math-analysis-xsin-x-x-2-2x-2-dx-xsin-x-x-1-2-1-dx-x-1-t-t-1-sin-t-1-t-2-1-dt-

Question Number 131969 by mnjuly1970 last updated on 10/Feb/21

\dots m a t h a n a l y s i s \dots

ϕ = \int_{- \infty}^{+ \infty} \frac{x s i n (x)}{x^{2} + 2 x + 2} d x = ?

ϕ = \int_{- \infty}^{+ \infty} \frac{x s i n (x)}{{(x + 1)}^{2} + 1} d x

\overset{x + 1 = t}{=} \int_{- \infty}^{+ \infty} \frac{(t - 1) s i n (t - 1)}{t^{2} + 1} d t

= \int_{- \infty}^{+ \infty} \frac{t s i n (t) c o s (1) - t c o s (t) s i n (1) - s i n (t) c o s (1) + c o s (t) s i n (1)}{t^{2} + 1} d t

= 2 c o s (1) \int_{0}^{\infty} \frac{t s i n (t)}{t^{2} + 1} d t + 2 s i n (1) \int_{0}^{\infty} \frac{c o s (t)}{t^{2} + 1} d t

= 2 c o s (1) . \frac{π}{2 e} + 2 s i n (1) . \frac{π}{2 e}

= \frac{π}{e} (c o s (1) + s i n (1)) \dots .

Answered by mathmax by abdo last updated on 10/Feb/21

another way Φ = \int_{- \infty}^{+ \infty} \frac{xsinx}{x^{2} + 2 x + 2} \Rightarrow Φ = Im (\int_{- \infty}^{+ \infty} \frac{{xe}^{ix}}{x^{2} + 2 x + 2})

let φ (z) = \frac{{ze}^{iz}}{z^{2} + 2 z + 2}, poles of φ

z^{2} + 2 z + 2 = 0 \to Δ^{^{'}} = - 1 \Rightarrow z_{1} = - 1 + i and z_{2} = - 1 - i

and \int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, z_{1}) but φ (z) = \frac{{ze}^{iz}}{(z - z_{1}) (z - z_{2})}

Res (φ, z_{1}) = \lim_{z \to z_{1}} (z - z_{1}) φ (z) = \lim_{z \to z_{1}} \frac{{ze}^{iz}}{z - z_{2}} = \frac{z_{1} e^{{iz}_{1}}}{z_{1} - z_{2}}

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

= \frac{e^{- 1}}{2 i} {- \cos 1 + isin 1 + icos 1 + \sin 1} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π . \frac{e^{- 1}}{2 i} {\sin 1 - \cos 1 + i (\cos 1 + \sin 1)} \Rightarrow

Φ = \frac{π}{e} (\cos (1) + \sin (1))