∫_(−∞) ^( +∞) ((cosx)/((x^2 +1)))dx don′t use feynmann trick


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Question Number 138315 by LUFFY last updated on 12/Apr/21

$\int_{- \infty}^{+ \infty} \frac{\cos x}{(x^{2} + 1)} d x$ ${don}^{'} t use feynmann trick$
	Answered by Ñï= last updated on 12/Apr/21
	$I(t)=∫_(−∞) ^(+∞) ((cos (tx))/(x^2 +1))dx I(t)′=−∫_(−∞) ^(+∞) ((xsin (tx))/(x^2 +1))dx=−∫_(−∞) ^(+∞) (((x^2 +1−1)sin (tx))/(x(x^2 +1)))dx =−∫_(−∞) ^(+∞) ((sin (tx))/x)−((sin (tx))/(x(x^2 +1)))dx=−π+∫_(−∞) ^(+∞) ((sin (tx))/(x(x^2 +1)))dx I(t)′′=∫_(−∞) ^(+∞) ((cos (tx))/(x^2 +1))dx=I(t) I(t)′′−I(t)=0 ⇒ I(t)=C_1 e^x +C_2 e^(−x) I(t)′=C_1 e^x −C_2 e^(−x) { ((I(0)=π)),((I(0)′=−π)) :} ⇒C_1 =0 C_2 =π ⇒I(t)=πe^(−x) ⇒∫_(−∞) ^(+∞) ((cos x)/(x^2 +1))dx=I(1)=(π/e)$
	$I (t) = \int_{- \infty}^{+ \infty} \frac{\cos (t x)}{x^{2} + 1} d x$ $I {(t)}^{'} = - \int_{- \infty}^{+ \infty} \frac{x \sin (t x)}{x^{2} + 1} d x = - \int_{- \infty}^{+ \infty} \frac{(x^{2} + 1 - 1) \sin (t x)}{x (x^{2} + 1)} d x$ $= - \int_{- \infty}^{+ \infty} \frac{\sin (t x)}{x} - \frac{\sin (t x)}{x (x^{2} + 1)} d x = - π + \int_{- \infty}^{+ \infty} \frac{\sin (t x)}{x (x^{2} + 1)} d x$ $I {(t)}^{″} = \int_{- \infty}^{+ \infty} \frac{\cos (t x)}{x^{2} + 1} d x = I (t)$ $I {(t)}^{″} - I (t) = 0$ $\Rightarrow I (t) = C_{1} e^{x} + C_{2} e^{- x}$ $I {(t)}^{'} = C_{1} e^{x} - C_{2} e^{- x}$ ${\begin{cases} I (0) = π \\ I {(0)}^{'} = - π \end{cases}$ $\Rightarrow C_{1} = 0 C_{2} = π$ $\Rightarrow I (t) = π e^{- x}$ $\Rightarrow \int_{- \infty}^{+ \infty} \frac{\cos x}{x^{2} + 1} d x = I (1) = \frac{π}{e}$
	Answered by Ñï= last updated on 12/Apr/21

	$I = \int_{- \infty}^{+ \infty} \frac{co s x}{x^{2} + 1} d x = ℜ \int_{- \infty}^{+ \infty} \frac{e^{i x}}{x^{2} + 1} d x = ℜ [2 π i R e s (\frac{e^{i x}}{x^{2} + 1}, i)]$ $= ℜ [2 π i \lim_{x \to i} \frac{x - i}{x^{2} + 1} e^{i x}] = ℜ [2 π i \cdot \frac{e^{- 1}}{2 i}] = \frac{π}{e}$
	Answered by Ñï= last updated on 12/Apr/21
	$I(t)=∫_0 ^(+∞) ((cos (tx))/(x^2 +1))dx L[I(t)]=∫_0 ^∞ dx∫_0 ^(+∞) ((cos (tx))/(x^2 +1))e^(−st) dt =∫_0 ^∞ ((L[cos (tx)(s)])/(x^2 +1))dx =∫_0 ^∞ (s/(s^2 +x^2 ))∙(dx/(x^2 +1)) =(s/(s^2 −1))[∫_0 ^∞ (dx/(x^2 +1))dx−∫_0 ^∞ ((sdx)/(x^2 +s^2 ))] =((πs)/(2(s^2 −1)))−(π/(2(s^2 −1))) I(t)=L^(−1) [((πs)/(2(s^2 −1)))−(π/(2(s^2 −1)))] =(π/2)L^(−1) {(s/(s^2 −1))}−(π/2)L^(−1) {(1/(s^2 −1))} =(π/2)(cosh t−sinh t) =(π/2)e^(−t) ⇒∫_(−∞) ^(+∞) ((cos x)/(x^2 +1))dx=2I(1)=(π/e)$
	$I (t) = \int_{0}^{+ \infty} \frac{\cos (t x)}{x^{2} + 1} d x$ $L [I (t)] = \int_{0}^{\infty} d x \int_{0}^{+ \infty} \frac{\cos (t x)}{x^{2} + 1} e^{- s t} d t$ $= \int_{0}^{\infty} \frac{L [\cos (t x) (s)]}{x^{2} + 1} d x$ $= \int_{0}^{\infty} \frac{s}{s^{2} + x^{2}} \cdot \frac{d x}{x^{2} + 1}$ $= \frac{s}{s^{2} - 1} [\int_{0}^{\infty} \frac{d x}{x^{2} + 1} d x - \int_{0}^{\infty} \frac{s d x}{x^{2} + s^{2}}]$ $= \frac{π s}{2 (s^{2} - 1)} - \frac{π}{2 (s^{2} - 1)}$ $I (t) = L^{- 1} [\frac{π s}{2 (s^{2} - 1)} - \frac{π}{2 (s^{2} - 1)}]$ $= \frac{π}{2} L^{- 1} {\frac{s}{s^{2} - 1}} - \frac{π}{2} L^{- 1} {\frac{1}{s^{2} - 1}}$ $= \frac{π}{2} (\cosh t - \sinh t)$ $= \frac{π}{2} e^{- t}$ $\Rightarrow \int_{- \infty}^{+ \infty} \frac{\cos x}{x^{2} + 1} d x = 2 I (1) = \frac{π}{e}$
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