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Question Number 66794 by mathmax by abdo last updated on 19/Aug/19
calculate ∫_0 ^∞ ((cos(2arctan(2x)))/(9+x^2 ))dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{2}{arctan}\left(\mathrm{2}{x}\right)\right)}{\mathrm{9}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 21/Aug/19
let A =∫_0 ^∞ ((cos(2arctan(2x)))/(9+x^2 ))dx we have cos(2arctan(2x)) =2cos^2 (arctan(2x))−1 =2((1/( (√(1+(2x)^2 )))))^2 −1 =(2/(1+4x^2 ))−1 =((2−1−4x^2 )/(4x^2 +1)) =((1−4x^2 )/(4x^2 +1)) ⇒ A =∫_0 ^∞ ((1−4x^2 )/((4x^2 +1)(x^2 +9)))dx =(1/4)∫_0 ^∞ ((1−4x^2 )/((x^2 +(1/4))(x^(2 ) +9)))dx =(1/8)∫_(−∞) ^(+∞) ((1−4x^2 )/((x^2 +(1/4))(x^2 +9)))dx let W(z) =((1−4z^2 )/((z^2 +(1/4))(z^2 +9))) ⇒ W(z) =((1−4z^2 )/((z−(i/2))(z+(i/2))(z−3i)(z+3i))) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ{ Res(W,(i/2))+Res(W,3i)} Res(W,(i/2)) =((1−4((i/2))^2 )/(i(−1+9))) =((1+1)/(8i)) =(1/(4i)) Res(W,3i) =((1−4(3i)^2 )/((−9+(1/4))(6i))) =((1+36)/((6i)(−35)))×4 =((4×37)/(−(35×6)i)) =−((2×37)/(3×35i)) =−((74)/(105i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ{(1/(4i))−((74)/(105i))} =(π/2)−((148π)/(105)) ⇒ A =(1/8)((π/2)−((148)/(105))π)
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left(\mathrm{2}{arctan}\left(\mathrm{2}{x}\right)\right)}{\mathrm{9}+{x}^{\mathrm{2}} }{dx}\:\:{we}\:{have}\: \\ $$$${cos}\left(\mathrm{2}{arctan}\left(\mathrm{2}{x}\right)\right)\:=\mathrm{2}{cos}^{\mathrm{2}} \left({arctan}\left(\mathrm{2}{x}\right)\right)−\mathrm{1} \\ $$$$=\mathrm{2}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\mathrm{2}{x}\right)^{\mathrm{2}} }}\right)^{\mathrm{2}} −\mathrm{1}\:=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }−\mathrm{1}\:=\frac{\mathrm{2}−\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:=\frac{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }{\left(\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{9}\right)}{dx}\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({x}^{\mathrm{2}\:} +\mathrm{9}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({x}^{\mathrm{2}} \:+\mathrm{9}\right)}{dx}\:\:{let}\:{W}\left({z}\right)\:=\frac{\mathrm{1}−\mathrm{4}{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)\left({z}^{\mathrm{2}} \:+\mathrm{9}\right)}\:\Rightarrow \\ $$$${W}\left({z}\right)\:=\frac{\mathrm{1}−\mathrm{4}{z}^{\mathrm{2}} }{\left({z}−\frac{{i}}{\mathrm{2}}\right)\left({z}+\frac{{i}}{\mathrm{2}}\right)\left({z}−\mathrm{3}{i}\right)\left({z}+\mathrm{3}{i}\right)}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left({W},\frac{{i}}{\mathrm{2}}\right)+{Res}\left({W},\mathrm{3}{i}\right)\right\} \\ $$$${Res}\left({W},\frac{{i}}{\mathrm{2}}\right)\:=\frac{\mathrm{1}−\mathrm{4}\left(\frac{{i}}{\mathrm{2}}\right)^{\mathrm{2}} }{{i}\left(−\mathrm{1}+\mathrm{9}\right)}\:=\frac{\mathrm{1}+\mathrm{1}}{\mathrm{8}{i}}\:=\frac{\mathrm{1}}{\mathrm{4}{i}} \\ $$$${Res}\left({W},\mathrm{3}{i}\right)\:=\frac{\mathrm{1}−\mathrm{4}\left(\mathrm{3}{i}\right)^{\mathrm{2}} }{\left(−\mathrm{9}+\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{6}{i}\right)}\:=\frac{\mathrm{1}+\mathrm{36}}{\left(\mathrm{6}{i}\right)\left(−\mathrm{35}\right)}×\mathrm{4}\:=\frac{\mathrm{4}×\mathrm{37}}{−\left(\mathrm{35}×\mathrm{6}\right){i}} \\ $$$$=−\frac{\mathrm{2}×\mathrm{37}}{\mathrm{3}×\mathrm{35}{i}}\:=−\frac{\mathrm{74}}{\mathrm{105}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{\mathrm{1}}{\mathrm{4}{i}}−\frac{\mathrm{74}}{\mathrm{105}{i}}\right\}\:=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{148}\pi}{\mathrm{105}}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{148}}{\mathrm{105}}\pi\right) \\ $$

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