Σ_(n=0) ^∞ arctan (((−1)^n )/(2n+1))=?


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Question Number 158822 by qaz last updated on 09/Nov/21

$\underset{n = 0}{\sum^{\infty}} \arctan \frac{{(- 1)}^{n}}{2 n + 1} = ?$
	Answered by mindispower last updated on 09/Nov/21
	$Σ_(n≥0) {arctan((1/(4n+1)))−arctan((1/(4n+3)))} =Σ_(n≥0) ∫_0 ^1 (((4n+1))/((4n+1)^2 +x^2 ))−(((4n+3))/((4n+3)^2 +x^2 ))dx =(1/2)Σ_(n≥0) ∫_0 ^1 {(dx/(4n+1+ix))+(dx/(4n+1−ix))}−(1/2){∫_0 ^1 (dx/(4n+3+ix))+(dx/(4n+3−ix))} =(1/8)∫_0 ^1 Σ_(n≥1) ((1/(n+((1+ix)/4)))−(1/n)+(1/(n+((1−ix)/4)))−(1/n))dx −(1/8)∫_0 ^1 Σ_(n≥1) ((1/(n+((3+ix)/4)))−(1/n)+(1/(n+((3−ix)/4)))−(1/n))dx +arctan(1)−arctan((1/3)) =(1/8)∫_0 ^1 (H_((1+ix)/4) +H_((1−ix)/4) −H_((3+ix)/4) +H_((3−ix)/4) )+arctan(1)−arctan((1/3)) =∫_0 ^1 ((Ψ(((1+ix)/4))−Ψ(((3−ix)/4))+Ψ(((1−ix)/4))−Ψ(((3+ix)/4)))/8)dx+(π/4)−arctan(3) =(1/(2i))(log(Γ(((1+i)/4)))−logΓ((1/4))+logΓ(((3−i)/4))−logΓ((3/4)) −logΓ(((1−i)/4))+logΓ((1/4))−logΓ(((3+i)/4))+logΓ((3/4)))+(π/4)−arctan 3 =(1/2)log(((Γ(((1+i)/4))Γ(((3−i)/4)))/(Γ(((1−i)/4))Γ(((3+i)/4)))))+(π/4)−arctan((1/3)) =(1/2)log((π/(sin(((π(1+i))/4)))).((sin(((π(1−i))/4)))/π))+(π/4)−arctan((1/3)) =(1/(2i))log(((1−i)/(1+i)))+(π/4)−arctan((1/3))=(1/(2i))log(−i)+(π/4)−arctan((1/3)) =−arctan((1/3))$
	$\sum_{n ⩾ 0} {a r c t a n (\frac{1}{4 n + 1}) - a r c t a n (\frac{1}{4 n + 3})}$ $= \sum_{n ⩾ 0} \int_{0}^{1} \frac{(4 n + 1)}{{(4 n + 1)}^{2} + x^{2}} - \frac{(4 n + 3)}{{(4 n + 3)}^{2} + x^{2}} d x$ $= \frac{1}{2} \sum_{n ⩾ 0} \int_{0}^{1} {\frac{d x}{4 n + 1 + i x} + \frac{d x}{4 n + 1 - i x}} - \frac{1}{2} {\int_{0}^{1} \frac{d x}{4 n + 3 + i x} + \frac{d x}{4 n + 3 - i x}}$ $= \frac{1}{8} \int_{0}^{1} \sum_{n ⩾ 1} (\frac{1}{n + \frac{1 + i x}{4}} - \frac{1}{n} + \frac{1}{n + \frac{1 - i x}{4}} - \frac{1}{n}) d x$ $- \frac{1}{8} \int_{0}^{1} \sum_{n ⩾ 1} (\frac{1}{n + \frac{3 + i x}{4}} - \frac{1}{n} + \frac{1}{n + \frac{3 - i x}{4}} - \frac{1}{n}) d x$ $+ a r c t a n (1) - a r c t a n (\frac{1}{3})$ $= \frac{1}{8} \int_{0}^{1} (H_{\frac{1 + i x}{4}} + H_{\frac{1 - i x}{4}} - H_{\frac{3 + i x}{4}} + H_{\frac{3 - i x}{4}}) + a r c t a n (1) - a r c t a n (\frac{1}{3})$ $= \int_{0}^{1} \frac{Ψ (\frac{1 + i x}{4}) - Ψ (\frac{3 - i x}{4}) + Ψ (\frac{1 - i x}{4}) - Ψ (\frac{3 + i x}{4})}{8} d x + \frac{π}{4} - a r c t a n (3)$ $= \frac{1}{2 i} (l o g (Γ (\frac{1 + i}{4})) - l o g Γ (\frac{1}{4}) + l o g Γ (\frac{3 - i}{4}) - l o g Γ (\frac{3}{4})$ $- l o g Γ (\frac{1 - i}{4}) + l o g Γ (\frac{1}{4}) - l o g Γ (\frac{3 + i}{4}) + l o g Γ (\frac{3}{4})) + \frac{π}{4} - a r c t a n 3$ $= \frac{1}{2} l o g (\frac{Γ (\frac{1 + i}{4}) Γ (\frac{3 - i}{4})}{Γ (\frac{1 - i}{4}) Γ (\frac{3 + i}{4})}) + \frac{π}{4} - a r c t a n (\frac{1}{3})$ $= \frac{1}{2} l o g (\frac{π}{s i n (\frac{π (1 + i)}{4})} . \frac{s i n (\frac{π (1 - i)}{4})}{π}) + \frac{π}{4} - a r c t a n (\frac{1}{3})$ $= \frac{1}{2 i} l o g (\frac{1 - i}{1 + i}) + \frac{π}{4} - a r c t a n (\frac{1}{3}) = \frac{1}{2 i} l o g (- i) + \frac{π}{4} - a r c t a n (\frac{1}{3})$ $= - a r c t a n (\frac{1}{3})$
	Commented by mindispower last updated on 10/Nov/21

	$p l e a s u r$ $t h e r e i s a m i s t a k i n c a l c u l u s i w i l l t c h e k i t$ $a f t e r s o r r y$
	Commented by qaz last updated on 10/Nov/21

	$thanks a lot . sir$
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