find-the-value-of-0-sinx-x-1-x-2-dx-

Question Number 26558 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} \frac{s i n x}{x (1 + x^{2})} d x

Commented by abdo imad last updated on 27/Dec/17

l e t p u t I = \int_{0}^{\infty} \frac{s i n x}{x (1 + x^{2})} d x w e k n o w t h a t \int_{0}^{\infty} \frac{c o s (α x)}{1 + x^{2}} d x = \frac{π}{2} e^{- α}

l e t t a k e α ⩾ 0 \Rightarrow \int_{0}^{1} \frac{π}{2} e^{- α} d α = - \frac{π}{2} {[e^{- α}]}_{0}^{1} = - \frac{π}{2} (e^{- 1} - 1)

= \frac{π}{2} (1 - e^{- 1}) a n d f r o m a n o t h e r s i d e

\int_{0}^{1} \frac{π}{2} e^{- α} d α = \int_{0}^{1} (\int_{0}^{\infty} \frac{c o s (α x)}{1 + x^{2}} d x) d α

= \int_{0}^{\infty} (\int_{0}^{1} c o s (α x) d α) \frac{d x}{1 + x^{2}} (b y f u b i n i t h e o r e m)

= \int_{0}^{\infty} ({[\frac{s i n (α x)}{x}]}_{α = 0}^{α = 1}) \frac{d x}{1 + x^{2}} = \int_{0}^{\infty} \frac{s i n x}{x} . \frac{d x}{1 + x^{2}}

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