let-I-0-e-tx-sint-dt-with-x-gt-0-find-the-value-of-I-

Question Number 35589 by abdo mathsup 649 cc last updated on 20/May/18

l e t I = \int_{0}^{\infty} e^{- t x} ∣ s i n t ∣ d t w i t h x > 0

f i n d t h e v a l u e o f I .

Commented by abdo mathsup 649 cc last updated on 21/May/18

I = \sum_{n = 0}^{\infty} \int_{n π}^{(n + 1) π} e^{- t x} ∣ s i n t ∣ d t c h a n g e m e n t

t = n π + u g i v e

I = \sum_{n = 0}^{\infty} \int_{0}^{π} e^{- n π x} e^{- x u} ∣ s i n u ∣ d u

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

b u t A (x) = \int_{0}^{π} e^{- x u} s i n (u) d u = I m (\int_{0}^{π} e^{- x u + i u} d u)

= I m (\int_{0}^{π} e^{(- x + i) u} d u) b u t

\int_{0}^{π} e^{(- x + i) u} d u = {[\frac{1}{- x + i} e^{(- x + i) u}]}_{0}^{π}

= \frac{- 1}{x - i} {e^{- x π + i π} - 1} = \frac{1 + e^{- π x}}{x - i}

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

\sum_{n = 0}^{\infty} e^{- n π x} = \sum_{n = 0}^{\infty} {(e^{- π x})}^{n} = \frac{1}{1 - e^{- π x}} s o

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