find-the-value-of-0-e-px-sinx-dx-with-p-gt-0-

Question Number 26570 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^{- p x} / s i n x / d x w i t h p > 0

Commented by abdo imad last updated on 02/Jan/18

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

I = {l i m}_{n - >\propto} \int_{0}^{n π} e^{- p x} / s i n x / d x b u t

\int_{0}^{n π} e^{- p x} / s i n x / d x = \sum_{k = 0}^{n - 1} \int_{k π}^{(k + 1) π} e^{- p x} / s i n x / d x

= \sum_{k = 0}^{n - 1} \int_{0}^{π} e^{- p (k π + t)} / s i n (k π + t) / d t (w e d o t h e c h a n g e m e n t x = k π + t)

= \sum_{k = 0}^{n - 1} e^{- p k π} \int_{0}^{π} e^{- p t} s i n t d t = \sum_{k = 0}^{n - 1} {(e^{- p π})}^{k} \int_{0}^{π} e^{- p t} s i n t d t

= \frac{1 - e^{- n p π}}{1 - e^{- p π}} \int_{0}^{π} e^{- p t} s i n t d t b u t

\int_{0}^{π} e^{- p t} s i n t d t = I m (\int_{0}^{π} e^{(i - p) t} d t)

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

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