calculate-n-0-1-4n-1-3-

Question Number 157823 by mnjuly1970 last updated on 28/Oct/21

c a l c u l a t e :

Ω := \underset{n = 0}{\sum^{\infty}} \frac{1}{{(4 n + 1)}^{3}} = ?

Answered by qaz last updated on 28/Oct/21

\underset{n = 0}{\sum^{\infty}} \frac{1}{{(4 n + 1)}^{3}}

= \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} x^{4 n} \ln^{2} xdx

= \frac{1}{2} \int_{0}^{1} \frac{\ln^{2} x}{1 - x^{4}} dx

= \frac{1}{4} \int_{0}^{1} (\frac{1}{1 - x^{2}} + \frac{1}{1 + x^{2}}) \ln^{2} xdx

= \frac{1}{4} \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} (1 + {(- 1)}^{n}) x^{2 n} \ln^{2} xdx

= \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{1 + {(- 1)}^{n}}{{(2 n + 1)}^{3}}

= \frac{1}{2} (1 - 2^{- 3}) ζ (3) + \frac{1}{2} β (3)

= \frac{7}{16} ζ (3) + \frac{π^{3}}{64}

Commented by mnjuly1970 last updated on 28/Oct/21

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