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Question Number 80230 by M±th+et£s last updated on 01/Feb/20

	Answered by mind is power last updated on 03/Feb/20
	$let f(z)=((πcot(πz))/(1+z^4 )),Main idee use Residus Th and evaluate Integrales in Two ways pols of f are n∈Z∪{e^(i(((1+2k)/4)}π) ,k∈{0,1,2,3}} Res(f,n)=(1/(1+n^4 )) Res(f,e^(i(((1+2k)/4))π) )=(π/(4e^(i3(((1+2k)/4))π) ))cot(πe^(i(((1+2k)/4))π) )=g(k) ∫_C_R f(z)dz=0 C_R =Re^(iθ) ,θ∈[0,2π] cause∣∫_C_R f(z)∣dz≤2πR.Max∣f(z)∣ since ∣cot(z)∣≤M ⇒Max(∣f(z)∣)≤((πM)/(R^4 −1)) ⇒∣∫_C_R f(z)dz∣≤((2πMR)/(R^4 −1))→0 as R→∞ Res th⇒∫_C_R f(z)dz=2iπΣ_(z∈C_R ) Res(f,z),R→∞ ⇒ ∫_C_R f(z)dz=2iπΣ_(n∈Z) (1/(1+n^4 ))+2iπΣ_(k=0) ^3 ((πcot(πe^(i(((1+2k)/4))π) ))/4)e^(−i3(((1+2k)/4))π) =0..1 ⇒Σ_(n∈Z) (1/(n^4 +1))=−Σ_(k=0) ^3 g(k) g(3)=g(1)^− g(2)=g(0)^� ⇒Σ_(k=0) ^3 g(k)=2Re{g(0)+g(1)} g(1)=g(0)^− ⇒Σ_(k=0) ^3 g(k)=4Re{g(0)} ⇒1⇔ Σ_(n∈Z) (1/(1+n^4 ))=−4Re{(π/4)cot(πe^(i(π/4)) )}e^((−3iπ)/4) =−πRe{cot((π/(√2))+i(π/(√2)))e^(−((3iπ)/4)) } =((π(√2))/2)Re{((cos((π/(√2)))ch((π/(√2)))−isin((π/(√2)))sh((π/(√2))))/(sin((π/(√2)))ch((π/(√2)))+icos((π/(√2)))sh((π/(√2)))))(1+i)} ⇔ multiply by (sin((π/(√2)))ch((π/(√2)))−icos((π/(√2)))sh((π/(√2)))) =((π(√2))/2){((((sin(π(√2)))/2)−i((sh(π(√2)))/2))/(sin^2 ((π/(√2)))+sh^2 ((π/(√2)))))(1+i)} =((π(√2))/2){((sin(π(√2))−ish(π(√2)))/(ch(π(√2))−cos(π(√2))))(1+i)} =((π(√2))/2) ((sin(π(√2))+sh(π(√2)))/(ch(π(√2))−cos(π(√2))))=Σ_(n∈Z) (1/(1+n^4 ))=2Σ_(n≥1) (1/(n^4 +1))+1 ⇒Σ_(n≥1) (1/(1+n^4 ))=(π/(2(√2))) ((sin(π(√2))+sh(π(√2)))/(ch(π(√2))−cos(π(√2))))−(1/2)$
	$l e t f (z) = \frac{π c o t (π z)}{1 + z^{4}}, M a i n i d e e u s e R e s i d u s T h$ $a n d e v a l u a t e I n t e g r a l e s i n T w o w a y s$ $p o l s o f f a r e n \in Z \cup {e^{i (\frac{1 + 2 k}{4}} π}, k \in {0, 1, 2, 3}}$ $R e s (f, n) = \frac{1}{1 + n^{4}}$ $R e s (f, e^{i (\frac{1 + 2 k}{4}) π}) = \frac{π}{4 e^{i 3 (\frac{1 + 2 k}{4}) π}} c o t (π e^{i (\frac{1 + 2 k}{4}) π}) = g (k)$ $\int_{C_{R}} f (z) d z = 0$ $C_{R} = {R e}^{i θ}, θ \in [0, 2 π]$ $c a u s e ∣ \int_{C_{R}} f (z) ∣ d z ⩽ 2 π R . M a x ∣ f (z) ∣$ $s i n c e ∣ c o t (z) ∣⩽ M$ $\Rightarrow M a x (∣ f (z) ∣) ⩽ \frac{π M}{R^{4} - 1}$ $\Rightarrow∣ \int_{C_{R}} f (z) d z ∣⩽ \frac{2 π M R}{R^{4} - 1} \to 0 a s R \to \infty$ $R e s t h \Rightarrow \int_{C_{R}} f (z) d z = 2 i π \sum_{z \in C_{R}} R e s (f, z), R \to \infty \Rightarrow$ $\int_{C_{R}} f (z) d z = 2 i π \sum_{n \in Z} \frac{1}{1 + n^{4}} + 2 i π \underset{k = 0}{\sum^{3}} \frac{π c o t (π e^{i (\frac{1 + 2 k}{4}) π})}{4} e^{- i 3 (\frac{1 + 2 k}{4}) π} = 0 . . 1$ $\Rightarrow \sum_{n \in Z} \frac{1}{n^{4} + 1} = - \underset{k = 0}{\sum^{3}} g (k)$ $Missing \left or extra \right$ $Missing \left or extra \right$ $\Rightarrow \underset{k = 0}{\sum^{3}} g (k) = 2 R e {g (0) + g (1)}$ $Missing \left or extra \right$ $\Rightarrow 1 \Leftrightarrow$ $\sum_{n \in Z} \frac{1}{1 + n^{4}} = - 4 R e {\frac{π}{4} c o t (π e^{i \frac{π}{4}})} e^{\frac{- 3 i π}{4}}$ $= - π R e {c o t (\frac{π}{\sqrt{2}} + i \frac{π}{\sqrt{2}}) e^{- \frac{3 i π}{4}}}$ $= \frac{π \sqrt{2}}{2} R e {\frac{c o s (\frac{π}{\sqrt{2}}) c h (\frac{π}{\sqrt{2}}) - i s i n (\frac{π}{\sqrt{2}}) s h (\frac{π}{\sqrt{2}})}{s i n (\frac{π}{\sqrt{2}}) c h (\frac{π}{\sqrt{2}}) + i c o s (\frac{π}{\sqrt{2}}) s h (\frac{π}{\sqrt{2}})} (1 + i)}$ $\Leftrightarrow m u l t i p l y b y (s i n (\frac{π}{\sqrt{2}}) c h (\frac{π}{\sqrt{2}}) - i c o s (\frac{π}{\sqrt{2}}) s h (\frac{π}{\sqrt{2}}))$ $= \frac{π \sqrt{2}}{2} {\frac{\frac{s i n (π \sqrt{2})}{2} - i \frac{s h (π \sqrt{2})}{2}}{{s i n}^{2} (\frac{π}{\sqrt{2}}) + {s h}^{2} (\frac{π}{\sqrt{2}})} (1 + i)}$ $= \frac{π \sqrt{2}}{2} {\frac{s i n (π \sqrt{2}) - i s h (π \sqrt{2})}{c h (π \sqrt{2}) - c o s (π \sqrt{2})} (1 + i)}$ $= \frac{π \sqrt{2}}{2} \frac{s i n (π \sqrt{2}) + s h (π \sqrt{2})}{c h (π \sqrt{2}) - c o s (π \sqrt{2})} = \sum_{n \in Z} \frac{1}{1 + n^{4}} = 2 \sum_{n ⩾ 1} \frac{1}{n^{4} + 1} + 1$ $\Rightarrow \sum_{n ⩾ 1} \frac{1}{1 + n^{4}} = \frac{π}{2 \sqrt{2}} \frac{s i n (π \sqrt{2}) + s h (π \sqrt{2})}{c h (π \sqrt{2}) - c o s (π \sqrt{2})} - \frac{1}{2}$
	Commented by M±th+et£s last updated on 03/Feb/20

	$g o d b l e s s y o u s i r . t h a n k y o u$
	Commented by mind is power last updated on 03/Feb/20

	$W i t h e p l e a s u r S i r$
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