find-the-value-of-0-e-x-sinxdx-in-that-x-E-x-

Question Number 26575 by abdo imad last updated on 26/Dec/17

f i n d t h e v a l u e o f \int_{0}^{\infty} e^{- [x]} s i n x d x i n t h a t [x] = E (x)

Commented by abdo imad last updated on 29/Dec/17

l e t p u t I = \int_{0}^{\propto} e^{- [x]} s i n x d x I_{n} = \int_{0}^{n} e^{- [x]} s i n x d x

w e h a v e I = {l i m}_{n - >\propto} I_{n} b u t

I_{n} = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} e^{- [x]} s i n x d x = \sum_{k = 0}^{k = n - 1} e^{- k} \int_{k}^{k + 1} s i n x d x

I_{n} = \sum_{k = 0}^{k = n - 1} e^{- k} {[- c o s x]}_{k}^{k + 1} = \sum_{k = 0}^{k = n - 1} e^{- k} (c o s k - c o s (k + 1))

= \sum_{k = 0}^{k = n - 1} e^{- k} c o s k - \sum_{k = 0}^{k = n - 1} e^{- k} c o s (k + 1)

= \sum_{k = 0}^{k = n - 1} e^{- k} c o s k - \sum_{k = 1}^{n} e^{- (k - 1)} c o s k

= (1 - e) \sum_{k = 0}^{n - 1} e^{- k} c o s k

b u t \sum_{k = 0}^{n - 1} e^{- k} c o s k = R e (\sum_{k = 0}^{n - 1} e^{(- 1 + i) k})

a n d \sum_{k = 0}^{k = n - 1} e^{(- 1 + i) k} = \frac{1 - e^{(- 1 + i) n}}{1 - e^{- 1 + i}}

l i m_{n - >\propto} I_{n} = (1 - e) \sum_{k = 0}^{\propto} e^{- k} c o s k = (1 - e) R e (\frac{1}{1 - e^{- 1 + i}})

b u t \frac{1}{1 - e^{- 1 + i}} = \frac{1}{1 - e^{- 1} (c o s (1) + i 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 {l i m}_{n - >\propto} I_{n} = \frac{1 - e^{- 1} c o s (1)}{{(1 - e^{- 1} c o s (1))}^{2} + e^{- 2} {s i n}^{2} (1)} = \int_{0}^{\infty} e^{- [x]} s i n x d x