let A(θ) = ∫_0 ^∞ (dx/((x^2 +3)(x^4 −e^(iθ) ))) with 0<θ<(π/2) 1) calculate A(θ) interms of θ 2) determine also ∫


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Question Number 67932 by mathmax by abdo last updated on 02/Sep/19

$l e t A (θ) = \int_{0}^{\infty} \frac{d x}{(x^{2} + 3) (x^{4} - e^{i θ})} w i t h 0 < θ < \frac{π}{2}$ $1) c a l c u l a t e A (θ) i n t e r m s o f θ$ $2) d e t e r m i n e a l s o \int_{0}^{\infty} \frac{d x}{(x^{2} + 3) {(x^{4} - e^{i θ})}^{2}}$
Commented byMJS last updated on 02/Sep/19

$I get (1) A (θ) = - \frac{π (2 \sqrt{3} e^{i \frac{3 θ}{4}} - {3 e}^{i \frac{θ}{2}} - 9)}{{12 e}^{i \frac{3 θ}{4}} (e^{i θ} - 9)}$ $can this be right ?$
Commented bymathmax by abdo last updated on 02/Sep/19
$1) A(θ)=∫_0 ^∞ (dx/((x^2 +3)(x^4 −e^(iθ) ))) ⇒2A(θ)=∫_(−∞) ^(+∞) (dx/((x^2 +3)(x^4 −e^(iθ) ))) let W(z) =(1/((z^2 +3)(z^4 −e^(iθ) ))) poles of W? z^4 −e^(iθ) =0 ⇒z^4 =e^(iθ) let z =r e^(iα) ⇒r=1 and 4α =θ +2kπ ⇒ α_k =((θ+2kπ)/4) ⇒ the roots are Z_k =e^(i((θ+2kπ)/4)) and k∈[[0,3]] Z_0 =e^((iθ)/4) ,Z_1 =e^(i(((θ+2π)/4))) =e^(((iθ)/4)+((iπ)/2)) , Z_2 =e^(i(((θ+4π)/4))) =e^(((iθ)/4)+iπ) Z_3 = e^(i(((θ+6π)/4))) =e^(((iθ)/4)+i((3π)/2)) =e^(((iθ)/4)+i(2π−(π/2))) =e^(i((θ/4)−(π/2))) Z_1 =i e^((iπ)/4) =i{cos((π/4))+isin((π/4))}=−sin(π/4)+icos((π/4))⇒im(z_1 )>0 im(Z_0 )>0 im(Z_2 )<0 im(z_3 )<0 z^2 +3 =z^2 −(i(√3))^2 =(z−i(√3))(z+i(√3)) so the roots are +^− i(√3) residu theorem ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ{Res(W,i(√3)) +Res(W,Z_0 )+Res(W,Z_1 )} we have W(z) =(1/((z−i(√3))(z+i(√3))(z−Z_0 )(z−Z_1 )(z−Z_2 )(z−Z_3 ))) ⇒ Res(W,i(√3)) =(1/((2i(√3))((i(√3))^4 −e^(iθ) ))) =(1/((2i(√3))(9−e^(iθ) ))) Res(W,Z_0 ) =(1/((Z_0 ^2 +3)(Z_0 −Z_1 )(Z_0 −Z_2 )(Z_0 −Z_3 ))) Res(W,Z_1 ) =(1/((Z_1 ^2 +3)(Z_1 −Z_0 )(Z_1 −Z_2 )(Z_1 −Z_3 ))) rest to finich the calculus ....$
$1) A (θ) = \int_{0}^{\infty} \frac{d x}{(x^{2} + 3) (x^{4} - e^{i θ})} \Rightarrow 2 A (θ) = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} + 3) (x^{4} - e^{i θ})}$ $l e t W (z) = \frac{1}{(z^{2} + 3) (z^{4} - e^{i θ})} p o l e s o f W ?$ $z^{4} - e^{i θ} = 0 \Rightarrow z^{4} = e^{i θ} l e t z = r e^{i α} \Rightarrow r = 1 a n d 4 α = θ + 2 k π \Rightarrow$ $α_{k} = \frac{θ + 2 k π}{4} \Rightarrow t h e r o o t s a r e Z_{k} = e^{i \frac{θ + 2 k π}{4}} a n d k \in [[0, 3]]$ $Z_{0} = e^{\frac{i θ}{4}}, Z_{1} = e^{i (\frac{θ + 2 π}{4})} = e^{\frac{i θ}{4} + \frac{i π}{2}}, Z_{2} = e^{i (\frac{θ + 4 π}{4})} = e^{\frac{i θ}{4} + i π}$ $Z_{3} = e^{i (\frac{θ + 6 π}{4})} = e^{\frac{i θ}{4} + i \frac{3 π}{2}} = e^{\frac{i θ}{4} + i (2 π - \frac{π}{2})} = e^{i (\frac{θ}{4} - \frac{π}{2})}$ $Z_{1} = i e^{\frac{i π}{4}} = i {c o s (\frac{π}{4}) + i s i n (\frac{π}{4})} = - s i n \frac{π}{4} + i c o s (\frac{π}{4}) \Rightarrow i m (z_{1}) > 0$ $i m (Z_{0}) > 0 i m (Z_{2}) < 0 i m (z_{3}) < 0$ $z^{2} + 3 = z^{2} - {(i \sqrt{3})}^{2} = (z - i \sqrt{3}) (z + i \sqrt{3}) s o t h e r o o t s a r e \bar{+} i \sqrt{3}$ $r e s i d u t h e o r e m \Rightarrow \int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, i \sqrt{3})$ $+ R e s (W, Z_{0}) + R e s (W, Z_{1})} w e h a v e$ $W (z) = \frac{1}{(z - i \sqrt{3}) (z + i \sqrt{3}) (z - Z_{0}) (z - Z_{1}) (z - Z_{2}) (z - Z_{3})} \Rightarrow$ $R e s (W, i \sqrt{3}) = \frac{1}{(2 i \sqrt{3}) ({(i \sqrt{3})}^{4} - e^{i θ})} = \frac{1}{(2 i \sqrt{3}) (9 - e^{i θ})}$ $R e s (W, Z_{0}) = \frac{1}{(Z_{0}^{2} + 3) (Z_{0} - Z_{1}) (Z_{0} - Z_{2}) (Z_{0} - Z_{3})}$ $R e s (W, Z_{1}) = \frac{1}{(Z_{1}^{2} + 3) (Z_{1} - Z_{0}) (Z_{1} - Z_{2}) (Z_{1} - Z_{3})}$ $r e s t t o f i n i c h t h e c a l c u l u s . . . .$
Commented bymathmax by abdo last updated on 02/Sep/19

$p e r h a p s b e c a u s e i d o n t c o m p l e t e t h e c a l c u l u s . .$
Commented bymathmax by abdo last updated on 02/Sep/19

$2) w e h a v e A^{^{'}} (θ) = \int_{0}^{\infty} \frac{e^{i θ}}{(x^{2} + 3) {(x^{4} - e^{i θ})}^{2}} d x \Rightarrow$ $\int_{0}^{\infty} \frac{d x}{(x^{2} + 3) {(x^{4} - e^{i θ})}^{2}} = e^{- i θ} A^{^{'}} (θ) r e s t t o c a l c u l a t e A^{^{'}} (θ) .$
	Answered by mind is power last updated on 02/Sep/19

	$l e t f (z) = \frac{d z}{(z^{2} + 3) (x^{4} - e^{i θ})}$ $p o l e o f f a r e \underline{+} i \sqrt{3}, e^{i \frac{θ + 2 k π}{4}} k ⩽ 4$ $w e s h o s e u p e r [h a l f i m (z) > 0 p l a n a n d u s e r e s i d u s t h e o r e m$ $R e s (f . i \sqrt{3}) = \frac{1}{2 i \sqrt{3} (9 - e^{i θ})}$ $r e s (f, e^{i θ}) = \frac{1}{4 e^{i 3 θ} (e^{2 i θ} + 3)}$ $r e s (f, e^{i (θ + \frac{π}{2})}) = \frac{1}{4 e^{i (3 θ + \frac{3 π}{2})} (- e^{i 2 θ} + 3)}$ $\int_{0}^{+ \infty} \frac{d z}{(z^{2} + 3) (z^{4} + e^{i θ})} = \frac{1}{2} \int_{- \infty}^{+ \infty} f (z) d z = i π \sum_{r e s > 0} f (z)$ $= i π [\frac{1}{2 i \sqrt{3} (9 - e^{i θ})} + \frac{1}{4 e^{3 i θ} (e^{2 i θ} + 3)} + \frac{1}{- 4 {i e}^{3 i θ} (3 - e^{2 i θ})}]$ $= \frac{i π}{} [\frac{- i (9 - e^{- i θ})}{2 \sqrt{3} (82 - 18 c o s (θ))} + \frac{e^{- 3 i θ} (3 + e^{- 2 i θ})}{4 (10 + 6 c o s (2 θ))} + \frac{{i e}^{- i 3 θ} (3 - e^{- 2 i θ})}{4 (10 - 6 c o s (2 θ))})$ $2) \frac{d f}{d θ} = \int \frac{e^{i θ}}{(z^{2} + 3) {(z^{4} - e^{i θ})}^{2}} d z = e^{i θ} \int \frac{1}{(z^{2} + 3) (z^{4} - e^{i θ})} d z$ $\Rightarrow \int \frac{d x}{(x^{2} + 3) (x^{4} - e^{i θ})} = f^{'} (θ) e^{- i θ}$
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