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#25 by Charles R Greathouse IV at Sat Oct 01 23:42:46 EDT 2022
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#24 by Charles R Greathouse IV at Sat Oct 01 23:42:43 EDT 2022
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<a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
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(PARI) Pi^3/8 \\ Charles R Greathouse IV, Oct 01 2022
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approved
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#23 by Joerg Arndt at Tue Jun 15 11:05:23 EDT 2021
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#22 by Joerg Arndt at Tue Jun 15 11:05:20 EDT 2021
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| FORMULA
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Equals 2*(-1)^n*Integral_{x=-1/e..0} W(n,x)*(1-W(n,x))*log(-W(n,x))^2/x/(1-W(n,x)^4) ) dx, where W=LambertW, for n=0 and n=-1. (End)
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reviewed
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#21 by Michel Marcus at Tue Jun 15 10:50:49 EDT 2021
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#20 by Gleb Koloskov at Tue Jun 15 08:53:37 EDT 2021
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Discussion
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| Tue Jun 15
| 10:50
| Michel Marcus: I checked the 2 first integrals
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#19 by Gleb Koloskov at Tue Jun 15 08:53:04 EDT 2021
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From Gleb Koloskov, Jun 15 2021: (Start)
Equals 2*Integral_{x=0..1} log(x)^2/(1+x^2) dx.
Equals 2*Integral_{x=1..oo} log(x)^2/(1+x^2) dx.
Equals 2*(-1)^n*Integral_{x=-1/e..0} W(n,x)*(1-W(n,x))*log(-W(n,x))^2/x/(1-W(n,x)^4) dx, where W=LambertW, for n=0 and n=-1. (End)
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approved
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#18 by Susanna Cuyler at Fri Aug 21 05:48:45 EDT 2020
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#17 by Amiram Eldar at Fri Aug 21 03:46:44 EDT 2020
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#16 by Amiram Eldar at Fri Aug 21 03:21:16 EDT 2020
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| COMMENTS
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Also, decimal expansion of:
integral_{x>0} log(x)^2/(1+x^2);
integral_{x=0..Pi/2} log(tan(x))^2;
integral_{x=0..Pi/2} log(sin(x)^3)*log(sin(x))-(3*Pi/2)*log(2)^2;
27/7 * sum_{k>=0} (binomial(2*k, k)/((2*k+1)^3*16^k);
27/7 * 4F3([1/2, 1/2, 1/2, 1/2], [3/2, 3/2, 3/2], 1/4), where pFq() is the generalized hypergeometric function.
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Equals Integral_{x>0} log(x)^2/(1+x^2) dx.
Equals Integral_{x=0..Pi/2} log(tan(x))^2 dx.
Equals Integral_{x=0..Pi/2} log(sin(x)^3)*log(sin(x))-(3*Pi/2)*log(2)^2 dx.
Equals 27/7 * sum_{k>=0} (binomial(2*k, k)/((2*k+1)^3*16^k);
Equals 27/7 * 4F3([1/2, 1/2, 1/2, 1/2], [3/2, 3/2, 3/2], 1/4), where pFq() is the generalized hypergeometric function.
From Amiram Eldar, Aug 21 2020: (Start)
Equals Integral_{x=0..oo} x^2/cosh(x) dx.
Equals 2 + Integral_{x=0..oo} x^2 * exp(-x) * tanh(x) dx. (End)
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