prove-that-cos-x-2-2x-x-2-2-dx-pi-e-cos-1-

Question Number 162073 by mnjuly1970 last updated on 26/Dec/21

p r o v e t h a t

Ω = \int_{- \infty}^{+ \infty} \frac{c o s (x)}{{(2 + 2 x + x^{2})}^{2}} d x = \frac{π}{e} c o s (1)

Answered by mindispower last updated on 26/Dec/21

Ω = \int_{- \infty}^{\infty} \frac{c o s (x)}{{(1 + {(1 + x)}^{2})}^{2}} d x

1 + x = t

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

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

f (a) = \int_{- \infty}^{\infty} \frac{c o s (t)}{a^{2} + t^{2}} = R e \int_{- \infty}^{\infty} \frac{e^{i t}}{a^{2} + t^{2}} = R e (2 i π . e^{- a} . \frac{1}{2 i a})

= \frac{π}{a} e^{- a} = f (a)

f^{'} (a) = \int_{- \infty}^{\infty} \frac{- 2 a c o s (t)}{{(a^{2} + t^{2})}^{2}} d t

f^{'} (1) = - 2 \int_{- \infty}^{\infty} \frac{c o s (t) d t}{{(1 + t^{2})}^{2}}

f^{'} (a) = (- \frac{π}{a^{2}} e^{- a} - \frac{π}{a} e^{- a})

\int_{- \infty}^{\infty} \frac{c o s (t) d t}{{(1 + t^{2})}^{2}} = \frac{1}{2} . \frac{2 π}{e} = \frac{π}{e}

\int_{- \infty}^{\infty} \frac{c o s t)}{{(2 + 2 t + t^{2})}^{2}} d t = c o s (1) \int \frac{c o s (x)}{{(1 + x^{2})}^{2}} d x = c o s (1) \frac{π}{e}

Commented by Tawa11 last updated on 26/Dec/21

Great sir

Commented by mnjuly1970 last updated on 26/Dec/21

e x c e l l e n t s o l u t i o n s i r p o w e r \dots

Answered by Mathspace last updated on 26/Dec/21

Υ = \int_{- \infty}^{+ \infty} \frac{c o s x}{{(x^{2} + 2 x + 2)}^{2}} d x

Υ = R e (\int_{- \infty}^{+ \infty} \frac{e^{i x}}{{(x^{2} + 2 x + 2)}^{2}} d x)

l e t f (z) = \frac{e^{i z}}{{(z^{2} + 2 z + 2)}^{2}} p o l e s o f f ?

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

z_{1} = - 1 + i a n d z_{2} = - 1 - i

\int_{R} f (z) d z = 2 i π R e s (f, z_{1})

z_{1} p o l e d o u b l e \Rightarrow

R e s (f, z_{1}) = {l i m}_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} f (z)}^{(1)}

= {l i m}_{z \to z_{1}} {\frac{e^{i z}}{{(z - z_{2})}^{2}}}^{(1)}

= {l i m}_{z \to z_{1}} \frac{{i e}^{i z} {(z - z_{2})}^{2} - 2 (z - z_{2}) e^{i z}}{{(z - z_{2})}^{4}}

= {l i m}_{z \to z_{1}} \frac{{i (z - z_{2}) - 2} e^{i z}}{{(z - z_{2})}^{3}}

= \frac{{i (z_{1} - z_{2}) - 2} e^{i z 1}}{{(z_{1} - z_{2})}^{3}}

= \frac{{i (2 i) - 2} e^{i (- 1 + i)}}{{(2 i)}^{3}}

= \frac{{- 2 - 2} e^{- 1} {c o s (1) - i s i n (1)}}{- 8 i}

= \frac{1}{4 i} e^{- 1} {c o s (1) - i s i n (1)} \Rightarrow

\int_{- \infty}^{+ \infty} f (z) d z = \frac{2 i π}{4 i} e^{- 1} {c o s (1) - i s i n (1)}

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

a n d Υ = R e (\dots)

= \frac{π}{2 e} c o s (1)

Commented by mnjuly1970 last updated on 26/Dec/21

t h a n k s a l o t m a s t e r

Answered by Ar Brandon last updated on 24/Mar/22

Ω = \int_{- \infty}^{+ \infty} \frac{\cos x}{{(x^{2} + 2 x + 2)}^{2}} d x

φ (z) = \frac{e^{i z}}{{(z^{2} + 2 z + 2)}^{2}}, Poles : z_{1} = \frac{- 2 + 2 i}{2} = \sqrt{2} e^{\frac{3 π}{4} i}, z_{2} = \sqrt{2} e^{- \frac{π}{4} i}

Ω = R e \int_{- \infty}^{+ \infty} φ (z) d z = R e (2 i π R e s (φ, z_{1}))

R e s (φ, z_{1}) = \lim_{z \to z_{1}} \frac{d}{d z} {{(z - z_{1})}^{2} φ (z)} = \lim_{z \to z_{1}} \frac{d}{d z} {\frac{e^{i z}}{{(z - z_{2})}^{2}}}

= \lim_{z \to z_{1}} {\frac{{i e}^{i z} {(z - z_{2})}^{2} - 2 e^{i z} (z - z_{2})}{{(z - z_{2})}^{4}}} = \frac{{i e}^{{i z}_{1}} {(z_{1} - z_{2})}^{2} - 2 e^{{i z}_{1}} (z_{1} - z_{2})}{{(z_{1} - z_{2})}^{4}}

= \frac{{i e}^{i (- 1 + i)} {(2 i)}^{2} - 2 e^{i (- 1 + i)} (2 i)}{16} = - \frac{4 {i e}^{- (1 + i)} + 4 {i e}^{- (1 + i)}}{16}

= - \frac{1}{2 e} \times e^{(\frac{π}{2} - 1) i} = - \frac{1}{2 e} (\sin (1) + i \cos (1))

\Rightarrow R e (2 i π R e s (φ, z_{1}) = \frac{π}{e} \cos (1)

\Rightarrow \begin{matrix} \int_{- \infty}^{+ \infty} \frac{\cos x}{{(x^{2} + 2 x + 2)}^{2}} d x = \frac{π}{e} \cos (1) \end{matrix}