calculate ∫_(−∞) ^(+∞) ((cos(e^x +e^(−x) ))/((x^2 +x+1)^2 ))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 77751 by abdomathmax last updated on 09/Jan/20

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{c o s (e^{x} + e^{- x})}{{(x^{2} + x + 1)}^{2}} d x$
Commented by msup trace by abdo last updated on 10/Jan/20
$let I=∫_(−∞) ^(+∞) ((cos(e^x +e^(−x) ))/((x^2 +x+1)^2 )) dx ⇒ I=Re(∫_(−∞) ^(+∞) (e^(i(e^x +e^(−x) )) /((x^2 +x+1)^2 ))dx) let ϕ(z)=(e^(i(e^z +e^(−z) )) /((z^2 +z+1)^2 )) poles of ϕ? z^2 +z+1=0 →Δ=−3 ⇒ z_1 =((−1+i(√3))/2) =e^(i((2π)/3)) and z_2 =((−1−i(√3))/2) =e^(−((i2π)/3)) ⇒ϕ(z)=(e^(i(e^z +e^(−z) )) /((z−e^((i2π)/3) )^2 (z−e^(−((i2π)/3)) )^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz = 2iπ{ Re(ϕ,e^((i2π)/3) )+Res(ϕ,e^(−((i2π)/3)) )} Res(ϕ,e^((i2π)/3) )=lim_(z→e^((i2π)/3) ) (1/((2−1)!)){(z−e^(i((2π)/3)) )ϕ(z)}^((1))$
$l e t I = \int_{- \infty}^{+ \infty} \frac{c o s (e^{x} + e^{- x})}{{(x^{2} + x + 1)}^{2}} d x \Rightarrow$ $I = R e (\int_{- \infty}^{+ \infty} \frac{e^{i (e^{x} + e^{- x})}}{{(x^{2} + x + 1)}^{2}} d x)$ $l e t φ (z) = \frac{e^{i (e^{z} + e^{- z})}}{{(z^{2} + z + 1)}^{2}} p o l e s o f φ ?$ $z^{2} + z + 1 = 0 \to Δ = - 3 \Rightarrow$ $z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{i \frac{2 π}{3}} a n d z_{2} = \frac{- 1 - i \sqrt{3}}{2}$ $= e^{- \frac{i 2 π}{3}} \Rightarrow φ (z) = \frac{e^{i (e^{z} + e^{- z})}}{{(z - e^{\frac{i 2 π}{3}})}^{2} {(z - e^{- \frac{i 2 π}{3}})}^{2}}$ $\int_{- \infty}^{+ \infty} φ (z) d z =$ $2 i π {R e (φ, e^{\frac{i 2 π}{3}}) + R e s (φ, e^{- \frac{i 2 π}{3}})}$ $R e s (φ, e^{\frac{i 2 π}{3}}) = {l i m}_{z \to e^{\frac{i 2 π}{3}}} \frac{1}{(2 - 1)!} {(z - e^{i \frac{2 π}{3}}) φ (z)}^{(1)}$
Commented by mathmax by abdo last updated on 10/Jan/20
$forgive ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((i2π)/3) ) and Res(ϕ,e^(i((2π)/3)) )=lim_(z→e^((i2π)/3) ) (1/((2−1)!)){(z−e^((i2π)/3) )ϕ(z)}^((1)) =lim_(z→e^((i2π)/3) ) {(e^(i( e^z +e^(−z) )) /((z−e^(−((i2π)/3)) )^2 ))} =lim_(z→e^((i2π)/3) ) { ((i(e^z −e^(−z) )e^(i(e^z +e^(−z) )) (z−e^(−((i2π)/3)) )^2 −2(z−e^(−((i2π)/3)) )e^(i(e^x +e^(−x) )) )/((z−e^(−((i2π)/3)) )^4 ))} =lim_(z→e^((i2π)/3) ) {{ i(e^z −e^(−z) )(z−e^(−((i2π)/3)) )−2}e^(i(e^z −e^(−z) )) ×(z−e^(−((i2π)/3)) )^(−3) =(({−2sin(((2π)/3))2 sh( e^((i2π)/3) )−2}e^(2ish(e^((i2π)/3) )) )/((2i sin(((2π)/3)))^3 )) =((−2{2sin(((2π)/3))sh(e^((i2π)/3) )+1} e^(2ish(e^((i2π)/3) )) )/(−8i(((√3)/2))^3 )) =((((√3)sh(e^((i2π)/3) )+1)e^(2ish(e^(((i2π)/3) ) )) )/(4i ×(3(√3)))) =(((1+(√3)sh(−(1/2)+i((√3)/2)))e^(2i sh(−(1/2)+((i(√3))/2))) )/(12i(√3))) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =(π/(6(√3)))(1+(√3)×((e^(−(1/2)+i((√3)/2)) −e^(−(−(1/2)+((i(√3))/2))) )/2)) e^(2i(((e^(−(1/2)+((i(√3))/2)) −e^(−(−(1/2)+((i(√3))/2))) )/2)) =(π/(6(√3)))( 1+((√3)/2)( e^(−(1/2)) (cos(((√3)/2))+isin(((√3)/2))−e^(1/2) (cos(((√3)/2))−isin(((√3)/2))) ×e^(i( e^(−(1/2)) (cos(((√3)/2))+isin(((√3)/2))−e^(1/2) (cos(((√3)/2))−isin(((√3)/2)))) rest to extract Re(of this quantity)...be continued...$
$f o r g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, e^{\frac{i 2 π}{3}}) a n d$ $R e s (φ, e^{i \frac{2 π}{3}}) = {l i m}_{z \to e^{\frac{i 2 π}{3}}} \frac{1}{(2 - 1)!} {(z - e^{\frac{i 2 π}{3}}) φ (z)}^{(1)}$ $= {l i m}_{z \to e^{\frac{i 2 π}{3}}} {\frac{e^{i (e^{z} + e^{- z})}}{{(z - e^{- \frac{i 2 π}{3}})}^{2}}}$ $= {l i m}_{z \to e^{\frac{i 2 π}{3}}} {\frac{i (e^{z} - e^{- z}) e^{i (e^{z} + e^{- z})} {(z - e^{- \frac{i 2 π}{3}})}^{2} - 2 (z - e^{- \frac{i 2 π}{3}}) e^{i (e^{x} + e^{- x})}}{{(z - e^{- \frac{i 2 π}{3}})}^{4}}}$ $= {l i m}_{z \to e^{\frac{i 2 π}{3}}} {{i (e^{z} - e^{- z}) (z - e^{- \frac{i 2 π}{3}}) - 2} e^{i (e^{z} - e^{- z})} \times {(z - e^{- \frac{i 2 π}{3}})}^{- 3}$ $= \frac{{- 2 s i n (\frac{2 π}{3}) 2 s h (e^{\frac{i 2 π}{3}}) - 2} e^{2 i s h (e^{\frac{i 2 π}{3}})}}{{(2 i s i n (\frac{2 π}{3}))}^{3}}$ $= \frac{- 2 {2 s i n (\frac{2 π}{3}) s h (e^{\frac{i 2 π}{3}}) + 1} e^{2 i s h (e^{\frac{i 2 π}{3}})}}{- 8 i {(\frac{\sqrt{3}}{2})}^{3}} = \frac{(\sqrt{3} s h (e^{\frac{i 2 π}{3}}) + 1) e^{2 i s h (e^{\frac{i 2 π}{3}})}}{4 i \times (3 \sqrt{3})}$ $= \frac{(1 + \sqrt{3} s h (- \frac{1}{2} + i \frac{\sqrt{3}}{2})) e^{2 i s h (- \frac{1}{2} + \frac{i \sqrt{3}}{2})}}{12 i \sqrt{3}}$ $\Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = \frac{π}{6 \sqrt{3}} (1 + \sqrt{3} \times \frac{e^{- \frac{1}{2} + i \frac{\sqrt{3}}{2}} - e^{- (- \frac{1}{2} + \frac{i \sqrt{3}}{2})}}{2}) e^{2 i (\frac{e^{- \frac{1}{2} + \frac{i \sqrt{3}}{2}} - e^{- (- \frac{1}{2} + \frac{i \sqrt{3}}{2})}}{2}}$ $= \frac{π}{6 \sqrt{3}} (1 + \frac{\sqrt{3}}{2} (e^{- \frac{1}{2}} (c o s (\frac{\sqrt{3}}{2}) + i s i n (\frac{\sqrt{3}}{2}) - e^{\frac{1}{2}} (c o s (\frac{\sqrt{3}}{2}) - i s i n (\frac{\sqrt{3}}{2}))$ $\times e^{i (e^{- \frac{1}{2}} (c o s (\frac{\sqrt{3}}{2}) + i s i n (\frac{\sqrt{3}}{2}) - e^{\frac{1}{2}} (c o s (\frac{\sqrt{3}}{2}) - i s i n (\frac{\sqrt{3}}{2}))}$ $r e s t t o e x t r a c t R e (o f t h i s q u a n t i t y) . . . b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum