Question-184264

Question Number 184264 by Noorzai last updated on 04/Jan/23

Answered by ARUNG_Brandon_MBU last updated on 04/Jan/23

I (t) = \int_{0}^{\infty} e^{- t x} \frac{\sin m x}{x} d x \Rightarrow I^{'} (t) = - \int_{0}^{\infty} e^{- t x} \sin m x d x

\Rightarrow I^{'} (t) = - {[\frac{e^{- t x}}{t^{2} + m^{2}} (- t \sin m x - m \cos m x)]}_{0}^{\infty}

= - \frac{m}{t^{2} + m^{2}} \Rightarrow I (t) = - \arctan (\frac{t}{m}) + C

\lim_{t \to + \infty} I (t) = 0 = - \frac{π}{2} + C \Rightarrow C = \frac{π}{2}

\Rightarrow I (t) = \frac{π}{2} - \arctan (\frac{t}{m})

\Rightarrow \int_{0}^{\infty} e^{x} \frac{\sin m x}{x} d x = I (- 1) = \frac{π}{2} + \arctan (\frac{1}{m})