cosx-x-2-1-dx-don-t-use-feynmann-trick-

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) = \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) = \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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