calculate-0-cos-sinx-sin-cosx-x-2-1-2-dx-

Question Number 136858 by mathmax by abdo last updated on 27/Mar/21

calculate \int_{0}^{\infty} \frac{\cos (sinx) - \sin (cosx)}{{(x^{2} + 1)}^{2}} dx

Answered by mathmax by abdo last updated on 27/Mar/21

Φ = \int_{0}^{\infty} \frac{\cos (sinx)}{{(x^{2} + 1)}^{2}} dx - \int_{0}^{\infty} \frac{\sin (cosx)}{{(x^{2} + 1)}^{2}} dx = I - J

2 I = \int_{- \infty}^{+ \infty} \frac{\cos (sinx)}{{(x^{2} + 1)}^{2}} dx = Re (\int_{- \infty}^{+ \infty} \frac{e^{isinx}}{{(x^{2} + 1)}^{2}} dx) let

W (z) = \frac{e^{isinz}}{{(z^{2} + 1)}^{2}} \Rightarrow w (z) = \frac{e^{isinz}}{{(z - i)}^{2} {(z + i)}^{2}}

\int_{R} w (z) dz = 2 i π Res (w, i)

Res (w, i) = \lim_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} w (z)}^{(1)}

= \lim_{z \to i} {\frac{e^{isinz}}{{(z + i)}^{2}}}^{(1)} = \lim_{z \to i} \frac{icosz e^{isinz} {(z + i)}^{2} - 2 (z + i) e^{isinz}}{{(z + i)}^{4}}

= \lim_{z \to i} \frac{(icosz (z + i) - 2) e^{isinz}}{{(z + i)}^{3}} = \frac{(- 2 cosi - 2) e^{isini}}{{(2 i)}^{3}}

= \frac{(- 2 cosi - 2) e^{isini}}{- 8 i} = \frac{(1 + \cos (i)) e^{isin (i)}}{4 i}

we have cosz = \frac{e^{iz} + e^{- iz}}{2} \Rightarrow \cos (i) = \frac{e^{- 1} + e}{2} = ch (1)

\sin (z) = \frac{e^{iz} - e^{- iz}}{2 i} \Rightarrow \sin (i) = \frac{e^{- 1} - e}{2 i} \Rightarrow isin (i) = \frac{e^{- 1} - e}{2} = - sh (1) \Rightarrow

Res (w, i) = \frac{(1 + ch (1)) e^{- sh (1)}}{4 i} \Rightarrow \int_{R} w (z) dz = 2 i π \frac{(1 + ch (1)) e^{- sh (1)}}{4 i}

= \frac{π}{2} (1 + ch (1)) sh (1) \Rightarrow I = \frac{π}{4} (1 + ch (1)) e^{- sh (1)}

2 J = \int_{- \infty}^{+ \infty} \frac{\sin (cosx)}{{(x^{2} + 1)}^{2}} dx = Im (\int_{- \infty}^{+ \infty} \frac{e^{icosx}}{{(x^{2} + 1)}^{2}} dx) let

φ (z) = \frac{e^{icosz}}{{(z^{2} + 1)}^{2}} \Rightarrow φ (z) = \frac{e^{icosz}}{{(z - i)}^{2} {(z + i)}^{2}}

\int_{R} φ (z) dz = 2 i π Res (φ, i)

Res (φ, i) = \lim_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}

= \lim_{z \to i} {\frac{e^{icosz}}{{(z + i)}^{2}}}^{(1)} = \lim_{z \to i} \frac{- {isinze}^{icosz} {(z + i)}^{2} - 2 (z + i) e^{icosz}}{{(z + i)}^{4}}

= \lim_{z \to i} \frac{(- isinz (z + i) - 2) e^{icosz}}{{(z + i)}^{3}} = \frac{(2 sini - 2) e^{icosi}}{- 8 i}

= \frac{(1 - sini) e^{icosi}}{4 i} \Rightarrow \int_{R} φ (z) dz = 2 i π \frac{(1 - \sin (i)) e^{icosi}}{4 i}

= \frac{π}{2} (1 - \sin (i) e^{icos (i)}

\cos (i) = ch (1) \Rightarrow icos (i) = ich (1) \Rightarrow e^{icos (i)} = e^{ich (1)} = \cos (ch 1) + isin (ch (1))

\sin (i) = \frac{e^{i (i)} - e^{- i (i)}}{2 i} = \frac{e^{- i} - e}{2 i} = - \frac{e - e^{- 1}}{2 i} = - sh (1) rest to extract

Im (\int φ) \dots .