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Question Number 110539 by Her_Majesty last updated on 29/Aug/20

$$itseemstoohardformanyheretopost$$$$theirquestionsasquestions,answersas$$$$answersandcommentsascomments...$$$$herebyIintroducethenextstep:I^{\prime }llpost$$$$answersandthetaskis,findquestions$$$$totheseanswers$$$$\left(1\right)\zeta \left(3\right)+\pi ln2$$$$\left(2\right)\pi H_0\left(7\right)$$$$\left(3\right)true\mathrm{\forall }x\in \mathbb{C}\mathrm{\setminus }\mathbb{Q}$$$$\left(4\right){}_2{F}_1(\frac{3}{2},\frac{1}{5},\frac{2}{3},{sin}^{-1}\frac{x+1}{x-1})$$

Commented by Eric002 last updated on 29/Aug/20

$$ithinkonlyprof.mathmindand$$$$andprof.\ast M\ast th+et+shaveknowledge$$$$of\left(4\right)hypergeometricfunctionandthey$$$${haven}^{\prime }tbeenaroundforalongtime$$$$$$

Commented by Dwaipayan Shikari last updated on 29/Aug/20

$$1)\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}+\left((\frac{22}{7}-{\int }_0^1\frac{x^4{(1-x)}^4}{1+x^2})\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n{2}^n}\right)$$$$Or\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}+\left((\frac{22}{7}-{\int }_0^1\frac{x^4{(1-x)}^4}{1+x^2})\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}}{n}\right)$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}+\frac{22}{7}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n{2}^n}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n{2}^n}\left({\int }_0^1\frac{x^4{(1-x)}^4}{1+x^2}\right)$$

Commented by Dwaipayan Shikari last updated on 29/Aug/20

Infinitely many questions ����

Commented by Her_Majesty last updated on 29/Aug/20

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