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Question Number 62455 by aliesam last updated on 21/Jun/19

Commented by mathmax by abdo last updated on 22/Jun/19
$S =Σ_(n=1) ^∞ A_n with A_n =∫_(n−(π/2)) ^(n+(π/2)) e^(−x) ∣cos(x)∣dx changement x =n−(π/2) +t give A_n = ∫_0 ^π e^(−(n−(π/2)+t))) ∣cos((π/2)−(n+t))∣dt = e^((π/2)−n) ∫_0 ^π e^(−t) ∣sin(n+t)∣ dt if we suppose sin(n+t)≥0 on[0,π]( that need a proof) ⇒A_n = e^((π/2)−n) ∫_0 ^π (sin(n)cos(t) +cos(n)sin(t))dt =sin(n) e^((π/2)−n) ∫_0 ^π cost dt+cos(n) e^((π/2)−n) ∫_0 ^π sint dt =0 +cos(n) e^((π/2)−n) [−cost]_0 ^π =2 cos(n) e^((π/2)−n) ⇒ S =2Σ_(n=1) ^∞ cos(n)e^((π/2)−n) =2 e^(π/2) Σ_(n=1) ^∞ e^(−n) cos(n) =2 e^(π/2) (Σ_(n=0) ^∞ e^(−n) cos(n)−1) but Σ_(n=0) ^∞ e^(−n) cosn =Re( Σ_(n=0) ^∞ e^(−n+in) ) =Re(Σ_(n=0) ^∞ (e^(−1+i) )^n ) we have ∣e^(−1 +i) ∣ =(1/e)<1 ⇒ Σ_(n=0) ^∞ (e^(−1+i) )^n =(1/(1−e^(−1+i) )) =(1/(1−e^(−1) (cos(1) +isin(1)))) =(1/(1−e^(−1) cos(1)−ie^(−1) sin(1))) =((1−e^(−1) cos(1)+e^(−1) sin(1))/((1−e^(−1) cos(1))^2 +e^(−2) sin^2 (1))) ⇒ Σ_(n=0) ^∞ e^(−n) cos(n) =((1−e^(−1) cos(1))/((1−e^(−1) cos(1))^2 +e^(−2) sin^2 (1))) ⇒ S =2 e^(π/2) { ((1−e^(−1) cos(1))/(1−2e^(−1) cos(1) +e^(−2) )) −1} .$
$S = \sum_{n = 1}^{\infty} A_{n} w i t h A_{n} = \int_{n - \frac{π}{2}}^{n + \frac{π}{2}} e^{- x} ∣ c o s (x) ∣ d x c h a n g e m e n t x = n - \frac{π}{2} + t g i v e$ $A_{n} = \int_{0}^{π} e^{- (n - \frac{π}{2} + t))} ∣ c o s (\frac{π}{2} - (n + t)) ∣ d t$ $= e^{\frac{π}{2} - n} \int_{0}^{π} e^{- t} ∣ s i n (n + t) ∣ d t i f w e s u p p o s e s i n (n + t) ⩾ 0 o n [0, π] (t h a t n e e d$ $a p r o o f) \Rightarrow A_{n} = e^{\frac{π}{2} - n} \int_{0}^{π} (s i n (n) c o s (t) + c o s (n) s i n (t)) d t$ $= s i n (n) e^{\frac{π}{2} - n} \int_{0}^{π} c o s t d t + c o s (n) e^{\frac{π}{2} - n} \int_{0}^{π} s i n t d t$ $= 0 + c o s (n) e^{\frac{π}{2} - n} {[- c o s t]}_{0}^{π} = 2 c o s (n) e^{\frac{π}{2} - n} \Rightarrow$ $S = 2 \sum_{n = 1}^{\infty} c o s (n) e^{\frac{π}{2} - n} = 2 e^{\frac{π}{2}} \sum_{n = 1}^{\infty} e^{- n} c o s (n)$ $= 2 e^{\frac{π}{2}} (\sum_{n = 0}^{\infty} e^{- n} c o s (n) - 1) b u t$ $\sum_{n = 0}^{\infty} e^{- n} c o s n = R e (\sum_{n = 0}^{\infty} e^{- n + i n}) = R e (\sum_{n = 0}^{\infty} {(e^{- 1 + i})}^{n}) w e h a v e ∣ e^{- 1 + i} ∣ = \frac{1}{e} < 1 \Rightarrow$ $\sum_{n = 0}^{\infty} {(e^{- 1 + i})}^{n} = \frac{1}{1 - e^{- 1 + i}} = \frac{1}{1 - e^{- 1} (c o s (1) + i s i n (1))}$ $= \frac{1}{1 - e^{- 1} c o s (1) - {i e}^{- 1} s i n (1)} = \frac{1 - e^{- 1} c o s (1) + e^{- 1} s i n (1)}{{(1 - e^{- 1} c o s (1))}^{2} + e^{- 2} {s i n}^{2} (1)} \Rightarrow$ $\sum_{n = 0}^{\infty} e^{- n} c o s (n) = \frac{1 - e^{- 1} c o s (1)}{{(1 - e^{- 1} c o s (1))}^{2} + e^{- 2} {s i n}^{2} (1)} \Rightarrow$ $S = 2 e^{\frac{π}{2}} {\frac{1 - e^{- 1} c o s (1)}{1 - 2 e^{- 1} c o s (1) + e^{- 2}} - 1} .$
Commented by aliesam last updated on 22/Jun/19

$thank you sir brilliant solution$
Commented by mathmax by abdo last updated on 22/Jun/19

$y o u a r e w e l c o m e s i r .$
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