Question

How to prove $\int_0^\infty \ln(1+\frac{z}{\cosh(x)})dx=\frac{\pi^2}{8}+\frac{(\cosh^{-1}(z))^2}{2},z\ge1$ and a closed form for $-1<z<1$?

score 258 · Answer 1 · 2018-12-22 00:13:10

258

Views

How to prove $\int_0^\infty \ln(1+\frac{z}{\cosh(x)})dx=\frac{\pi^2}{8}+\frac{(\cosh^{-1}(z))^2}{2},z\ge1$ and a closed form for $-1<z<1$?

Published on 22 Dec 2018 - 0:13

score 310 · Answer 2 · 2026-03-25 04:39:27

310

Views

Closed form expression or asymptotic expansion for (periodic) generalized harmonic numbers?

Published on 25 Mar 2026 - 4:39

score 199 · Answer 3 · 2018-12-22 19:40:42

199

Views

What is the inverse of simply composited elementary functions?

Published on 22 Dec 2018 - 19:40

score 3.6k · Answer 4 · 2026-03-25 06:06:54

3.6k

Views

Proving that $\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$

Published on 25 Mar 2026 - 6:06

score 377 · Answer 5 · 2018-12-26 11:41:48

377

Views

Is there a closed form for the integral $\int_0^\infty \frac{e^{-x^2} I_0 \left(\beta x \right) d x}{\sqrt{ \alpha^2+x^2}}$

Published on 26 Dec 2018 - 11:41

score 246 · Answer 6 · 2018-12-27 19:15:31

246

Views

Closed form of this type $\sum_{j=0}^{\infty}\frac{2^jj^n}{(2j+1)(2j+3){2j \choose j}}$

Published on 27 Dec 2018 - 19:15

score 101 · Answer 7 · 2018-12-27 20:17:45

101

Views

A Closed form for the $\sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{1}{(25k^2+25k+4)(n-k+1)^3}$

Published on 27 Dec 2018 - 20:17

score 578 · Answer 8 · 2026-03-25 01:18:20

578

Views

Number of possible bishop moves on an $n \times m$ chessboard

Published on 25 Mar 2026 - 1:18

score 91 · Answer 9 · 2018-12-29 09:19:33

91

Views

Is there any solution (numeric or closed) to integration of ${\sin x}^{\cos x}$?

Published on 29 Dec 2018 - 9:19

score 404 · Answer 10 · 2026-03-25 14:26:02

404

Views

Evaluate $\int_0^1 \frac{\operatorname{arctanh}^3(x)}{x}dx$

Published on 25 Mar 2026 - 14:26

score 180 · Answer 11 · 2018-12-31 11:57:19

180

Views

Sum of $\sum_{n=1}^\infty\frac{\ln(n) -\ln(n+1)}{n+1}$

Published on 31 Dec 2018 - 11:57

score 85 · Answer 12 · 2019-01-01 04:22:58

85

Views

$F_k(a_1,b_1,c_1;a_2,b_2,c_2)=\int_0^k\frac{a_1x^2+b_1x+c_1}{a_2x^2+b_2x+c_2}\mathrm{d}x$

Published on 01 Jan 2019 - 4:22

score 152 · Answer 13 · 2019-01-01 23:17:25

152

Views

Is the closed-form of $\sum_{k=1}^{\infty}\frac{e^{-k}}{k^2}$ known?

Published on 01 Jan 2019 - 23:17

score 148 · Answer 14 · 2026-03-25 06:05:41

148

Views

Closed expression for alternating harmonically wrapped harmonic series $\sum_{n=1}^\infty (-1)^{n+1} H_{\frac{1}{n}}$

Published on 25 Mar 2026 - 6:05

score 262 · Answer 15 · 2026-03-25 06:05:54

262

Views

Generalizing $\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)}=2$

Published on 25 Mar 2026 - 6:05

List Question

How to prove $\int_0^\infty \ln(1+\frac{z}{\cosh(x)})dx=\frac{\pi^2}{8}+\frac{(\cosh^{-1}(z))^2}{2},z\ge1$ and a closed form for $-1<z<1$?

Closed form expression or asymptotic expansion for (periodic) generalized harmonic numbers?

What is the inverse of simply composited elementary functions?

Proving that $\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$

Is there a closed form for the integral $\int_0^\infty \frac{e^{-x^2} I_0 \left(\beta x \right) d x}{\sqrt{ \alpha^2+x^2}}$

Closed form of this type $\sum_{j=0}^{\infty}\frac{2^jj^n}{(2j+1)(2j+3){2j \choose j}}$

A Closed form for the $\sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{1}{(25k^2+25k+4)(n-k+1)^3}$

Number of possible bishop moves on an $n \times m$ chessboard

Is there any solution (numeric or closed) to integration of ${\sin x}^{\cos x}$?

Evaluate $\int_0^1 \frac{\operatorname{arctanh}^3(x)}{x}dx$

Sum of $\sum_{n=1}^\infty\frac{\ln(n) -\ln(n+1)}{n+1}$

$F_k(a_1,b_1,c_1;a_2,b_2,c_2)=\int_0^k\frac{a_1x^2+b_1x+c_1}{a_2x^2+b_2x+c_2}\mathrm{d}x$

Is the closed-form of $\sum_{k=1}^{\infty}\frac{e^{-k}}{k^2}$ known?

Closed expression for alternating harmonically wrapped harmonic series $\sum_{n=1}^\infty (-1)^{n+1} H_{\frac{1}{n}}$

Generalizing $\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)}=2$

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