let A_ = ∫_0 ^∞ e^(−x) cos[x]dx and B = ∫_0 ^∞ e^(−[x]) cosxdx calculate A−B .


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Question Number 34298 by math khazana by abdo last updated on 03/May/18

$l e t A_{} = \int_{0}^{\infty} e^{- x} c o s [x] d x a n d B = \int_{0}^{\infty} e^{- [x]} c o s x d x$ $c a l c u l a t e A - B .$
Commented by math khazana by abdo last updated on 08/May/18

$A = {l i m}_{n \to + \infty} \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} e^{- x} c o s k d x b u t$ $\sum_{k = 0}^{n - 1} \int_{k}^{k + 1} e^{- x} c o s k d x = \sum_{k = 0}^{n - 1} c o s k {[- e^{- x}]}_{k}^{k + 1}$ $= \sum_{k = 0}^{n - 1} c o s k (e^{- k} - e^{- (k + 1)})$ $= \sum_{k = 0}^{n - 1} e^{- k} c o s k - \sum_{k = 0}^{n - 1} e^{- (k + 1)} c o s k \Rightarrow$ $A = \sum_{k = 0}^{\infty} e^{- k} c o s k - \sum_{k = 0}^{\infty} e^{- (k + 1)} c o s k$ $= \sum_{k = 0}^{\infty} e^{- k} c o s k - e^{- 1} \sum_{k = 1}^{\infty} e^{- k} c o s k$ $= \sum_{k = 0}^{\infty} e^{- k} c o s k - e^{- 1} (\sum_{k = 0}^{\infty} e^{- k} c o s k - 1)$ $= (1 - e^{- 1}) \sum_{k = 0}^{\infty} e^{- k} c o s k + \frac{1}{e}$ $\sum_{k = 0}^{\infty} e^{- k} c o s k = R e (\sum_{k = 0}^{\infty} e^{- k + i k})$ $\sum_{k = 0}^{\infty} e^{(- 1 + i) k} = \sum_{k = 0}^{\infty} {(e^{- 1 + i})}^{k} = \frac{1}{1 - e^{- 1 + i}}$ $b e c a u s e ∣ e^{- 1 + i} ∣< 1$ $= \frac{1}{1 - e^{- 1} (c o s (1) + i s i n 1)} = \frac{1}{1 - e^{- 1} c o s 1 - i e^{- 1} s i n (1)}$ $= \frac{1 - e^{- 1} c o s (1) + i e^{- 1} s i n (1)}{{(1 - e^{- 1} c o s (1))}^{2} + e^{- 2} {s i n}^{2} (1)} \Rightarrow$ $\sum_{k = 0}^{\infty} e^{- k} c o s (k) = \frac{1 - e^{- 1} c o s (1)}{1 - 2 e^{- 1} c o s (1) + e^{- 2}} \Rightarrow$ $A = \frac{(1 - e^{- 1}) (1 - e^{- 1} c o s (1))}{1 - 2 e^{- 1} c o s (1) + e^{- 2}} + \frac{1}{e} .$
Commented by abdo mathsup 649 cc last updated on 08/May/18

$B = {l i m}_{n \to + \infty} \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} e^{- k} c o s x d x$ $= \sum_{k = 0}^{n - 1} e^{- k} {[s i n x]}_{k}^{k + 1} = \sum_{k = 0}^{n - 1} e^{- k} (s i n (k + 1) - s i n k)$ $= \sum_{k = 0}^{n - 1} e^{- k} s i n (k + 1) - \sum_{k = 0}^{n - 1} e^{- k} s i n k$ $= \sum_{k = 1}^{n} e^{- (k - 1)} s i n k - \sum_{k = 0}^{n - 1} e^{- k} s i n k$ $= e (\sum_{k = 1}^{n} e^{- k} s i n k) - \sum_{k = 0}^{n - 1} e^{- k} s i n k$ $\Rightarrow B = (e - 1) \sum_{k = 0}^{\infty} e^{- k} s i n k$ $= (e - 1) I m (\sum_{k = 0}^{\infty} e^{- k + i k})$ $= (e - 1) I m (\sum_{k = 0}^{\infty} e^{(- 1 + i) k})$ $\sum_{k = 0}^{\infty} e^{(- 1 + i) k} = \frac{1}{1 - e^{- 1 + i}}$ $= \frac{1}{1 - e^{- 1} (c o s 1 + i s i n (1))}$ $= \frac{1}{1 - e^{- 1} c o s (1) - {i e}^{- 1} s i n (1)}$ $= \frac{1 - e^{- 1} c o s (1) + i e^{- 1} s i n (1)}{{(1 - e^{- 1} c o s (1))}^{2} + e^{- 2} {s i n}^{2} (1)} \Rightarrow$ $B = \frac{(e - 1) e^{- 1} s i n (1)}{{(1 - e^{- 1} c o s (1))}^{2} + e^{- 2} {s i n}^{2} (1)}$ $s o t h e v a l u e o f A - B i s k n o w n .$
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