let-put-F-x-0-e-tx-sint-t-dt-with-x-0-we-accept-that-F-is-class-C-1-on-0-calculate-F-x-and-find-F-x-then-find-the-value-of-0-sint-t-dt-

Question Number 26559 by abdo imad last updated on 26/Dec/17

l e t p u t F (x) = \int_{0}^{\infty} e^{- t x} \frac{s i n t}{t} d t w i t h x ⩾ 0

w e a c c e p t t h a t F i s c l a s s C^{1} o n [0, \propto [

c a l c u l a t e \frac{\partial F}{\partial x} a n d f i n d F (x)

t h e n f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{s i n t}{t} d t

Commented by prakash jain last updated on 27/Dec/17

\frac{d F}{d x} = \int_{0}^{\infty} \frac{\partial}{\partial x} (e^{- t x} \frac{\sin t}{t}) d t

= - \int_{0}^{\infty} e^{- t x} \sin t d t

= {[\frac{e^{- t x} (x \sin t + \cos t)}{x^{2} + 1}]}_{t = 0}^{t = \infty}

= - \frac{1}{x^{2} + 1}

F (x) = - \tan^{- 1} x + C

F (\infty) = 0

\Rightarrow C = \frac{π}{2}

F (x) = - \tan^{- 1} x + \frac{π}{2}

F (0) = \int_{0}^{\infty} \frac{\sin t}{t} d t = \frac{π}{2}