calculate ∫_0 ^∞ ((cos(artan(2x)))/((3+x^2 )^2 ))dx


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Question Number 73337 by mathmax by abdo last updated on 10/Nov/19

$c a l c u l a t e \int_{0}^{\infty} \frac{c o s (a r t a n (2 x))}{{(3 + x^{2})}^{2}} d x$
Commented by mathmax by abdo last updated on 11/Nov/19
$let I =∫_0 ^∞ ((cos(arctan(2x)))/((x^2 +3)^2 ))dx ⇒I =_(x=(√3)t) ∫_0 ^∞ ((cos(arctan(2(√3)t)))/(9(t^2 +1)^2 ))(√3)dt ⇒2 I =((√3)/9)∫_(−∞) ^(+∞) ((cos(arctan(2(√3)t))/((t^2 +1)^2 ))dt =((√3)/9) Re(∫_(−∞) ^(+∞) (e^(iarctan(2(√3)t)) /((t^2 +1)^2 ))dt) let W(z)=(e^(iarctan(2(√3)z)) /((z^2 +1)^2 )) ⇒W(z)=(e^(iarctan(2(√3)z)) /((z−i)^2 (z+i)^2 )) and ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) Res(W,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 W(z)}^((1)) =lim_(z→i) {(e^(iarctan(2(√3)z)) /((z+i)^2 ))}^((1)) =lim_(z→i) ((i((2(√3))/(1+12z^2 ))(z+i)^2 e^(iarctan(2(√3)z)) −2(z+i)e^(iarctan(2(√3)z)) )/((z+i)^4 )) =lim_(z→i) (({((2i(√3)(z+i))/(1+12z^2 ))−2}e^(iarctan(2(√3)z)) )/((z+i)^3 )) =(((((−4(√3))/(−11))−2}e^(i arctan(2(√3)i)) )/(−8i)) =(((4(√3)+22)e^(i arctan(2(√3)i)) )/(88i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ×(((4(√3)+22)e^(i arctan(2(√3)i)) )/(88i)) =(π/(44)){ cos(arctan(2(√3)i)+isin(arctan(2(√3)i))} ⇒ I =((√3)/(18))×(π/(44)) cos(arctan(2(√3)i)$
$l e t I = \int_{0}^{\infty} \frac{c o s (a r c t a n (2 x))}{{(x^{2} + 3)}^{2}} d x \Rightarrow I =_{x = \sqrt{3} t} \int_{0}^{\infty} \frac{c o s (a r c t a n (2 \sqrt{3} t))}{9 {(t^{2} + 1)}^{2}} \sqrt{3} d t$ $\Rightarrow 2 I = \frac{\sqrt{3}}{9} \int_{- \infty}^{+ \infty} \frac{c o s (a r c t a n (2 \sqrt{3} t)}{{(t^{2} + 1)}^{2}} d t = \frac{\sqrt{3}}{9} R e (\int_{- \infty}^{+ \infty} \frac{e^{i a r c t a n (2 \sqrt{3} t)}}{{(t^{2} + 1)}^{2}} d t)$ $l e t W (z) = \frac{e^{i a r c t a n (2 \sqrt{3} z)}}{{(z^{2} + 1)}^{2}} \Rightarrow W (z) = \frac{e^{i a r c t a n (2 \sqrt{3} z)}}{{(z - i)}^{2} {(z + i)}^{2}} a n d$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i)$ $R e s (W, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} W (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{e^{i a r c t a n (2 \sqrt{3} z)}}{{(z + i)}^{2}}}^{(1)}$ $= {l i m}_{z \to i} \frac{i \frac{2 \sqrt{3}}{1 + 12 z^{2}} {(z + i)}^{2} e^{i a r c t a n (2 \sqrt{3} z)} - 2 (z + i) e^{i a r c t a n (2 \sqrt{3} z)}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{{\frac{2 i \sqrt{3} (z + i)}{1 + 12 z^{2}} - 2} e^{i a r c t a n (2 \sqrt{3} z)}}{{(z + i)}^{3}}$ $= \frac{(\frac{- 4 \sqrt{3}}{- 11} - 2} e^{i a r c t a n (2 \sqrt{3} i)}}{- 8 i} = \frac{(4 \sqrt{3} + 22) e^{i a r c t a n (2 \sqrt{3} i)}}{88 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π \times \frac{(4 \sqrt{3} + 22) e^{i a r c t a n (2 \sqrt{3} i)}}{88 i}$ $= \frac{π}{44} {c o s (a r c t a n (2 \sqrt{3} i) + i s i n (a r c t a n (2 \sqrt{3} i))} \Rightarrow$ $I = \frac{\sqrt{3}}{18} \times \frac{π}{44} c o s (a r c t a n (2 \sqrt{3} i)$
Commented by mathmax by abdo last updated on 11/Nov/19

$f o r g i v e w e h a v e \int_{- \infty}^{+ \infty} W (z) d z = \frac{π}{44} (4 \sqrt{3} + 22) e^{i a r c t a n (2 \sqrt{3} i)} \Rightarrow$ $I = \frac{\sqrt{3}}{18} \times \frac{π}{22} (2 \sqrt{3} + 11) c o s (a r c t a n (2 \sqrt{3} i)$
Commented by mathmax by abdo last updated on 11/Nov/19
$we have arctan(z)=(1/(2i))ln(((1+iz)/(1−iz))) ⇒arctan(2(√3)i) =(1/(2i))ln(((1−2(√3))/(1+2(√3)))) =(1/(2i))ln(−((2(√3)−1)/(2(√3)+1))) =(1/(2i))ln(−1)+(1/(2i))ln(((2(√3)−1)/(2(√3)+1))) =(1/(2i))(iπ) +(1/(2i))ln(((2(√3)−1)/(2(√3)+1))) =(π/2) +(1/(2i))ln(((2(√3)−1)/(2(√3)+1))) ⇒cos(arctan(2(√3)i))=−sin(−(i/2)ln(((2(√3)−1)/(2(√3)+1))) =sin((i/2)ln(((2(√3)−1)/(2(√3)+1)))) =sh(−(1/2)ln(((2(√3)−1)/(2(√3)+1)))) =((e^(−(1/2)ln(((2(√3)−1)/(2(√3)+1)))) −e^((1/2)ln(((2(√3)−1)/(2(√3)+1)))) )/2) =−(1/2){(√((2(√3)−1)/(2(√3)+1)))−(1/(√((2(√3)−1)/(2(√3)+1))))} ⇒ I=((π(√3))/(18×22))(2(√3)+11){(((2(√3)−1)/(2(√3)+1)))^(−(1/2)) −(((2(√3)−1)/(2(√3)+1)))^(1/2) }$
$w e h a v e a r c t a n (z) = \frac{1}{2 i} l n (\frac{1 + i z}{1 - i z}) \Rightarrow a r c t a n (2 \sqrt{3} i)$ $= \frac{1}{2 i} l n (\frac{1 - 2 \sqrt{3}}{1 + 2 \sqrt{3}}) = \frac{1}{2 i} l n (- \frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1})$ $= \frac{1}{2 i} l n (- 1) + \frac{1}{2 i} l n (\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1}) = \frac{1}{2 i} (i π) + \frac{1}{2 i} l n (\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1})$ $= \frac{π}{2} + \frac{1}{2 i} l n (\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1}) \Rightarrow c o s (a r c t a n (2 \sqrt{3} i)) = - s i n (- \frac{i}{2} l n (\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1})$ $= s i n (\frac{i}{2} l n (\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1})) = s h (- \frac{1}{2} l n (\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1}))$ $= \frac{e^{- \frac{1}{2} l n (\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1})} - e^{\frac{1}{2} l n (\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1})}}{2} = - \frac{1}{2} {\sqrt{\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1}} - \frac{1}{\sqrt{\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1}}}} \Rightarrow$ $I = \frac{π \sqrt{3}}{18 \times 22} (2 \sqrt{3} + 11) {{(\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1})}^{- \frac{1}{2}} - {(\frac{2 \sqrt{3} - 1}{2 \sqrt{3} + 1})}^{\frac{1}{2}}}$
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