calculate-0-cosx-x-2-1-x-2-2-x-2-3-dx-

Question Number 146898 by mathmax by abdo last updated on 16/Jul/21

calculate \int_{0}^{\infty} \frac{cosx}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)} dx

Answered by qaz last updated on 16/Jul/21

L (\int_{0}^{\infty} \frac{\cos (tx)}{x^{2} + a^{2}} dx) (s) \dots \dots \dots (a > 0)

= \int_{0}^{\infty} \frac{s}{(x^{2} + a^{2}) (s^{2} + x^{2})} dx

= \frac{s}{s^{2} - a^{2}} \int_{0}^{\infty} (\frac{1}{x^{2} + a^{2}} - \frac{1}{s^{2} + x^{2}}) dx

= \frac{s}{s^{2} - a^{2}} (\frac{1}{a} \tan^{- 1} \frac{x}{a} - \frac{1}{s} \tan^{- 1} \frac{x}{s}) ∣_{0}^{\infty}

= \frac{π}{2 a (s + a)}

\int_{0}^{\infty} \frac{\cos x}{x^{2} + a^{2}} dx = L^{- 1} (\frac{π}{2 a (s + a)}) (t = 1) = \frac{π}{2 a} e^{- at} ∣_{t = 1} = \frac{π}{2 a} e^{- a}

\int_{0}^{\infty} \frac{\cos x}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)} dx

= \int_{0}^{\infty} (\frac{\cos x}{2 (x^{2} + 1)} - \frac{\cos x}{x^{2} + 2} + \frac{\cos x}{2 (x^{2} + 3)}) dx

= \frac{π}{4} e^{- 1} - \frac{π}{2 \sqrt{2}} e^{- \sqrt{2}} + \frac{π}{4 \sqrt{3}} e^{- \sqrt{3}}

Commented by mathmax by abdo last updated on 16/Jul/21

thank sir answer correct . .

Answered by mathmax by abdo last updated on 16/Jul/21

Υ = \int_{0}^{\infty} \frac{cosx}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)} dx \Rightarrow 2 Υ = Re (\int_{- \infty}^{+ \infty} \frac{e^{ix}}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)})

let Λ (z) = \frac{e^{iz}}{(z^{2} + 1) (z^{2} + 2) (z^{2} + 3)} \Rightarrow Λ (z) = \frac{e^{iz}}{(z - i) (z + i) (z - \sqrt{2} i) (z + \sqrt{2} i) (z - i \sqrt{3}) (z + i \sqrt{3})}

\int_{R} Λ (z) dz = 2 i π {Res (Λ, i) + Res (Λ, i \sqrt{2}) + Res (Λ, i \sqrt{3})}

Res (Λ, i) = \frac{e^{- 1}}{(2 i) (- 1 + 2) (- 1 + 3)} = \frac{e^{- 1}}{4 i}

Res (Λ, i \sqrt{2}) = \frac{e^{- \sqrt{2}}}{(2 \sqrt{2}) i (- 2 + 1) (- 2 + 3)} = - \frac{e^{- \sqrt{2}}}{2 \sqrt{2} i}

Res (Λ, i \sqrt{3}) = \frac{e^{- \sqrt{3}}}{(2 i \sqrt{3}) (- 3 + 1) (- 3 + 2)} = \frac{e^{- \sqrt{3}}}{4 i \sqrt{3}} \Rightarrow

\int_{R} Λ (z) dz = 2 i π {\frac{e^{- 1}}{4 i} - \frac{e^{- \sqrt{2}}}{2 \sqrt{2} i} + \frac{e^{- \sqrt{3}}}{4 i \sqrt{3}}}

= \frac{π}{2} e^{- 1} - \frac{π}{\sqrt{2}} e^{- \sqrt{2}} + \frac{π}{2 \sqrt{3}} e^{- \sqrt{3}} \Rightarrow

Υ = \frac{π}{4} e^{- 1} - \frac{π}{2 \sqrt{2}} e^{- \sqrt{2}} + \frac{π}{4 \sqrt{3}} e^{- \sqrt{3}}