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Question Number 121739 by Dwaipayan Shikari last updated on 11/Nov/20

\frac{1^{13}}{e^{2 π} - 1} + \frac{2^{13}}{e^{4 π} - 1} + \frac{3^{13}}{e^{6 π} - 1} + \dots .

Answered by mindispower last updated on 13/Nov/20

f (z) = \frac{z^{13}}{e^{2 π z} - 1} \to 0, z \to 0, f^{*} (z) = {\begin{cases} f (z), z \neq 0 \\ 0 z = 0 \end{cases}

\sum_{n ⩾ 1} f (n) = \sum_{n ⩾ 0} f^{*} (n)

a b e l p l a n a f o r m u l a

\sum_{n ⩾ 0} \overset{*}{f} (n) = \int_{0}^{\infty} f (z) d z + \frac{f (0)}{2} + i \int_{0}^{\infty} \frac{f (i t) - f (- i t)}{e^{2 π t} - 1} d t

= \int_{0}^{\infty} \frac{z^{13}}{e^{2 π z} - 1} d z + i \int_{0}^{\infty} 2 i I m (f (i t)) . \frac{d t}{e^{2 π t} - 1}

\frac{{i t}^{13}}{e^{2 i π t} - 1} + \frac{{i t}^{13}}{e^{- 2 i π t} - 1} = f (i t) - f (- i t) = \frac{{i t}^{13}}{e^{2 π t} - 1} (1 - e^{2 π i t}) = - {i t}^{13}

= 2 \int_{0}^{\infty} \frac{z^{13}}{e^{2 π z} - 1} d z

2 π z = t \Rightarrow d z = \frac{d t}{2 π}, = 2 \int_{0}^{\infty} {(\frac{t}{2 π})}^{13} . \frac{d t}{2 π (e^{t} - 1)}

= \frac{1}{π^{14} . 2^{13}} \int_{0}^{\infty} \frac{t^{13}}{e^{t} - 1} d t = S

\int_{0}^{\infty} \frac{t^{s}}{e^{t} - 1} d t = \sum_{n ⩾ 0} \int_{0}^{\infty} t^{s} e^{- (1 + n) t} d t = \sum_{n ⩾ 0} Γ (s + 1) \frac{1}{{(n + 1)}^{s + 1}}

= Γ (s + 1) ζ (s + 1)

S = \frac{1}{π^{14} 2^{13}} ζ (14) Γ (14) = \frac{1}{π^{14} . 2^{13}} . \frac{2 π^{14}}{18243225} . (13)!

= \frac{1}{2^{12}} . \frac{(13)!}{25.13.7 . 9^{3} . 11} = \frac{1}{2^{12}} . (12.10.8.6.4.3.2) . \frac{1}{5 . 9^{2}})

= \frac{1}{2^{12}} (3.2.2.5 . 2^{3} . 2.3 . 2^{3} . 3) . \frac{1}{5 . 3^{4}} = \frac{1}{2^{12}} . (2^{9}) . \frac{1}{3} = \frac{1}{8.3}

= \frac{1}{24}

\sum_{n ⩾ 1} \frac{n^{13}}{e^{2 π n} - 1} = \frac{1}{24} I t r i e d m a n y w a y s t o g e t i t

Commented by Dwaipayan Shikari last updated on 14/Nov/20

G r e a t s i r! T h a n k i n g y o u