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-

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

\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) = \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