find-f-t-0-sin-x-e-t-x-dx-with-t-gt-0-

Question Number 39035 by maxmathsup by imad last updated on 01/Jul/18

f i n d f (t) = \int_{0}^{\infty} s i n (x) e^{- t [x]} d x w i t h t > 0

Commented by maxmathsup by imad last updated on 02/Jul/18

w e h a v e f (t) = \sum_{n = 0}^{\infty} \int_{n}^{n + 1} s i n (x) e^{- n t} d x

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

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

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

= 1 + (1 - e^{t}) \sum_{n = 1}^{\infty} e^{- n t} c o s (n) b u t

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

= R e (\sum_{n = 0}^{\infty} e^{- n t + i n}) = R e (\sum_{n = 0}^{\infty} {(e^{- t + i})}^{n}) b u t

\sum_{n = 0}^{\infty} {(e^{- t + i})}^{n} = \frac{1}{1 - e^{- t + i}} = \frac{1}{1 - e^{- t} (c o s 1 + i s i n (1))}

= \frac{1}{1 - e^{- t} c o s (1) - i e^{- t} s i n (1)} = \frac{1 - e^{- t} c o s (1) + i e^{- t} s i n (1)}{{(1 - e^{- t} c o s (1))}^{2} + e^{- 2 t} {s i n}^{2} (1)} \Rightarrow

f (t) = 1 + (1 - e^{t}) {\frac{1 - e^{- t} c o s (1)}{{(1 - e^{- t} c o s (1))}^{2} + e^{- 2 t} {s i n}^{2} (1)} - 1}