we give for t>0 ∫_0 ^∞ ((sinx)/x) e^(−tx) dx =(π/2) −arctant use this result to find the value of ∫


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Question Number 33979 by abdo imad last updated on 28/Apr/18

$w e g i v e f o r t > 0 \int_{0}^{\infty} \frac{s i n x}{x} e^{- t x} d x = \frac{π}{2} - a r c t a n t$ $u s e t h i s r e s u l t t o f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{(1 - e^{- x}) s i n x}{x^{2}} d x .$
Commented by abdo mathsup 649 cc last updated on 03/May/18

$w e k n o w t h a t \int_{0}^{\infty} \frac{s i n x}{x} e^{- t x} d x = \frac{π}{2} - a r c t a n t \Rightarrow$ $\int_{0}^{1} (\frac{π}{2} - a r c t a n t) d t = \int_{0}^{1} (\int_{0}^{\infty} \frac{s i n x}{x} e^{- t x} d x)$ $= \int_{0}^{\infty} (\int_{0}^{1} e^{- t x} d t) \frac{s i n x}{x} d x (b y f u b i n i)$ $= \int_{0}^{\infty} ({[- \frac{1}{x} e^{- t x}]}_{t = 0}^{t = 1}) \frac{s i n x}{x} d x$ $= \int_{0}^{\infty} \frac{1}{x^{2}} (1 - e^{- x}) s i n x d x b u t$ $\int_{0}^{1} (\frac{π}{2} - a r c t a n t) d t = \frac{π}{2} - \int_{0}^{1} a r c t a n t d t$ $b y p a r t s \int_{0}^{1} a r c t a n t d t = {[t a r c t a n t]}_{0}^{1} - {\int^{1}}_{0} \frac{t}{1 + t^{2}} d t$ $= \frac{π}{4} - \frac{1}{2} {[l n (1 + t^{2})]}_{0}^{1} = \frac{π}{4} - \frac{1}{2} l n (2) \Rightarrow$ $\int_{0}^{\infty} \frac{(1 - e^{- x}) s i n x}{x^{2}} d x = \frac{π}{4} + \frac{1}{2} l n (2) .$
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