1-calculate-A-n-0-e-n-x-sin-x-dx-with-n-integr-and-n-1-2-find-nature-of-n-1-A-n-

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 (\dots) = \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} = \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 + \dots + \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})