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Question Number 97707 by Power last updated on 09/Jun/20

	Answered by smridha last updated on 09/Jun/20

	$= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{e^{i x}}{x^{2} + 1^{2}} d x + \frac{1}{2} \int_{+ \infty}^{- \infty} \frac{e^{- (- i x)}}{{(- x)}^{2} + 1^{2}} d (- x)$ $= \int_{- \infty}^{+ \infty} \frac{e^{i x}}{x^{2} + 1^{2}} d x = \oint_{C} \frac{e^{i z}}{z^{2} + 1^{2}} d z$ $= \oint_{C} \frac{e^{i z}}{(z + i) (z - i)} d z$ $n o w t i m e f o r c a l c u l a t e t h e r e s i d u e s$ $r e s i d u e a t z_{1} = i$ $\lim_{z \to i} \frac{(z - i) e^{i z}}{(z - i) (z + i)} = \frac{e^{- 1}}{2 i}$ $b u t z_{2} = - i t h i s p o l e d o e s n o t e n c l o s e d$ $b y t h e c a n t o u r (C)$ $n o w :$ $I = \oint_{C} f (z) d z = 2 π i Σ e n c l o s e d r e s u d u e b y t h e c a n t o u r$ $= 2 π i . \frac{e^{- 1}}{2 i} = π e^{- 1}$
	Answered by smridha last updated on 09/Jun/20

	$M e t h o d (2)$ $l e t F (t) = 2 \int_{0}^{+ \infty} \frac{c o s (t x)}{x^{2} + 1} d x [g i v e n i n t e g r a n d i s e v e n]$ $n o w b y L a p l a c e T r a n s f o r m a t i o n$ $f (s) = L [f (t)] . . . . . (i)$ $s o f (s) = \int_{0}^{\infty} e^{- s t} [2 \int_{0}^{\infty} \frac{c o s (t x)}{x^{2} + 1} d x] d t$ $= 2 \int_{0}^{\infty} \frac{1}{x^{2} + 1} [\int_{0}^{\infty} e^{- s t} c o s (t x) d t] d x$ $= 2 \int_{0}^{\infty} \frac{s}{(x^{2} + 1) (s^{2} + x^{2})} d x$ $= \frac{2 s}{(s^{2} - 1)} [\int_{0}^{\infty} (\frac{1}{(x^{2} + 1)} - \frac{1}{(s^{2} + x^{2})}) d x]$ $= \frac{2 s}{(s^{2} - 1)} {[\tan^{- 1} (x) - \frac{1}{s} \tan^{- 1} (\frac{x}{s})]}_{0}^{\infty}$ $= \frac{π}{(s - (- 1))}$ $n o w f r o m (i) F (t) = L^{- 1} [f (s)]$ $= π . L^{- 1} [\frac{1}{(s - (- 1))}]$ $= π e^{- 1}$ $n o w I = F (1) = \frac{π}{e}$
	Commented by Power last updated on 09/Jun/20

	$thanks sir$
	Commented by smridha last updated on 09/Jun/20

	$a l s o t h a n k f u l b e c a u s e t h i s i s$ $v e r y s i g n i f i c a n t i n t e g r a l .$
	Answered by abdomathmax last updated on 09/Jun/20

	$A = \int_{- \infty}^{+ \infty} \frac{cosx}{x^{2} + 1} dx \Rightarrow A = Re (\int_{- \infty}^{+ \infty} \frac{e^{ix}}{x^{2} + 1} dx)$ $let φ (z) = \frac{e^{iz}}{z^{2} + 1} \Rightarrow φ (z) = \frac{e^{iz}}{(z - i) (z + i)}$ $residus tbeorem give$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, i) = 2 i π \times \frac{e^{- 1}}{2 i}$ $= π e^{- 1} \Rightarrow A = \frac{π}{e}$
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