Question

Simplification of $G_{2,4}^{4,2}\left(\frac18,\frac12\middle|\begin{array}{c}\frac12,\frac12\\0,0,\frac12,\frac12\\\end{array}\right)$

score 847 · Answer 1 · 2026-04-07 22:59:00

847

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Simplification of $G_{2,4}^{4,2}\left(\frac18,\frac12\middle|\begin{array}{c}\frac12,\frac12\\0,0,\frac12,\frac12\\\end{array}\right)$

Published on 07 Apr 2026 - 22:59

score 1.6k · Answer 2 · 2026-04-07 22:58:53

1.6k

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A closed form for $\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx$

Published on 07 Apr 2026 - 22:58

score 1k · Answer 3 · 2026-04-07 23:01:26

1k

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Integral $\int_0^\infty{_1F_2}\left(\begin{array}{c}\tfrac12\\1,\tfrac32\end{array}\middle|-x\right)\frac{dx}{1+4\,x}$

Published on 07 Apr 2026 - 23:01

score 324 · Answer 4 · 2026-03-26 09:48:02

324

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Monstrous Moonshine for $M_{24}$?

Published on 26 Mar 2026 - 9:48

score 1.6k · Answer 5 · 2026-04-07 22:59:04

1.6k

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Derivative of the Meijer G-function with respect to one of its parameters

Published on 07 Apr 2026 - 22:59

score 141 · Answer 6 · 2013-11-24 16:03:53

141

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How to speedup evaluation of hypergeometric ${}_3 F_2(1)$?

Published on 24 Nov 2013 - 16:03

score 357 · Answer 7 · 2026-04-11 11:38:29

357

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Definite Integral of Modified Bessel function representation

Published on 11 Apr 2026 - 11:38

score 8.4k · Answer 8 · 2013-11-25 12:51:59

8.4k

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Generalizing $\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{\operatorname dx}{x^{2}+1} = \frac{5\pi^{2}}{96}$

Published on 25 Nov 2013 - 12:51

score 117 · Answer 9 · 2026-03-27 00:00:19

117

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Solving Some Transcendental Equations

Published on 27 Mar 2026 - 0:00

score 1.1k · Answer 10 · 2026-04-12 06:55:02

1.1k

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Are elementary and generalized hypergeometric functions sufficient to express all algebraic numbers?

Published on 12 Apr 2026 - 6:55

score 774 · Answer 11 · 2013-11-28 18:26:46

774

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Using the Maclaurin series of $\arctan(x)$ to evaluate $\int_{0}^{\frac{\pi}{2}} \arctan( a \sin x) \, \mathrm dx$

Published on 28 Nov 2013 - 18:26

score 6.6k · Answer 12 · 2026-04-11 15:41:29

6.6k

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Bessel function with complex argument

Published on 11 Apr 2026 - 15:41

score 182 · Answer 13 · 2013-11-29 08:42:49

182

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About large z behavior of hypergeometric function $_2F_1(1/2,1/2,1;z)$

Published on 29 Nov 2013 - 8:42

score 1.8k · Answer 14 · 2026-03-31 17:51:42

1.8k

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Different methods to prove $\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{s\pi}{2}\right) \Gamma (1-s) \zeta (1-s)$.

Published on 31 Mar 2026 - 17:51

score 1.7k · Answer 15 · 2026-03-30 05:13:55

1.7k

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Conjecture $_2F_1\left(\frac14,\frac34;\,\frac23;\,\frac13\right)=\frac1{\sqrt{\sqrt{\frac4{\sqrt{2-\sqrt[3]4}}+\sqrt[3]{4}+4}-\sqrt{2-\sqrt[3]4}-2}}$

Published on 30 Mar 2026 - 5:13

List Question

Simplification of $G_{2,4}^{4,2}\left(\frac18,\frac12\middle|\begin{array}{c}\frac12,\frac12\\0,0,\frac12,\frac12\\\end{array}\right)$

A closed form for $\int_0^1{_2F_1}\left(-\frac{1}{4},\frac{5}{4};\,1;\,\frac{x}{2}\right)^2dx$

Integral $\int_0^\infty{_1F_2}\left(\begin{array}{c}\tfrac12\\1,\tfrac32\end{array}\middle|-x\right)\frac{dx}{1+4\,x}$

Monstrous Moonshine for $M_{24}$?

Derivative of the Meijer G-function with respect to one of its parameters

How to speedup evaluation of hypergeometric ${}_3 F_2(1)$?

Definite Integral of Modified Bessel function representation

Generalizing $\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{\operatorname dx}{x^{2}+1} = \frac{5\pi^{2}}{96}$

Solving Some Transcendental Equations

Are elementary and generalized hypergeometric functions sufficient to express all algebraic numbers?

Using the Maclaurin series of $\arctan(x)$ to evaluate $\int_{0}^{\frac{\pi}{2}} \arctan( a \sin x) \, \mathrm dx$

Bessel function with complex argument

About large z behavior of hypergeometric function $_2F_1(1/2,1/2,1;z)$

Different methods to prove $\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{s\pi}{2}\right) \Gamma (1-s) \zeta (1-s)$.

Conjecture $_2F_1\left(\frac14,\frac34;\,\frac23;\,\frac13\right)=\frac1{\sqrt{\sqrt{\frac4{\sqrt{2-\sqrt[3]4}}+\sqrt[3]{4}+4}-\sqrt{2-\sqrt[3]4}-2}}$

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