3.288)$ tenemos

$$\cot x\ln(\cos x)=\sum_{n=1}^\infty(-1)^n\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}\right)-\frac1n\right)\sin(2nx)$$

$$=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\sin(2nx),\quad 0<x<\frac{\pi}{2}$$

Así,

$$\int_0^{\pi/2}x^2\cot x\ln(\cos x)dx=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\left(\int_0^{\pi/2}x^2\sin(2nx)dx\right)$$

$$=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\left(\frac{\cos(n\pi)}{4n^3}-\frac{3\zeta(2)\cos(n\pi)}{4n}-\frac{1}{4n^3}\right)$$

$$=\sum_{n=1}^\infty(-1)^n\left(\int_0^1\frac{1-t}{1+t}t^{n-1}dt\right)\left(\frac{(-1)^n}{4n^3}-\frac{3\zeta(2)(-1)^n}{4n}-\frac{1}{4n^3}\right)$$

$$=\frac14\int_0^1\frac{1-t}{t(1+t)}\left(\sum_{n=1}^\infty\frac{t^n}{n^3}-\frac{3\zeta(2)t^n}{n}-\frac{(-t)^n}{n^3}\right)dt$$

$$=\frac14\int_0^1\left(\frac1t-\frac2{1+t}\right)\left(\text{Li}_3(t)+3\zeta(2)\ln(1-t)-\text{Li}_3(-t)\right)dt$$

$$=\frac14\underbrace{\int_0^1\frac{\text{Li}_3(t)-\text{Li}_3(-t)}{t}dt}_{\mathcal{I}_1}-\frac12\underbrace{\int_0^1\frac{\text{Li}_3(t)-\text{Li}_3(-t)}{1+t}dt}_{\mathcal{I}_2}$$ $$+\frac34\zeta(2)\underbrace{\int_0^1\frac{\ln(1-t)}{t}dt}_{\mathcal{I}_3}-\frac32\zeta(2)\underbrace{\int_0^1\frac{\ln(1-t)}{1+t}dt}_{\mathcal{I}_4}$$

$$\mathcal{I}_1=\text{Li}_4(1)-\text{Li}_4(-1)=\zeta(4)+\frac78\zeta(4)=\boxed{\frac{15}{8}\zeta(4)}$$

Por integración por partes tenemos

$$\mathcal{I}_2=\frac74\ln(2)\zeta(3)-\int_0^1\frac{\ln(1+t)\text{Li}_2(t)}{t}dt+\int_0^1\frac{\ln(1+t)\text{Li}_2(-t)}{t}dt$$

$$=\frac74\ln(2)\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1 t^{n-1}\text{Li}_2(t)dt-\frac12\text{Li}_2^2(-t)|_0^1$$

$$=\frac74\ln(2)\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n}{n} \left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)-\frac5{16}\zeta(4)$$ $$=\frac74\ln(2)\zeta(3)-\frac54\zeta(4)-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\frac5{16}\zeta(4)$$ sustituir

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$$

obtenemos

$$\mathcal{I}_2=\boxed{-2\operatorname{Li_4}\left(\frac12\right)-\frac{25}{16}\zeta(4)+\frac12\ln^22\zeta(2)-\frac{1}{12}\ln^42}$$

$$\mathcal{I}_3=-\text{Li}_2(1)=\boxed{-\zeta(2)}$$

$$\mathcal{I}_4=\int_0^1\frac{\ln(1-t)}{1+t}dt=\int_0^1\frac{\ln x}{2-x}dx=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln xdx$$ $$=-\sum_{n=1}^\infty\frac{1}{n^22^n}=-\text{Li}_2\left(\frac12\right)=\boxed{\frac12\ln^22-\frac12\zeta(2)}$$

Combine todos los resultados en caja que finalmente obtenemos

$$\int_0^{\pi/2}x^2\cot x\ln(\cos x)dx=\text{Li}_4\left(\frac12\right)-\frac18\zeta(4)-\ln^2(2)\zeta(2)+\frac{1}{24}\ln^4(2)$$

que nos da

$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \ begin {align} & \ bbox [5px, # ffd] {\ int_ {0} ^ {\ pi / 2} x ^ {2} \ cot \ pars {x} \ cos \ pars {2nx} \, \ dd x} \\ [5 mm] & = \ izquierda. \ Re \ int_ {x \ = \ 0} ^ {x \ = \ \ pi / 2} \ bracks {- \ ic \ ln \ pars {z}} ^ {\, 2} \, \ pars {{z ^ {2} + 1 \ sobre z ^ {2} - 1} \, \ ic} z ^ {2n} \, {\ dd z \ sobre \ ic z} \, \ right \ vert _ {\ z \ = \ \ exp \ pars {\ ic x}} \\ [5mm] = & \ \ left. - \, \ Re \ int_ {x \ = \ 0} ^ {x \ = \ \ pi / 2} z ^ {2n - 1} \ ln ^ {2} \ pars {z} \, {z ^ {2 } + 1 \ sobre z ^ {2} - 1} \, \ dd z \, \ right \ vert _ {\ z \ = \ \ exp \ pars {\ ic x}} \\ [5mm] = & \ \ Re \ int_ {1} ^ {0} \ pars {-1} ^ {n + 1} \, \ ic \, y ^ {2n - 1} \ bracks {\ ln \ pars {y} + {\ pi \ over 2} \, \ ic} ^ {\, 2} \, {-y ^ {2} + 1 \ over -y ^ {2} - 1} \, \ ic \, \ dd y \\ [2mm] & \ + \ int_ {0} ^ {1} x ^ {2n - 1} \ ln ^ {2} \ pars {x} \, {x ^ {2} + 1 \ over x ^ {2} - 1} \ , \ dd x \\ [5 mm] = & \ \ pars {-1} ^ {n} \ int_ {0} ^ {1} y ^ {2n - 1} \ bracks {\ ln ^ {2} \ pars { y} - {\ pi ^ {2} \ over 4}} {1 - y ^ {2} \ over 1 + y ^ {2}} \, \ dd y \\ [2mm] & \ - \ int_ {0 } ^ {1} x ^ {2n - 1} \ ln ^ {2} \ pars {x} \, {1 + x ^ {2} \ over 1 - x ^ {2}} \, \ dd x \\ [5 mm] & = \ pars {-1} ^ {n} \, \ mathcal {I} '' \ pars {2n - 1} - \ pars {-1} ^ {n} \, {\ pi ^ {2 } \ over 4} \, \ mathcal {I} \ pars {2n - 1} - \ mathcal {J} '' \ pars {2n - 1} \\ & \ \ mbox {donde} \ quad \ left \ {\ begin {array} {rcl} \ ds {\ mathcal {I} \ pars {\ nu}} & \ ds {\ equiv} & \ ds {\ int_ {0} ^ {1} y ^ {\ nu} \, {1 - y ^ {2} \ over 1 + y ^ {2}} \, \ dd y} \\ [2mm] \ ds {\ mathcal {J} \ pars {\ nu}} & \ ds {\ equiv } & \ ds {\ int_ {0} ^ {1} \ pars {y ^ {\ nu} - 1} \, {1 + y ^ {2} \ over 1 - y ^ {2}} \, \ dd y} \ end {matriz} \ right. \ end {align} Evaluemos algunas integrales que necesitamos para evaluar nuestro resultado principal:
$\ds{\Large\mathcal{I}\pars{\nu}:\ ?.}$ \ begin {align} \ mathcal {I} \ pars {\ nu} & \ equiv \ int_ {0} ^ {1} y ^ {\ nu} \, {1 - y ^ {2} \ over 1 + y ^ {2}} \, \ dd y = \ int_ {0} ^ {1} {y ^ {\ nu} - 2y ^ {\ nu + 2} + y ^ {\ nu + 4} \ over 1 - y ^ {4}} \, \ dd y \\ [5 mm] & = {1 \ over 4} \ int_ {0} ^ {1} {y ^ {\ nu / 4 - 3/4} - 2y ^ {\ nu / 4 - 1/4} + y ^ {\ nu / 4 + 1/4} \ over 1 - y} \, \ dd y \\ [5mm] & = {1 \ over 4} \ bracks {% 2 \ int_ {0} ^ {1} {1 - y ^ {\ nu / 4 - 1/4} \ over 1 - y} \, \ dd y - \ int_ {0} ^ {1} {1 - y ^ { \ nu / 4 - 3/4} \ over 1 - y} \, \ dd y - \ int_ {0} ^ {1} {1 - y ^ {\ nu / 4 + 1/4} \ over 1 - y } \, \ dd y} \\ [5 mm] & = {1 \ over 4} \ bracks {% 2 \ Psi \ pars {{\ nu \ over 4} + {3 \ over 4}} - \ Psi \ pars {{\ nu \ over 4} + {1 \ over 4}} - \ Psi \ pars {{\ nu \ over 4} + {5 \ over 4}}} \\ [5mm] & = - \, {1 \ over \ nu + 1} - {1 \ over 2} \, \ Psi \ pars {\ nu + 1 \ over 4} + {1 \ over 2} \, \ Psi \ pars {\ nu + 3 \ over 4 } \\ [1cm] & \ \ left \ {\ begin {array} {rcl} \ ds {\ mathcal {I} \ pars {2n - 1}} & \ ds {=} & \ ds {- \, { 1 \ sobre 2n} - {1 \ sobre 2} \, \ Psi \ pars {n \ sobre 2} + {1 \ sobre 2} \, \ Psi \ pars {n + 1 \ sobre 2}} \\ [2 mm ] \ ds {\ mathcal {I} '' \ pars {2n - 1}} & \ ds {=} & \ ds {- \, {1 \ over 4 n ^ {3}} - {1 \ over 32} \, \ Psi \, '' \ pars {n \ over 2} + {1 \ over 32} \, \ Psi \, '' \ pars {n + 1 \ over 2}} \ end {array} \ right. \ end {align}

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$\ds{\Large\mathcal{J}\pars{\nu}:\ ?.}$ \ begin {align} \ mathcal {J} \ pars {\ nu} & \ equiv \ int_ {0} ^ {1} \ pars {y ^ {\ nu} - 1} \, {1 + y ^ {2} \ over 1 - y ^ {2}} \, \ dd y = {1 \ over 2} \ int_ {0} ^ {1} {- y ^ {- 1/2} - y ^ {1/2} + y ^ {\ nu / 2 - 1/2} + y ^ {\ nu / 2 + 1/2} \ over 1 - y} \, \ dd y \\ [5mm] & = {1 \ over 2} \ bracks {\ Psi \ pars {1 \ over 2} + \ Psi \ pars {3 \ over 2} - \ Psi \ pars {{\ nu \ over 2} + {1 \ over 2}} - \ Psi \ pars { {\ nu \ over 2} + {3 \ over 2}}} \\ [5 mm] & = 1 - \ gamma - 2 \ ln \ pars {2} - {1 \ over 2} \, \ Psi \ pars { \ nu + 1 \ over 2} - {1 \ over 2} \, \ Psi \ pars {\ nu + 3 \ over 2} \\ [1cm] \ mathcal {J} \, '' \ pars {2n - 1 } & = - \, {1 \ over 8} \, \ Psi \, '' \ pars {n} - \, {1 \ over 8} \, \ Psi \, '' \ pars {n + 1} = - \, {1 \ over 4} \, \ Psi \, '' \ pars {n} - {1 \ over 4n ^ {3}} \ end {align}