Calculate. ∫_0 ^( ∞) ((sin(z))/(z(z^2 +1)))dz method 1. using Laplace Transform. method 2. using Contour Integral. method 3. using Feymann′s parametirc trick HELP!!!!!!!


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Question Number 195421 by MathedUp last updated on 02/Aug/23

$Calculate . \int_{0}^{\infty} \frac{\sin (z)}{z (z^{2} + 1)} d z$ $method 1 . using Laplace Transform .$ $method 2 . using Contour Integral .$ $method 3 . using {Feymann}^{'} s parametirc trick$ $HELP!!!!!!!$
	Answered by witcher3 last updated on 02/Aug/23


	Commented by witcher3 last updated on 02/Aug/23

	$Small Ω Circle radius ϵ$ $Big Γ radius R$ $L = [- R, - ϵ], L^{'} = [ϵ, R]$ $Γ = {Re}^{ia}, a \in [0, π]$ $c = ϵ e^{ia}, a \in [0, π]$ $D = Ω \cup Γ \cup L \cup L^{'}$ $f (z) = \frac{e^{iz}}{z (1 + z^{2})}, hase Twos pols in D$ $z^{2} + 1 = 0, z = i$ $\int_{D} f (z) dz = 2 i π Res (f, i,) = 2 i π . \frac{e^{- 1}}{i (2 i)} = \frac{π}{ie}$ $\int_{- R}^{- ϵ} f (z) dz + \int_{π}^{0} f (ϵ e^{ia}) . i ϵ e^{ia} da + \int_{ϵ}^{R} f (z) dz + \int_{0}^{π} f ({Re}^{ia}) {iRe}^{ia} da$ $\int_{- R}^{- ϵ} f (z) dz + \int_{ϵ}^{R} f (z) dz = \int_{ϵ}^{R} (f (z) + f (- z)) dz$ $= 2 i \int_{ϵ}^{R} \frac{\sin (z)}{z (1 + z^{2})} dz$ $\lim_{ϵ \to 0} \int_{π}^{0} f (ϵ e^{ia}) i ϵ e^{ia} da = \underset{ϵ \to 0}{\lim i} \int_{π}^{0} \frac{e^{i ϵ e^{ia}}}{(1 + ϵ^{2} e^{2 ia})} da$ $integral Cv uniformaly$ $= i \int_{π}^{0} \lim_{ϵ \to 0} \frac{e^{i ϵ e^{ia}}}{1 + ϵ^{2} e^{2 ia}} da = - i π$ $\lim_{ϵ \to 0} \int_{Ω} f (z) dz = - i π$ $\int_{0}^{π} f ({Re}^{ia}) {iRe}^{ia} da = i \int_{0}^{π} \frac{e^{{iRe}^{ia}}}{1 + R^{2} e^{2 ia}} da$ $= i \int_{0}^{π} \frac{e^{iRcos (a)} . e^{- Rsin (a)}}{1 + R^{2} e^{2 ia}}$ $p$ $∣ \frac{e^{iRcos (a)} . e^{- Rsin (a)}}{1 + R^{2} e^{2 ia}} ∣⩽ \frac{e^{- Rsin (a)}}{R^{2} - 1} \Rightarrow$ $= \lim_{R \to \infty} ∣ i \int_{0}^{π} \frac{e^{iRcos (a)} . e^{- Rsin (a)}}{1 + R^{2} e^{2 ia}} ∣ da ⩽ \int_{0}^{π} \lim_{R \to \infty} \frac{e^{- Rsin (a)}}{R^{2} - 1} da$ $= 0$ $\lim_{R \to \infty} \int_{Γ} f (z) dz = 0$ $\lim_{ϵ \to 0, R \to \infty} \int_{D} f (z) dz = 2 i \int_{0}^{\infty} \frac{\sin (z)}{z (1 + z^{2})} dz - i π + 0 = \frac{π}{ie}$ $\int_{0}^{\infty} \frac{\sin (x)}{x (1 + x^{2})} dx = \frac{π}{2} - \frac{π}{2 e} = \frac{π}{2} (\frac{e - 1}{e})$
	Answered by senestro last updated on 02/Aug/23

	$\frac{π}{2} (1 + \frac{1}{e})$
	Answered by witcher3 last updated on 02/Aug/23
	$Laplace ∫_0 ^∞ ((sin(zt))/(z(1+z^2 )))dz=f(t) L(f(t))=∫_0 ^∞ f(t)e^(−ts) dt=∫_0 ^∞ ∫_0 ^∞ ((sin(zt))/(z(1+z^2 )))dze^(−ts) dt =∫_0 ^∞ (1/(z(1+z^2 )))∫_0 ^∞ sin(zt)e^(−ts) dtdz =∫_0 ^∞ (1/(z(1+z^2 )))L{sin(zt)}dz =∫_0 ^∞ (1/(z(1+z^2 ))).(z/(z^2 +s^2 ))dz=∫_0 ^∞ (dz/((1+z^2 )(z^2 +s^2 ))) =(1/2)∫_(−∞) ^∞ (dz/((1+z^2 )(z^2 +s^2 )))=iπ.(1/(2i(s^2 −1)))+iπ.(1/(2is(1−s^2 ))) =(π/(2s(1+s)))=(π/(2s))−(π/(2(1+s))) f(t)=L^− (L(f(t))=L^− ((π/2)((1/s)−(1/(s+1)))) =(π/2)L^− ((1/s))−(π/2)L^− ((1/(s+1)))=(π/2)−(π/2)e^(−t) f(t)=∫_0 ^∞ ((sin(xt))/(x(1+x^2 )))dx=(π/2)(1−e^(−t) ) ∫_0 ^∞ ((sin(x))/(x(1+x^2 )))dx=f(1)=(π/2)(1−(1/e))$
	$Laplace$ $\int_{0}^{\infty} \frac{\sin (zt)}{z (1 + z^{2})} dz = f (t)$ $L (f (t)) = \int_{0}^{\infty} f (t) e^{- ts} dt = \int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin (zt)}{z (1 + z^{2})} {dze}^{- ts} dt$ $= \int_{0}^{\infty} \frac{1}{z (1 + z^{2})} \int_{0}^{\infty} \sin (zt) e^{- ts} dtdz$ $= \int_{0}^{\infty} \frac{1}{z (1 + z^{2})} L {\sin (zt)} dz$ $= \int_{0}^{\infty} \frac{1}{z (1 + z^{2})} . \frac{z}{z^{2} + s^{2}} dz = \int_{0}^{\infty} \frac{dz}{(1 + z^{2}) (z^{2} + s^{2})}$ $= \frac{1}{2} \int_{- \infty}^{\infty} \frac{dz}{(1 + z^{2}) (z^{2} + s^{2})} = i π . \frac{1}{2 i (s^{2} - 1)} + i π . \frac{1}{2 is (1 - s^{2})}$ $= \frac{π}{2 s (1 + s)} = \frac{π}{2 s} - \frac{π}{2 (1 + s)}$ $f (t) = L^{-} (L (f (t)) = L^{-} (\frac{π}{2} (\frac{1}{s} - \frac{1}{s + 1}))$ $= \frac{π}{2} L^{-} (\frac{1}{s}) - \frac{π}{2} L^{-} (\frac{1}{s + 1}) = \frac{π}{2} - \frac{π}{2} e^{- t}$ $f (t) = \int_{0}^{\infty} \frac{\sin (xt)}{x (1 + x^{2})} dx = \frac{π}{2} (1 - e^{- t})$ $\int_{0}^{\infty} \frac{\sin (x)}{x (1 + x^{2})} dx = f (1) = \frac{π}{2} (1 - \frac{1}{e})$
	Answered by witcher3 last updated on 02/Aug/23
	$methode (3) y(t)=∫_0 ^∞ ((sin(xt))/(x(1+x^2 )))dx,t>0 y(0)=0 y′(t)=∫_0 ^∞ ((cos(xt))/(1+x^2 ))dx,y′(0)=(π/2) y′′(t)=∫_0 ^∞ ((−x^2 sin(xt))/(x(1+x^2 )))dx Y′′(t)−Y(t)=∫_0 ^∞ ((−x^2 sin(xt))/(x(1+x^2 )))−((sin(xt))/(x(1+x^2 )))dx =−∫_0 ^∞ ((sin(xt))/x)dx=−∫_0 ^∞ ((sin(xt))/(xt))d(xt) =−∫_0 ^∞ ((sin(z))/z)dz=−(π/2) y′′(t)−y(t)=−(π/2) y(t)=ae^t +be^(−t) +(π/2) y(0)=0,y′(0)=(π/2)⇒ { ((a+b=−(π/2))),((a−b=(π/2))) :}⇒a=0,b=−(π/2) y(t)=(π/2)(1−(1/e^t )),y(1)=∫_0 ^∞ ((sin(x))/(x(1+x^2 )))=(π/2)(1−(1/e))$
	$methode (3)$ $y (t) = \int_{0}^{\infty} \frac{\sin (xt)}{x (1 + x^{2})} dx, t > 0$ $y (0) = 0$ $y^{'} (t) = \int_{0}^{\infty} \frac{\cos (xt)}{1 + x^{2}} dx, y^{'} (0) = \frac{π}{2}$ $y^{″} (t) = \int_{0}^{\infty} \frac{- x^{2} \sin (xt)}{x (1 + x^{2})} dx$ $Y^{″} (t) - Y (t) = \int_{0}^{\infty} \frac{- x^{2} \sin (xt)}{x (1 + x^{2})} - \frac{\sin (xt)}{x (1 + x^{2})} dx$ $= - \int_{0}^{\infty} \frac{\sin (xt)}{x} dx = - \int_{0}^{\infty} \frac{\sin (xt)}{xt} d (xt)$ $= - \int_{0}^{\infty} \frac{\sin (z)}{z} dz = - \frac{π}{2}$ $y^{″} (t) - y (t) = - \frac{π}{2}$ $y (t) = {ae}^{t} + {be}^{- t} + \frac{π}{2}$ $y (0) = 0, y^{'} (0) = \frac{π}{2} \Rightarrow$ ${\begin{cases} a + b = - \frac{π}{2} \\ a - b = \frac{π}{2} \end{cases} \Rightarrow a = 0, b = - \frac{π}{2}$ $y (t) = \frac{π}{2} (1 - \frac{1}{e^{t}}), y (1) = \int_{0}^{\infty} \frac{\sin (x)}{x (1 + x^{2})} = \frac{π}{2} (1 - \frac{1}{e})$
	Commented by MathedUp last updated on 03/Aug/23

	$i t h i n k y o u a r e g o d o f m a t h L O L$
	Commented by witcher3 last updated on 03/Aug/23

	$no Just i just orck Hard and the secret is always$ $worck hard bro do your best and stop social media!$ $evrey one withe enough worck progress in evrey$ $steps of life,$ $in this forum they are many people Good$ $in Maths & Physics God bless You Sir$ $‘ ‘ please Dont use God for a person^{″}$
	Commented by MM42 last updated on 03/Aug/23

	$p e a c e b e u p o n y o u . y o u a r e a$ $p e r s o n w i t h a p e r s o n a l i t y .$
	Commented by witcher3 last updated on 09/Aug/23

	$thank You God bless You$
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