Question

Prove $\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{x(1-x^2)}dx=\frac{17}{16}\zeta(4)$

score 170 · Answer 1 · 2026-03-25 17:41:57

170

Views

Prove $\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{x(1-x^2)}dx=\frac{17}{16}\zeta(4)$

Published on 25 Mar 2026 - 17:41

score 65 · Answer 2 · 2019-09-06 07:14:54

65

Views

Algebraic or polynomial approximation of weighted harmonic mean

Published on 06 Sep 2019 - 7:14

score 757 · Answer 3 · 2026-03-25 16:01:42

757

Views

Challenging sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}$

Published on 25 Mar 2026 - 16:01

score 461 · Answer 4 · 2026-03-25 17:42:07

461

Views

Very advanced sums: Compute $\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^{(2)}}{(2 n)^2}$ and $\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^2}{(2 n)^2}$

Published on 25 Mar 2026 - 17:42

score 839 · Answer 5 · 2026-03-25 11:13:29

839

Views

Integrating $\int_0^1\frac{\ln^2x\ln(1+x)}{1+x^2} dx$ using real methods

Published on 25 Mar 2026 - 11:13

score 110 · Answer 6 · 2019-09-15 09:09:10

110

Views

Estimating the quantity

Published on 15 Sep 2019 - 9:09

score 445 · Answer 7 · 2026-03-25 16:01:43

445

Views

Compute $\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n+1)^3}$

Published on 25 Mar 2026 - 16:01

score 283 · Answer 8 · 2019-09-19 04:05:58

283

Views

Discussion about $\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)}$

Published on 19 Sep 2019 - 4:05

score 150 · Answer 9 · 2019-09-19 20:12:04

150

Views

$a_{n+1}\ge a_n$ for $a_n=\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}$

Published on 19 Sep 2019 - 20:12

score 521 · Answer 10 · 2026-03-25 06:07:15

521

Views

The closed-form of $\sum_{n=0}^{\infty}\frac{(-1)^n H^{(2)}_{n}}{(2n+1)^2} $

Published on 25 Mar 2026 - 6:07

score 1.6k · Answer 11 · 2026-03-25 17:37:42

1.6k

Views

A group of important generating functions involving harmonic number.

Published on 25 Mar 2026 - 17:37

score 277 · Answer 12 · 2026-03-25 17:40:10

277

Views

Is there a closed form for the integral $ \int_{0}^{1}\frac{\mathrm{Li}_2(-x^2)}{x^2+1} \mathrm{d}x $

Published on 25 Mar 2026 - 17:40

score 364 · Answer 13 · 2026-03-25 17:34:08

364

Views

Arithmetic mean greater than equal to harmonic mean inequality problem

Published on 25 Mar 2026 - 17:34

score 433 · Answer 14 · 2026-03-25 16:02:26

433

Views

Advanced Sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}$

Published on 25 Mar 2026 - 16:02

score 490 · Answer 15 · 2026-03-25 22:30:42

490

Views

How to calculate $\int_0^1\frac{\ln^2x\ln^2(1-x^2)}{1-x^2}\ dx$?

Published on 25 Mar 2026 - 22:30

List Question

Prove $\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{x(1-x^2)}dx=\frac{17}{16}\zeta(4)$

Algebraic or polynomial approximation of weighted harmonic mean

Challenging sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}$

Very advanced sums: Compute $\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^{(2)}}{(2 n)^2}$ and $\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^2}{(2 n)^2}$

Integrating $\int_0^1\frac{\ln^2x\ln(1+x)}{1+x^2} dx$ using real methods

Estimating the quantity

Compute $\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n+1)^3}$

Discussion about $\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)}$

$a_{n+1}\ge a_n$ for $a_n=\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}$

The closed-form of $\sum_{n=0}^{\infty}\frac{(-1)^n H^{(2)}_{n}}{(2n+1)^2} $

A group of important generating functions involving harmonic number.

Is there a closed form for the integral $ \int_{0}^{1}\frac{\mathrm{Li}_2(-x^2)}{x^2+1} \mathrm{d}x $

Arithmetic mean greater than equal to harmonic mean inequality problem

Advanced Sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}$

How to calculate $\int_0^1\frac{\ln^2x\ln^2(1-x^2)}{1-x^2}\ dx$?

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