let I = ∫_0 ^∞ e^(−[x]) cos^2 (2πx)dx and J =∫_0 ^∞ e^(−[x]) sin^2 (2πx) dx 1) calculate I +J and I −J 2) find the value of I and J .


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Question Number 37884 by prof Abdo imad last updated on 18/Jun/18

$l e t I = \int_{0}^{\infty} e^{- [x]} {c o s}^{2} (2 π x) d x a n d$ $J = \int_{0}^{\infty} e^{- [x]} {s i n}^{2} (2 π x) d x$ $1) c a l c u l a t e I + J a n d I - J$ $2) f i n d t h e v a l u e o f I a n d J .$
Commented by math khazana by abdo last updated on 20/Jun/18
$1) I+J = ∫_0 ^∞ e^(−[x]) dx=Σ_(n=0) ^∞ ∫_n ^(n+1) e^(−n) dx =Σ_(n=0) ^∞ (e^(−1) )^n = (1/(1−e^(−1) )) = (e/(e−1)) . I −J = ∫_0 ^∞ e^(−[x]) cos(4πx)dx =Σ_(n=0) ^∞ ∫_n ^(n+1) e^(−n) cos(4πx)dx =Σ_(n=0) ^∞ e^(−n) ∫_n ^(n+1) cos(4πx)dx = Σ_(n=0) ^∞ e^(−n) [ (1/(4π))sin(4πx)]_n ^(n+1) = (1/(4π)) Σ_(n=0) ^∞ e^(−n) { sin(4π(n+1))−sin(4πn)} =0 ⇒ I =J = (e/(2(e−1))) .$
$1) I + J = \int_{0}^{\infty} e^{- [x]} d x = \sum_{n = 0}^{\infty} \int_{n}^{n + 1} e^{- n} d x$ $= \sum_{n = 0}^{\infty} {(e^{- 1})}^{n} = \frac{1}{1 - e^{- 1}} = \frac{e}{e - 1} .$ $I - J = \int_{0}^{\infty} e^{- [x]} c o s (4 π x) d x$ $= \sum_{n = 0}^{\infty} \int_{n}^{n + 1} e^{- n} c o s (4 π x) d x$ $= \sum_{n = 0}^{\infty} e^{- n} \int_{n}^{n + 1} c o s (4 π x) d x$ $= \sum_{n = 0}^{\infty} e^{- n} {[\frac{1}{4 π} s i n (4 π x)]}_{n}^{n + 1}$ $= \frac{1}{4 π} \sum_{n = 0}^{\infty} e^{- n} {s i n (4 π (n + 1)) - s i n (4 π n)}$ $= 0 \Rightarrow I = J = \frac{e}{2 (e - 1)} .$
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