calculate-0-cos-sinx-sin-cosx-x-2-8-dx-

Question Number 137693 by Mathspace last updated on 05/Apr/21

c a l c u l a t e \int_{0}^{\infty} \frac{c o s (s i n x) - s i n (c o s x)}{x^{2} + 8} d x

Answered by mathmax by abdo last updated on 07/Apr/21

\int_{0}^{\infty} \frac{\cos (sinx) - \sin (cosx)}{x^{2} + 8} dx = \int_{0}^{\infty} \frac{\cos (sinx)}{x^{2} + 8} dx - \int_{0}^{\infty} \frac{\sin (cosx)}{x^{2} + 8} dx

= I - J

I =_{x = 2 \sqrt{2} t} \int_{0}^{\infty} \frac{\cos (\sin (2 \sqrt{2} t)}{8 (t^{2} + 1)} (2 \sqrt{2}) t dt

= \frac{\sqrt{2}}{4} \int_{0}^{\infty} \frac{\cos (\sin (2 \sqrt{2} t))}{t^{2} + 1} dt = \frac{\sqrt{2}}{8} \int_{- \infty}^{+ \infty} \frac{\cos (\sin (2 \sqrt{2} t)}{t^{2} + 1} dt

= \frac{\sqrt{2}}{8} Re (\int_{- \infty}^{+ \infty} \frac{e^{isin (2 \sqrt{2}) t}}{t^{2} + 1} dt) let φ (z) = \frac{e^{isin (2 \sqrt{2} z)}}{z^{2} + 1}

\int_{R} φ (z) dz = 2 i π Res (φ, i) = 2 i π . \frac{e^{isin (2 i \sqrt{2})}}{2 i} = π e^{isin (i 2 \sqrt{2})}

we know sinz = \frac{e^{iz} - e^{- iz}}{2 i} \Rightarrow \sin (i 2 \sqrt{2}) = \frac{e^{i (2 i \sqrt{2})} - e^{- i (2 i \sqrt{2})}}{2 i}

= \frac{e^{- 2 \sqrt{2}} - e^{2 \sqrt{2}}}{2 i} = - \frac{i}{2} (e^{- 2 \sqrt{2}} - e^{2 \sqrt{2})} = ish (2 \sqrt{2}) \Rightarrow

\int_{R} φ (z) dz = π e^{i (ish (2 \sqrt{2})} = π e^{- sh (2 \sqrt{2})} \Rightarrow I = \frac{π \sqrt{2}}{8} e^{- sh (2 \sqrt{2})}

the same changement give J = \frac{\sqrt{2}}{8} \int_{- \infty}^{+ \infty} \frac{\sin (\cos (2 \sqrt{2} t))}{t^{2} + 1} dt

= \frac{\sqrt{2}}{8} Im (\int_{- \infty}^{+ \infty} \frac{e^{icos (2 \sqrt{2} t)}}{t^{2} + 1} dt) let Φ (z) = \frac{e^{icos (2 \sqrt{2} z)}}{z^{2} + 1} dt

\int_{R} Φ (z) dz = 2 i π Res (Φ, i) = 2 i π . \frac{e^{icos (2 \sqrt{2} i)}}{2 i} = π e^{ich (2 \sqrt{2})}

= π {\cos (ch (2 \sqrt{2}) + isin (ch (2 \sqrt{2})} \Rightarrow

J = \frac{π \sqrt{2}}{8} \sin (ch (2 \sqrt{2})) \Rightarrow

\int_{0}^{\infty} \frac{\cos (sinx) - \sin (cosx)}{x^{2} + 8} dx = \frac{π \sqrt{2}}{8} (e^{- sh (2 \sqrt{2})} - \sin (ch (2 \sqrt{2}))