1) calculate A_n =∫_0 ^∞ e^(−n[x]) sin(x)dx with n integr and n≥1 2) find nature of Σ_(n=1) ^∞ A


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Question Number 49636 by maxmathsup by imad last updated on 08/Dec/18

$1) c a l c u l a t e A_{n} = \int_{0}^{\infty} e^{- n [x]} s i n (x) d x w i t h n i n t e g r a n d n ⩾ 1$ $2) f i n d n a t u r e o f \sum_{n = 1}^{\infty} A_{n}$
Commented by Abdo msup. last updated on 10/Dec/18

$1) A_{n} = \sum_{p = 0}^{\infty} \int_{p}^{p + 1} e^{- n p} s i n (x) d x = \sum_{p = 0}^{\infty} e^{- n p} A_{p}$ $A_{p} = \int_{p}^{p + 1} s i n x d x = {[- c o s x]}_{p}^{p + 1} = c o s p - c o s (p + 1) \Rightarrow$ $A_{n} = \sum_{p = 0}^{\infty} e^{- n p} c o s p - \sum_{p = 0}^{\infty} e^{- n p} c o s (p + 1)$ $= \sum_{p = 0}^{\infty} e^{- n p} c o s p - \sum_{p = 1}^{\infty} e^{- n (p - 1)} c o s p$ $= 1 + (1 - e^{n}) \sum_{p = 1}^{\infty} e^{- n p} c o s p b u t$ $\sum_{p = 1}^{\infty} e^{- n p} c o s p = R e (\sum_{p = 1}^{\infty} e^{- n p + i p})$ $\sum_{p = 1}^{\infty} e^{(- n + i) p} = \sum_{p = 0}^{\infty} e^{(- n + i) p} - 1$ $= \frac{1}{1 - e^{- n + i}} - 1 = \frac{1}{1 - e^{- n} (c o s 1 + i s i n 1)} - 1$ $= \frac{1 - e^{- n} c o s (1) + i s i n (1)}{{(1 - e^{- n} c o s (1))}^{2} + e^{- 2 n} {s i n}^{2} (1)} - 1 \Rightarrow$ $R e (. . .) = \frac{1 - e^{- n} c o s (1)}{1 - 2 e^{- n} c o s (1) + e^{- 2 n}} - 1 \Rightarrow$ $A_{n} = 1 + (1 - e^{- n}) (\frac{1 - e^{- n} c o s (1)}{1 - 2 e^{- n} c o s (1) + e^{- 2 n}} - 1) .$
Commented by Abdo msup. last updated on 10/Dec/18

$2) w e h s v e {l i m}_{n \to + \infty} A_{n} = 1 \Rightarrow Σ A_{n} d i v e r g e s .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Dec/18
	$P_n =∫_0 ^∞ e^(−n[x]) cosxdx Q_n =∫_0 ^∞ e^(−n[x]) sinx A_n =Q_n P_n +iQ_n =∫_0 ^∞ e^(−n[x]) e^(ix) dx =∫_0 ^∞ e^(−n[x]+ix) dx =∫_0 ^1 e^(−n×0+ix) dx+∫_1 ^2 e^(−n×1+ix) dx+...+∫_r ^(r+1) e^(−n×r+ix) dx+..upto ∞ =Σ_(r=0) ^∞ ∫_r ^(r+1) e^(−nr+ix) dx =Σ_(r=0) ^∞ e^(−nr) ∫_r ^(r+1) e^(ix) dx =Σ_(r=0) ^∞ e^(−nr) ×∣(e^(ix) /i)∣_r ^(r+1) =Σ_(r=0) ^∞ e^(−nr) ×(−i)(e^(i(r+1)) −e^(ir) ) =Σ_(r=0) ^∞ e^(−nr) ×i×(e^(ir) −e^(i(r+1)) ) =Σ_(r=0) ^∞ e^(−nr) ×i×[{cosr+isinr}−{cos(r+1)+isin(r+1)}] =Σ_(r=0) ^∞ e^(−nr) ×i×[2sin(r+(1/2))sin((1/2))+i{2cos(r+(1/2))sin((1/2))] =Σ_(r=0) ^∞ e^(−nr) ×2sin((1/2))[−cos(r+(1/2))+isin(r+(1/2)) =(−2)sin((1/2))Σ_(r=0) ^∞ e^(−nr) ×[cos(r+(1/2))−isin(r+(1/2))] Q_n =A_n =2sin((1/2))Σ_(r=0) ^∞ e^(−nr) sin(r+(1/2))$
	$P_{n} = \int_{0}^{\infty} e^{- n [x]} c o s x d x$ $Q_{n} = \int_{0}^{\infty} e^{- n [x]} s i n x$ $A_{n} = Q_{n}$ $P_{n} + {i Q}_{n} = \int_{0}^{\infty} e^{- n [x]} e^{i x} d x$ $= \int_{0}^{\infty} e^{- n [x] + i x} d x$ $= \int_{0}^{1} e^{- n \times 0 + i x} d x + \int_{1}^{2} e^{- n \times 1 + i x} d x + . . . + \int_{r}^{r + 1} e^{- n \times r + i x} d x + . . u p t o \infty$ $= \underset{r = 0}{\sum^{\infty}} \int_{r}^{r + 1} e^{- n r + i x} d x$ $= \underset{r = 0}{\sum^{\infty}} e^{- n r} \int_{r}^{r + 1} e^{i x} d x$ $= \underset{r = 0}{\sum^{\infty}} e^{- n r} \times ∣ \frac{e^{i x}}{i} ∣_{r}^{r + 1}$ $= \underset{r = 0}{\sum^{\infty}} e^{- n r} \times (- i) (e^{i (r + 1)} - e^{i r})$ $= \underset{r = 0}{\sum^{\infty}} e^{- n r} \times i \times (e^{i r} - e^{i (r + 1)})$ $= \underset{r = 0}{\sum^{\infty}} e^{- n r} \times i \times [{c o s r + i s i n r} - {c o s (r + 1) + i s i n (r + 1)}]$ $= \underset{r = 0}{\sum^{\infty}} e^{- n r} \times i \times [2 s i n (r + \frac{1}{2}) s i n (\frac{1}{2}) + i {2 c o s (r + \frac{1}{2}) s i n (\frac{1}{2})]$ $= \underset{r = 0}{\sum^{\infty}} e^{- n r} \times 2 s i n (\frac{1}{2}) [- c o s (r + \frac{1}{2}) + i s i n (r + \frac{1}{2})$ $= (- 2) s i n (\frac{1}{2}) \underset{r = 0}{\sum^{\infty}} e^{- n r} \times [c o s (r + \frac{1}{2}) - i s i n (r + \frac{1}{2})]$ $Q_{n} = A_{n}$ $= 2 s i n (\frac{1}{2}) \underset{r = 0}{\sum^{\infty}} e^{- n r} s i n (r + \frac{1}{2})$
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