let-x-gt-0-calculate-f-x-0-e-t-sin-xt-dt-

Question Number 42188 by maxmathsup by imad last updated on 19/Aug/18

l e t x > 0 c a l c u l a t e f (x) = \int_{0}^{+ \infty} e^{- t} ∣ s i n (x t) ∣ d t

Commented by maxmathsup by imad last updated on 23/Aug/18

c h a n g e m e n t x t = u g i v e f (x) = \int_{0}^{\infty} e^{- \frac{u}{x}} ∣ s i n (u) ∣ \frac{d u}{x}

= \frac{1}{x} \int_{0}^{\infty} e^{- \frac{u}{x}} ∣ s i n (u) ∣ d u \Rightarrow x f (x) = \sum_{n = 0}^{\infty} \int_{n π}^{(n + 1) π} e^{- \frac{u}{x}} ∣ s i n u ∣ d u

=_{u = n π + t} \sum_{n = 0}^{\infty} \int_{0}^{π} e^{- \frac{n π + t}{x}} ∣ s i n (n π + t) ∣ d t

= \sum_{n = 0}^{\infty} \int_{0}^{π} e^{- \frac{n π}{x}} e^{\frac{t}{x}} s i n t d t = \sum_{n = 0}^{\infty} e^{- \frac{n π}{x}} \int_{0}^{π} e^{\frac{t}{x}} s i n t d t b u t

\int_{0}^{π} e^{\frac{t}{x}} s i n t d t = I m (\int_{0}^{π} e^{\frac{t}{x} + i t} d t) a n d \int_{0}^{π} e^{(\frac{1}{x} + i) t} d t

= {[\frac{1}{\frac{1}{x} + i} e^{(\frac{1}{x} + i) t}]}_{0}^{π} = \frac{x}{1 + x i} {e^{\frac{π}{x} + i π} - 1} = \frac{x (1 - x i)}{1 + x^{2}} {- e^{\frac{π}{x}} - 1}

= \frac{x (- 1 + x i) (1 + e^{\frac{π}{x}})}{1 + x^{2}} = \frac{1 + e^{\frac{π}{x}}}{1 + x^{2}} (- x + x^{2} i) \Rightarrow \int_{0}^{π} e^{\frac{t}{x}} s i n t d t = \frac{x^{2}}{1 + x^{2}} (1 + e^{\frac{π}{x}}) \Rightarrow

x f (x) = \frac{x^{2}}{1 + x^{2}} (1 + e^{\frac{π}{x}}) \sum_{n = 0}^{\infty} {(e^{- \frac{π}{x}})}^{n} = \frac{x^{2}}{1 + x^{2}} (1 + e^{\frac{π}{x}}) \frac{1}{1 - e^{- \frac{π}{x}}} \Rightarrow

x f (x) = \frac{x^{2}}{1 + x^{2}} \frac{1 + e^{\frac{π}{x}}}{1 - e^{- \frac{π}{x}}} \Rightarrow f (x) = \frac{x}{1 + x^{2}} \frac{1 + e^{\frac{π}{x}}}{1 - e^{- \frac{π}{x}}} .