Prove that Σ_(p=0) ^∞ (((−1)^p )/(4p+1)) = ((π−argcoth((√2) ))/(4(√2))) and Σ_(p=0) ^(∞ ) (((−1)^p )/(4p+3)) = ((π+argcoth((√2) ))/(4(√2) ))


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Question Number 69231 by ~ À ® @ 237 ~ last updated on 21/Sep/19

$P r o v e t h a t \underset{p = 0}{\sum^{\infty}} \frac{{(- 1)}^{p}}{4 p + 1} = \frac{π - a r g c o t h (\sqrt{2})}{4 \sqrt{2}} a n d$ $\underset{p = 0}{\sum^{\infty}} \frac{{(- 1)}^{p}}{4 p + 3} = \frac{π + a r g c o t h (\sqrt{2})}{4 \sqrt{2}}$
Commented by mathmax by abdo last updated on 21/Sep/19
$let s(x)=Σ_(n=0) ^∞ (((−1)^n )/(4n+1))x^(4n+1) with ∣x∣<1 ⇒s^′ (x)=Σ_(n=0) ^∞ (−1)^n x^(4n) =Σ_(n=0) ^∞ (−x^4 )^n =(1/(1+x^4 )) ⇒s(x) =∫_0 ^x (dt/(1+t^4 )) +c c=s(0)=0 ⇒s(x) =∫_0 ^x (dt/(1+t^4 )) and Σ_(n=0) ^∞ (((−1)^n )/(4n+1)) =s(1)=∫_0 ^1 (dt/(t^4 +1)) let decompose F(t)=(1/(t^4 +1)) ⇒F(t)=(1/((t^2 +1)^2 −2t^2 )) =(1/((t^2 +1+(√2)t)(t^2 +1−(√2)t))) =((at +b)/(t^2 +(√2)t +1)) +((ct+d)/(t^2 −(√2)t +1)) F(−t)=F(t) ⇒((−at+b)/(t^2 −(√2)t +1)) +((−ct+d)/(t^2 +(√2)t +1)) =F(t) ⇒ c=−a and d=b ⇒ F(t) =((at+b)/(t^2 +(√2)t +1)) +((−at+b)/(t^2 −(√2)t +1)) F(o) =1 =2b ⇒b=(1/2) F(1) =((a+b)/(2+(√2))) +((−a+b)/(2−(√2))) =(1/2) ⇒(((2−(√2))(a+b)+(2+(√2))(−a+b))/2)=(1/2) ⇒(2−(√2))a +(2−(√2))b −(2+(√2))a+(2+(√2))b =1 ⇒ −2(√2)a +2 =1 ⇒a =(1/(2(√2))) ⇒ F(t) =(((1/(2(√2)))t +(1/2))/(t^2 +(√2)t +1)) +((−(1/(2(√2)))t+(1/2))/(t^2 −(√2)t +1)) =(1/(2(√2))){ ((t+(√2))/(t^2 +(√2)t+1)) −((t−(√2))/(t^2 −(√2)t +1))} ⇒ ∫_0 ^1 F(t)dt =(1/(4(√2))) ∫_0 ^1 ((2t+(√2)+(√2))/(t^2 +(√2)t +1))dt −(1/(4(√2))) ∫_0 ^1 ((2t−(√2)−(√2))/(t^2 −(√2)t +1))dt =(1/(4(√2)))[ln(t^2 +(√2)t +1)]_0 ^(1 ) +(1/4) ∫_0 ^1 (dt/(t^2 +(√2)t +1)) −(1/(4(√2))) [ln(t^2 −(√2)t +1)]_0 ^1 +(1/4) ∫_0 ^1 (dt/(t^2 −(√2)t +1)) =(1/(4(√2))){ln(2+(√2))−ln(2−(√2)) +(1/4) { ∫_0 ^1 (...)dt +∫_0 ^1 (....)dt} ∫_0 ^1 (dt/(t^2 +(√2)t +1)) =∫_0 ^1 (dt/(t^2 +2((√2)/2)t +(1/2)+(1/2))) =∫_0 ^1 (dt/((t+(1/(√2)))^(2 ) +(1/2))) =_(t+(1/(√2))=(u/(√2))) 2∫_1 ^(1+(√2)) (1/(u^2 +1)) (du/(√2)) =(√2)[arctan(u)]_1 ^(1+(√2)) =(√2){arctan(1+(√2))−(π/4)} =(√2){(π/2) −arctan((√2)−1)−(π/4)} =(√2){(π/4)−(π/8)} =(√2)×(π/8) ∫_0 ^1 (dt/(t^2 −(√2)t +1)) =∫_0 ^1 (dt/((t−(1/(√2)))^2 +(1/2))) =_(t−(1/(√2))=(u/(√2))) 2 ∫_(−1) ^((√2)−1) (1/(u^2 +1))(du/(√2)) =(√2)[arctan(u)]_(−1) ^((√2)−1) =(√2){(π/8) +(π/4)} =(√2)×((3π)/8) ⇒ ∫_0 ^1 F(t)dt =(1/(4(√2)))ln(((2+(√2))/(2−(√2)))) +(1/4){((π(√2))/8) +((3π(√2))/8)} =(1/(4(√2)))ln(((4+4(√2)+2)/2))+(1/4)((π(√2))/2) =(1/(4(√2)))ln(3+2(√2))+((π(√2))/8) ⇒ Σ_(n=0) ^∞ (((−1)^n )/(4n+1)) =(1/(4(√2)))ln(3+2(√2)) +((π(√2))/8)$
$l e t s (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} x^{4 n + 1} w i t h ∣ x ∣< 1 \Rightarrow s^{^{'}} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{4 n}$ $= \sum_{n = 0}^{\infty} {(- x^{4})}^{n} = \frac{1}{1 + x^{4}} \Rightarrow s (x) = \int_{0}^{x} \frac{d t}{1 + t^{4}} + c$ $c = s (0) = 0 \Rightarrow s (x) = \int_{0}^{x} \frac{d t}{1 + t^{4}} a n d \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} = s (1) = \int_{0}^{1} \frac{d t}{t^{4} + 1}$ $l e t d e c o m p o s e F (t) = \frac{1}{t^{4} + 1} \Rightarrow F (t) = \frac{1}{{(t^{2} + 1)}^{2} - 2 t^{2}}$ $= \frac{1}{(t^{2} + 1 + \sqrt{2} t) (t^{2} + 1 - \sqrt{2} t)} = \frac{a t + b}{t^{2} + \sqrt{2} t + 1} + \frac{c t + d}{t^{2} - \sqrt{2} t + 1}$ $F (- t) = F (t) \Rightarrow \frac{- a t + b}{t^{2} - \sqrt{2} t + 1} + \frac{- c t + d}{t^{2} + \sqrt{2} t + 1} = F (t) \Rightarrow$ $c = - a a n d d = b \Rightarrow$ $F (t) = \frac{a t + b}{t^{2} + \sqrt{2} t + 1} + \frac{- a t + b}{t^{2} - \sqrt{2} t + 1}$ $F (o) = 1 = 2 b \Rightarrow b = \frac{1}{2}$ $F (1) = \frac{a + b}{2 + \sqrt{2}} + \frac{- a + b}{2 - \sqrt{2}} = \frac{1}{2} \Rightarrow \frac{(2 - \sqrt{2}) (a + b) + (2 + \sqrt{2}) (- a + b)}{2} = \frac{1}{2}$ $\Rightarrow (2 - \sqrt{2}) a + (2 - \sqrt{2}) b - (2 + \sqrt{2}) a + (2 + \sqrt{2}) b = 1 \Rightarrow$ $- 2 \sqrt{2} a + 2 = 1 \Rightarrow a = \frac{1}{2 \sqrt{2}} \Rightarrow$ $F (t) = \frac{\frac{1}{2 \sqrt{2}} t + \frac{1}{2}}{t^{2} + \sqrt{2} t + 1} + \frac{- \frac{1}{2 \sqrt{2}} t + \frac{1}{2}}{t^{2} - \sqrt{2} t + 1}$ $= \frac{1}{2 \sqrt{2}} {\frac{t + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} - \frac{t - \sqrt{2}}{t^{2} - \sqrt{2} t + 1}} \Rightarrow$ $\int_{0}^{1} F (t) d t = \frac{1}{4 \sqrt{2}} \int_{0}^{1} \frac{2 t + \sqrt{2} + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} d t - \frac{1}{4 \sqrt{2}} \int_{0}^{1} \frac{2 t - \sqrt{2} - \sqrt{2}}{t^{2} - \sqrt{2} t + 1} d t$ $= \frac{1}{4 \sqrt{2}} {[l n (t^{2} + \sqrt{2} t + 1)]}_{0}^{1} + \frac{1}{4} \int_{0}^{1} \frac{d t}{t^{2} + \sqrt{2} t + 1}$ $- \frac{1}{4 \sqrt{2}} {[l n (t^{2} - \sqrt{2} t + 1)]}_{0}^{1} + \frac{1}{4} \int_{0}^{1} \frac{d t}{t^{2} - \sqrt{2} t + 1}$ $= \frac{1}{4 \sqrt{2}} {l n (2 + \sqrt{2}) - l n (2 - \sqrt{2}) + \frac{1}{4} {\int_{0}^{1} (. . .) d t + \int_{0}^{1} (. . . .) d t}$ $\int_{0}^{1} \frac{d t}{t^{2} + \sqrt{2} t + 1} = \int_{0}^{1} \frac{d t}{t^{2} + 2 \frac{\sqrt{2}}{2} t + \frac{1}{2} + \frac{1}{2}} = \int_{0}^{1} \frac{d t}{{(t + \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}}$ $=_{t + \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}}} 2 \int_{1}^{1 + \sqrt{2}} \frac{1}{u^{2} + 1} \frac{d u}{\sqrt{2}} = \sqrt{2} {[a r c t a n (u)]}_{1}^{1 + \sqrt{2}}$ $= \sqrt{2} {a r c t a n (1 + \sqrt{2}) - \frac{π}{4}} = \sqrt{2} {\frac{π}{2} - a r c t a n (\sqrt{2} - 1) - \frac{π}{4}}$ $= \sqrt{2} {\frac{π}{4} - \frac{π}{8}} = \sqrt{2} \times \frac{π}{8}$ $\int_{0}^{1} \frac{d t}{t^{2} - \sqrt{2} t + 1} = \int_{0}^{1} \frac{d t}{{(t - \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}} =_{t - \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}}} 2 \int_{- 1}^{\sqrt{2} - 1} \frac{1}{u^{2} + 1} \frac{d u}{\sqrt{2}}$ $= \sqrt{2} {[a r c t a n (u)]}_{- 1}^{\sqrt{2} - 1} = \sqrt{2} {\frac{π}{8} + \frac{π}{4}} = \sqrt{2} \times \frac{3 π}{8} \Rightarrow$ $\int_{0}^{1} F (t) d t = \frac{1}{4 \sqrt{2}} l n (\frac{2 + \sqrt{2}}{2 - \sqrt{2}}) + \frac{1}{4} {\frac{π \sqrt{2}}{8} + \frac{3 π \sqrt{2}}{8}}$ $= \frac{1}{4 \sqrt{2}} l n (\frac{4 + 4 \sqrt{2} + 2}{2}) + \frac{1}{4} \frac{π \sqrt{2}}{2} = \frac{1}{4 \sqrt{2}} l n (3 + 2 \sqrt{2}) + \frac{π \sqrt{2}}{8} \Rightarrow$ $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} = \frac{1}{4 \sqrt{2}} l n (3 + 2 \sqrt{2}) + \frac{π \sqrt{2}}{8}$
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