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

$. . . A d v a n c e d . . . . . . M a t h e m a t i c s . . .$ $E v a l u a t e ::$ $ϕ := \underset{n = 1}{\sum^{\infty}} \frac{c o t h (π n)}{n^{3}} = ?$
	Answered by mindispower last updated on 16/Jun/21
	$let f(z)=((πcoth(πz)cot(πz))/z^3 ) pols of f are {ik,k;k∈Z} let C_R :{Re^(iθ) ,θ∈[0,2π[} we use the reisidus theorem withe the fact that coth,cot are bounded ⇒ lim_(R→∞) ∫_C_R f(z)dz=0 Residue therem⇒ΣRes(f)=0 Res(f,k)=lim_(x→k) (x−k).π((coth(πx)cot(πx))/x^3 )=((coth(πk))/k^3 ),k≠0 Res(f,ik)=((coth(πk))/k^3 ),k≠0 ⇒Σ_(k∈Z−{0}) .((coth(πk))/k^3 )+Σ_(k∈Z−{0}) ((coth(πk))/k^3 )+Res(f,0)=0 ⇒4Σ_(k≥1) ((coth(πk))/k^3 )=−Res(f,0) Res(f,0) cotan(πx)=(1/(πx))−((πx)/3)−((π^3 x^3 )/(45))+0(x^3 ) coth(πx)=icot(iπx)=(1/(πx))+((πx)/3)−(π^3 /(45))x^3 Res(f,0),coeficuent of (1/x)in π((coth(πx)cot(πx))/x^3 ) =(π/x^3 )(−(π^2 /9)−2(π^2 /(45)))x^2 =−((7π^3 )/(45)) 4Σ_(n≥1) ((coth(πn))/n^3 )=−.−((7π^3 )/(45)) Σ_(n≥1) ((coth(πn))/n^3 )=((7π^3 )/(180))$
	$l e t f (z) = \frac{π c o t h (π z) c o t (π z)}{z^{3}}$ $p o l s o f f a r e {i k, k; k \in Z}$ $l e t C_{R} : {{R e}^{i θ}, θ \in [0, 2 π [}$ $w e u s e t h e r e i s i d u s t h e o r e m$ $w i t h e t h e f a c t t h a t c o t h, c o t a r e b o u n d e d$ $\Rightarrow$ $\lim_{R \to \infty} \int_{C_{R}} f (z) d z = 0$ $R e s i d u e t h e r e m \Rightarrow Σ R e s (f) = 0$ $R e s (f, k) = \lim_{x \to k} (x - k) . π \frac{c o t h (π x) c o t (π x)}{x^{3}} = \frac{c o t h (π k)}{k^{3}}, k \neq 0$ $R e s (f, i k) = \frac{c o t h (π k)}{k^{3}}, k \neq 0$ $\Rightarrow \sum_{k \in Z - {0}} . \frac{c o t h (π k)}{k^{3}} + \sum_{k \in Z - {0}} \frac{c o t h (π k)}{k^{3}} + R e s (f, 0) = 0$ $\Rightarrow 4 \sum_{k ⩾ 1} \frac{c o t h (π k)}{k^{3}} = - R e s (f, 0)$ $R e s (f, 0)$ $c o t a n (π x) = \frac{1}{π x} - \frac{π x}{3} - \frac{π^{3} x^{3}}{45} + 0 (x^{3})$ $c o t h (π x) = i c o t (i π x) = \frac{1}{π x} + \frac{π x}{3} - \frac{π^{3}}{45} x^{3}$ $R e s (f, 0), c o e f i c u e n t o f \frac{1}{x} i n π \frac{c o t h (π x) c o t (π x)}{x^{3}}$ $= \frac{π}{x^{3}} (- \frac{π^{2}}{9} - 2 \frac{π^{2}}{45}) x^{2} = - \frac{7 π^{3}}{45}$ $4 \sum_{n ⩾ 1} \frac{c o t h (π n)}{n^{3}} = - . - \frac{7 π^{3}}{45}$ $\sum_{n ⩾ 1} \frac{c o t h (π n)}{n^{3}} = \frac{7 π^{3}}{180}$
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