find-the-value-of-0-cos-2x-x-4-x-2-1-dx-

Question Number 114933 by mathmax by abdo last updated on 22/Sep/20

find the value of \int_{0}^{\infty} \frac{\cos (2 x)}{x^{4} + x^{2} + 1} dx

Answered by mathdave last updated on 22/Sep/20

s o l u t i o n

l e t

I = \int_{0}^{\infty} \frac{\cos (2 x)}{1 + x^{2} + x^{4}} d x = \frac{1}{2} \int_{- \infty}^{\infty} \frac{\cos 2 x}{1 + x^{2} + x^{4}} d x

2 I = \int_{- \infty}^{\infty} \frac{e^{2 i x} + e^{- 2 i x}}{1 + x^{2} + x^{4}} d x = \int_{- \infty}^{\infty} \frac{e^{2 i x}}{1 + x^{2} + x^{4}} d x

l e t 1 + x^{2} + x^{4} = (x^{2} - x + 1) (x^{2} + x + 1)

l e t t h e r o o t o f x^{2} - x + 1 = 0 a r e

z_{1} = \frac{1}{2} + i \frac{\sqrt{3}}{2} a n d z_{2} = \frac{1}{2} - i \frac{\sqrt{3}}{2}

l e t t h e r o o t o f x^{2} + x + 1 = 0 a r e

z_{3} = - \frac{1}{2} + i \frac{\sqrt{3}}{2} a n d z_{4} = - \frac{1}{2} - i \frac{\sqrt{3}}{2}

b y u s i n g t h e i n f i n i t e s e m i c y c l e o n t h e

u p p e r h a l f p l a n e a s t h e c o n t o u r, w e h a v e

t h e p o l e s h a s z_{1} = \frac{1}{2} + i \frac{\sqrt{3}}{2} a n d z_{3} = - \frac{1}{2} + i \frac{\sqrt{3}}{2}

t h e n a p p l y r e s i d u e t h e o r e m

l e t

f (z) = \frac{e^{2 i z}}{1 + x^{2} + x^{4}} = \frac{e^{2 i z}}{(z - z_{1}) (z - z_{2}) (z - z_{3}) (z - z_{4})}

b y r e s i d u e t h e o r e m

R e s [f (z) : z_{1}] = \lim_{z \to z_{1}} [(z - z_{1}) f (z)] = \frac{e^{2 {i z}_{1}}}{(z_{1} - z_{2}) (z_{1} - z_{3}) (z_{1} - z_{4})}

= \frac{e^{- \sqrt{3} + i}}{- 3 + i \sqrt{3}} = - \frac{1}{12} (3 + i \sqrt{3}) e^{- \sqrt{3} + i} \dots \dots (1) a n d

R e s [f (z) : z_{3}] = \lim_{z \to z_{3}} [(z - z_{3}) f (z)] = \frac{e^{2 {i z}_{3}}}{(z_{3} - z_{1}) (z_{3} - z_{2}) (z_{3} - z_{4})}

= \frac{e^{- \sqrt{3} - i}}{3 + i \sqrt{3}} = \frac{1}{12} (3 - i \sqrt{3}) e^{- \sqrt{3} - i} \dots \dots \dots (2) b u t

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

2 I = 2 π i [\frac{1}{12} (3 - i \sqrt{3}) e^{- \sqrt{3} - i} - \frac{1}{12} (3 + i \sqrt{3}) e^{- \sqrt{3} + i}]

2 I = \frac{π e^{- \sqrt{3}}}{6} [(3 i + \sqrt{3}) e^{- i} - (3 i - \sqrt{3}) e^{i}]

2 I = \frac{π e^{- \sqrt{3}}}{6} [- 3 i (e^{i} - e^{- i}) + \sqrt{3} (e^{i} + e^{- i})]

2 I = \frac{π e^{- \sqrt{3}}}{6} [- \frac{6 i \times i}{2 i} (e^{i} - e^{- i}) + \frac{2 \sqrt{3}}{2} (e^{i} + e^{- i})]

2 I = \frac{π e^{- \sqrt{3}}}{6} [6 (\frac{e^{i} - e^{- i}}{2 i}) + 2 \sqrt{3} (\frac{e^{i} + e^{- i}}{2})]

2 I = π e^{- \sqrt{3}} (\sin (1) + \frac{1}{\sqrt{3}} \cos (1))

∵ I = \frac{1}{2} π e^{- \sqrt{3}} [\sin (1) + \frac{\sqrt{3}}{3} \cos (1)]

\int_{0}^{\infty} \frac{\cos (2 x)}{1 + x^{2} + x^{4}} d x = \frac{π e^{- \sqrt{3}}}{6} [3 \sin (1) + \sqrt{3} \cos (1)]

b y m a t h d a v e (22 / 09 / 2020)

Commented by mathmax by abdo last updated on 23/Sep/20

thank you sir

Commented by mathdave last updated on 23/Sep/20

u r e w e l c o m e

Commented by Tawa11 last updated on 06/Sep/21

great sir

Answered by mathmax by abdo last updated on 27/Sep/20

A = \int_{0}^{\infty} \frac{\cos (2 x)}{x^{4} + x^{2} + 1} dx \Rightarrow A = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{\cos (2 x)}{x^{4} + x^{2} + 1} dx

\Rightarrow 2 A = Re (\int_{- \infty}^{+ \infty} \frac{e^{2 ix}}{x^{4} + x^{2} + 1} dx) let φ (z) = \frac{e^{2 iz}}{z^{4} + z^{2} + 1} polesof φ ?

u^{2} + u + 1 = 0 withu = z^{2}

Δ = - 3 \Rightarrow u_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{i \frac{2 π}{3}} and u_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}}

\Rightarrow φ (z) = \frac{e^{2 iz}}{(z^{2} - e^{\frac{i 2 π}{3}}) (z^{2} - e^{- \frac{i 2 π}{3}})} = \frac{e^{2 iz}}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{- \frac{i π}{3}})}

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{3}}) + Res (φ, - e^{- \frac{i π}{3}})}

Res (φ, e^{\frac{i π}{3}}) = \frac{e^{{2 ie}^{\frac{i π}{3}}}}{{2 e}^{\frac{i π}{3}} (2 isin (\frac{2 π}{3}))} = \frac{e^{2 i (\frac{1}{2} + i \frac{\sqrt{3}}{2})}}{4 i . \frac{\sqrt{3}}{2}} \times e^{- \frac{i π}{3}}

= \frac{e^{i} . e^{- \sqrt{3}}}{2 i \sqrt{3}} . e^{- \frac{i π}{3}}

Res (φ, - e^{- \frac{i π}{3}}) = \frac{e^{- 2 i e^{- \frac{i π}{3}}}}{- {2 e}^{- \frac{i π}{3}} (- 2 isin (\frac{2 π}{3}))} = \frac{e^{- 2 i (\frac{1}{2} - \frac{i \sqrt{3}}{2})}}{4 i . \frac{\sqrt{3}}{2}} \times e^{\frac{i π}{3}}

= \frac{e^{- i} . e^{- \sqrt{3}}}{2 i \sqrt{3}} \times e^{\frac{i π}{3}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {\frac{e^{- \sqrt{3}}}{2 i \sqrt{3}} (e^{i - \frac{i π}{3}} + e^{- i + \frac{i π}{3}})}

= \frac{π}{\sqrt{3}} e^{- \sqrt{3}} {e^{(1 - \frac{π}{3}) i} + e^{- (1 - \frac{π}{3}) i}} = \frac{2 π}{\sqrt{3}} e^{- \sqrt{3}} \cos (1 - \frac{π}{3}) \Rightarrow

2 A = \frac{2 π}{\sqrt{3}} e^{- \sqrt{3}} \cos (1 - \frac{π}{3}) \Rightarrow ★ A = \frac{π}{\sqrt{3}} e^{- \sqrt{3}} \cos (1 - \frac{π}{3}) ★