Advanced-Mathematics-Evaluate-n-1-coth-pin-n-3-

Question Number 143428 by mnjuly1970 last updated on 14/Jun/21

\dots A d v a n c e d \dots \dots M a t h e m a t i c s \dots

E v a l u a t e ::

ϕ := \underset{n = 1}{\sum^{\infty}} \frac{c o t h (π n)}{n^{3}} = ?

Answered by mindispower last updated on 16/Jun/21

l e t f (z) = \frac{π c o t h (π z) c o t (π z)}{z^{3}}

p o l s o f f a r e {i k, k; k \in Z}

l e t C_{R} : {{R e}^{i θ}, θ \in [0, 2 π [}

w e u s e t h e r e i s i d u s t h e o r e m

w i t h e t h e f a c t t h a t c o t h, c o t a r e b o u n d e d

\Rightarrow

\lim_{R \to \infty} \int_{C_{R}} f (z) d z = 0

R e s i d u e t h e r e m \Rightarrow Σ R e s (f) = 0

R e s (f, k) = \lim_{x \to k} (x - k) . π \frac{c o t h (π x) c o t (π x)}{x^{3}} = \frac{c o t h (π k)}{k^{3}}, k \neq 0

R e s (f, i k) = \frac{c o t h (π k)}{k^{3}}, k \neq 0

\Rightarrow \sum_{k \in Z - {0}} . \frac{c o t h (π k)}{k^{3}} + \sum_{k \in Z - {0}} \frac{c o t h (π k)}{k^{3}} + R e s (f, 0) = 0

\Rightarrow 4 \sum_{k ⩾ 1} \frac{c o t h (π k)}{k^{3}} = - R e s (f, 0)

R e s (f, 0)

c o t a n (π x) = \frac{1}{π x} - \frac{π x}{3} - \frac{π^{3} x^{3}}{45} + 0 (x^{3})

c o t h (π x) = i c o t (i π x) = \frac{1}{π x} + \frac{π x}{3} - \frac{π^{3}}{45} x^{3}

R e s (f, 0), c o e f i c u e n t o f \frac{1}{x} i n π \frac{c o t h (π x) c o t (π x)}{x^{3}}

= \frac{π}{x^{3}} (- \frac{π^{2}}{9} - 2 \frac{π^{2}}{45}) x^{2} = - \frac{7 π^{3}}{45}

4 \sum_{n ⩾ 1} \frac{c o t h (π n)}{n^{3}} = - . - \frac{7 π^{3}}{45}

\sum_{n ⩾ 1} \frac{c o t h (π n)}{n^{3}} = \frac{7 π^{3}}{180}