find u_n = ∫_0 ^∞ e^(−n[x]) cos(nx)dx and v_n =∫_0 ^∞ e^(n[x]) sin(nx)dx 2) find nature of Σ u_n v_n and Σ (u_n /v


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Question Number 45976 by maxmathsup by imad last updated on 19/Oct/18

$f i n d u_{n} = \int_{0}^{\infty} e^{- n [x]} c o s (n x) d x a n d v_{n} = \int_{0}^{\infty} e^{n [x]} s i n (n x) d x$ $2) f i n d n a t u r e o f Σ u_{n} v_{n} a n d Σ \frac{u_{n}}{v_{n}}$
Commented by maxmathsup by imad last updated on 20/Oct/18
$we have u_n +iv_n = ∫_0 ^∞ e^(−n[x]) e^(inx) dx = ∫_0 ^∞ e^(−n[x] +inx) dx = Σ_(k=0) ^∞ ∫_k ^(k+1) e^(−kn +inx) dx =Σ_(k=0) ^∞ e^(−nk) ∫_k ^(k+1) e^(inx) dx =Σ_(k=0) ^∞ e^(−nk) (1/(in))[e^(inx) ]_k ^(k+1) =(1/(in)) Σ_(k=0) ^∞ e^(−nk) { e^(in(k+1)) −e^(ink) } =(1/(in)) e^(in) Σ_(k=0) ^∞ e^(−nk+ink) −(1/(in)) Σ_(k=0) ^∞ e^(−nk+ink) =(1/(in)) e^(in) Σ_(k=0) ^∞ (e^(−n+in) )^k −(1/(in)) Σ_(k=0) ^∞ (e^(−n+in) )^k =(1/(in)) e^(in) (1/(1−e^(−n+in) )) −(1/(in)) (1/(1−e^(−n+in) )) =((e^(in) −1)/(in)) (1/(1−e^(−n+in) )) =((cos(n)+isin(n)−1)/(in)) (1/(1−e^(−n) (cosn +isin(n)))) =((−i(cos(n)+isin(n)−1))/n) (1/(1−e^(−n) cos(n)−ie^(−n) sin(n))) =((sin(n)−i cos(n) +i)/n) ((1−e^(−n) cos(n)+i e^(−n) sin(n))/((1−e^(−n) cos(n))^2 +e^(−2n) sin^2 (n))) af ter we separate Re (..) and Im(...)...$
$w e h a v e u_{n} + {i v}_{n} = \int_{0}^{\infty} e^{- n [x]} e^{i n x} d x = \int_{0}^{\infty} e^{- n [x] + i n x} d x$ $= \sum_{k = 0}^{\infty} \int_{k}^{k + 1} e^{- k n + i n x} d x$ $= \sum_{k = 0}^{\infty} e^{- n k} \int_{k}^{k + 1} e^{i n x} d x = \sum_{k = 0}^{\infty} e^{- n k} \frac{1}{i n} {[e^{i n x}]}_{k}^{k + 1}$ $= \frac{1}{i n} \sum_{k = 0}^{\infty} e^{- n k} {e^{i n (k + 1)} - e^{i n k}}$ $= \frac{1}{i n} e^{i n} \sum_{k = 0}^{\infty} e^{- n k + i n k} - \frac{1}{i n} \sum_{k = 0}^{\infty} e^{- n k + i n k}$ $= \frac{1}{i n} e^{i n} \sum_{k = 0}^{\infty} {(e^{- n + i n})}^{k} - \frac{1}{i n} \sum_{k = 0}^{\infty} {(e^{- n + i n})}^{k}$ $= \frac{1}{i n} e^{i n} \frac{1}{1 - e^{- n + i n}} - \frac{1}{i n} \frac{1}{1 - e^{- n + i n}} = \frac{e^{i n} - 1}{i n} \frac{1}{1 - e^{- n + i n}}$ $= \frac{c o s (n) + i s i n (n) - 1}{i n} \frac{1}{1 - e^{- n} (c o s n + i s i n (n))}$ $= \frac{- i (c o s (n) + i s i n (n) - 1)}{n} \frac{1}{1 - e^{- n} c o s (n) - {i e}^{- n} s i n (n)}$ $= \frac{s i n (n) - i c o s (n) + i}{n} \frac{1 - e^{- n} c o s (n) + i e^{- n} s i n (n)}{{(1 - e^{- n} c o s (n))}^{2} + e^{- 2 n} {s i n}^{2} (n)}$ $a f t e r w e s e p a r a t e R e (. .) a n d I m (. . .) . . .$
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