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Question Number 146621 by 777316 last updated on 14/Jul/21

∫_0 ^( ∞) ((xsin(4x))/(9 + 4x^2 ))dx =

$$\int_{\mathrm{0}} ^{\:\infty} \:\frac{{xsin}\left(\mathrm{4}{x}\right)}{\mathrm{9}\:+\:\mathrm{4}{x}^{\mathrm{2}} }{dx}\:=\: \\ $$

Answered by mathmax by abdo last updated on 14/Jul/21

Ψ=∫_0 ^∞ ((xsin(4x))/(4x^2 +9)) ⇒Ψ=_(2x=3z) ∫_0 ^∞ ((3zsin(4((3z)/2)))/(2.9(1+z^2 ))) (3/2)dz =(1/4)∫_0 ^∞ ((zsin(6z))/(z^2 +1))dz =(1/8)∫_(−∞) ^(+∞) ((zsin(6z))/(z^2 +1))dz =(1/8)Im(∫_(−∞) ^(+∞) ((ze^(6iz) )/(z^2 )1))dz) let ϕ(z)=((ze^(6iz) )/(z^2 +1)) ⇒ϕ(z)=((ze^(6iz) )/((z−i)(z+i))) residus th.⇒∫_R ϕ(z)dz=2iπRes(ϕ,i) Res(ϕ,i)=((ie^(−6) )/(2i)) ⇒∫_R ϕ(z)dz=2iπ.((ie^(−6) )/(2i))=((iπ)/e^6 ) ⇒ Ψ=(1/8)×(π/e^6 )=(π/(8e^6 ))

$$\Psi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{xsin}\left(\mathrm{4x}\right)}{\mathrm{4x}^{\mathrm{2}} \:+\mathrm{9}}\:\Rightarrow\Psi=_{\mathrm{2x}=\mathrm{3z}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{3zsin}\left(\mathrm{4}\frac{\mathrm{3z}}{\mathrm{2}}\right)}{\mathrm{2}.\mathrm{9}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{zsin}\left(\mathrm{6z}\right)}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{8}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{zsin}\left(\mathrm{6z}\right)}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{ze}^{\mathrm{6iz}} }{\left.\mathrm{z}^{\mathrm{2}} \:\right)\mathrm{1}}\mathrm{dz}\right) \\ $$$$\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{6iz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{6iz}} }{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)} \\ $$$$\mathrm{residus}\:\mathrm{th}.\Rightarrow\int_{\mathrm{R}} \varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\mathrm{Res}\left(\varphi,\mathrm{i}\right) \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\right)=\frac{\mathrm{ie}^{−\mathrm{6}} }{\mathrm{2i}}\:\Rightarrow\int_{\mathrm{R}} \varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi.\frac{\mathrm{ie}^{−\mathrm{6}} }{\mathrm{2i}}=\frac{\mathrm{i}\pi}{\mathrm{e}^{\mathrm{6}} }\:\Rightarrow \\ $$$$\Psi=\frac{\mathrm{1}}{\mathrm{8}}×\frac{\pi}{\mathrm{e}^{\mathrm{6}} }=\frac{\pi}{\mathrm{8e}^{\mathrm{6}} } \\ $$

Answered by KINMATICS last updated on 14/Jul/21

∫ ((xsin4x)/(9+4x^2 ))dx=∫sin4x.(x/((2x)^2 -(3i)^2 ))dx ∫ ((xsin4x)/(9+4x^2 ))dx=∫sin4x.(1/4)((1/((2x+3i)))+(1/(2x-3i)))dx ∫ ((xsin4x)/(9+4x^2 ))dx=(1/2)∫((sin4x)/((4x+6i)))dx+(1/2)∫((sin4x)/(4x-6i))dx For ∫((sin4x)/(4x+6i))dx, let 4x+6i=u,4x=u-6i, dx=(du/4) for ∫((sin4x)/(4x-6i))dx, let 4x-6i=y, 4x=y+6i, dx=(dy/4) ⇒∫((sin4xdx)/(4x^2 +9))=(1/8)∫((sin(u-6i))/u)du+(1/8)∫((sin(y+6i))/y)dy I=(1/8)∫((sin(u)cos(6i)-cos(u)sin(6i))/u)du+ (1/8)∫((sin(y)cos(6i)+cos(y)sin(6i))/y)dy I=((cos(6i))/8)∫((sin(u))/u)du-((sin(6i))/8)∫((cos(u))/u)du+ ((cos(6i))/8)∫((sin(y))/y)dy+((sin(6i))/8)∫((cos(y))/y)dy I= ((cos(6i))/8)Si(u)−((sin(6i))/8)Ci(u)+ ((cos(6i))/8)Si(y)+((sin(6i))/8)Ci(y)+C I=((cos(6i))/8)Si(4x+6i)-((sin(6i))/8)Ci(4x+6i) ((cos(6i))/8)Si(4x-6i)+((sin(6i))/8)Ci(4x-6i)+C By: Kin Tom contact www.hppt://kintom077∂gmail.com

$$\int\:\frac{\mathrm{xsin4x}}{\mathrm{9}+\mathrm{4x}^{\mathrm{2}} }\mathrm{dx}=\int\mathrm{sin4x}.\frac{\mathrm{x}}{\left(\mathrm{2x}\right)^{\mathrm{2}} -\left(\mathrm{3i}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$\int\:\frac{\mathrm{xsin4x}}{\mathrm{9}+\mathrm{4x}^{\mathrm{2}} }\mathrm{dx}=\int\mathrm{sin4x}.\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\left(\mathrm{2x}+\mathrm{3i}\right)}+\frac{\mathrm{1}}{\mathrm{2x}-\mathrm{3i}}\right)\mathrm{dx} \\ $$$$\int\:\frac{\mathrm{xsin4x}}{\mathrm{9}+\mathrm{4x}^{\mathrm{2}} }\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{sin4x}}{\left(\mathrm{4x}+\mathrm{6i}\right)}\mathrm{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{sin4x}}{\mathrm{4x}-\mathrm{6i}}\mathrm{dx} \\ $$$$\mathrm{For}\:\int\frac{\mathrm{sin4x}}{\mathrm{4x}+\mathrm{6i}}\mathrm{dx},\:\mathrm{let}\:\mathrm{4x}+\mathrm{6i}=\mathrm{u},\mathrm{4x}=\mathrm{u}-\mathrm{6i},\:\mathrm{dx}=\frac{\mathrm{du}}{\mathrm{4}} \\ $$$$\mathrm{for}\:\int\frac{\mathrm{sin4x}}{\mathrm{4x}-\mathrm{6i}}\mathrm{dx},\:\mathrm{let}\:\:\mathrm{4x}-\mathrm{6i}=\mathrm{y},\:\:\mathrm{4x}=\mathrm{y}+\mathrm{6i},\:\mathrm{dx}=\frac{\mathrm{dy}}{\mathrm{4}} \\ $$$$\Rightarrow\int\frac{\mathrm{sin4xdx}}{\mathrm{4x}^{\mathrm{2}} +\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{sin}\left(\mathrm{u}-\mathrm{6i}\right)}{\mathrm{u}}\mathrm{du}+\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{sin}\left(\mathrm{y}+\mathrm{6i}\right)}{\mathrm{y}}\mathrm{dy} \\ $$$$ \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{sin}\left(\mathrm{u}\right)\mathrm{cos}\left(\mathrm{6i}\right)-\mathrm{cos}\left(\mathrm{u}\right)\mathrm{sin}\left(\mathrm{6i}\right)}{\mathrm{u}}\mathrm{du}+ \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{sin}\left(\mathrm{y}\right)\mathrm{cos}\left(\mathrm{6i}\right)+\mathrm{cos}\left(\mathrm{y}\right)\mathrm{sin}\left(\mathrm{6i}\right)}{\mathrm{y}}\mathrm{dy} \\ $$$$ \\ $$$$\mathrm{I}=\frac{\mathrm{cos}\left(\mathrm{6i}\right)}{\mathrm{8}}\int\frac{\mathrm{sin}\left(\mathrm{u}\right)}{\mathrm{u}}\mathrm{du}-\frac{\mathrm{sin}\left(\mathrm{6i}\right)}{\mathrm{8}}\int\frac{\mathrm{cos}\left(\mathrm{u}\right)}{\mathrm{u}}\mathrm{du}+ \\ $$$$\:\:\:\:\:\frac{\mathrm{cos}\left(\mathrm{6i}\right)}{\mathrm{8}}\int\frac{\mathrm{sin}\left(\mathrm{y}\right)}{\mathrm{y}}\mathrm{dy}+\frac{\mathrm{sin}\left(\mathrm{6i}\right)}{\mathrm{8}}\int\frac{\mathrm{cos}\left(\mathrm{y}\right)}{\mathrm{y}}\mathrm{dy} \\ $$$$\: \\ $$$$\mathrm{I}=\:\frac{\mathrm{cos}\left(\mathrm{6i}\right)}{\mathrm{8}}\boldsymbol{\mathrm{S}}\mathrm{i}\left(\mathrm{u}\right)−\frac{\mathrm{sin}\left(\mathrm{6i}\right)}{\mathrm{8}}\mathrm{Ci}\left(\mathrm{u}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{cos}\left(\mathrm{6i}\right)}{\mathrm{8}}\mathrm{Si}\left(\mathrm{y}\right)+\frac{\mathrm{sin}\left(\mathrm{6i}\right)}{\mathrm{8}}\mathrm{Ci}\left(\mathrm{y}\right)+\mathrm{C} \\ $$$$\: \\ $$$$\mathrm{I}=\frac{\mathrm{cos}\left(\mathrm{6i}\right)}{\mathrm{8}}\mathrm{Si}\left(\mathrm{4x}+\mathrm{6i}\right)-\frac{\mathrm{sin}\left(\mathrm{6i}\right)}{\mathrm{8}}\mathrm{Ci}\left(\mathrm{4x}+\mathrm{6i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{cos}\left(\mathrm{6i}\right)}{\mathrm{8}}\mathrm{Si}\left(\mathrm{4x}-\mathrm{6i}\right)+\frac{\mathrm{sin}\left(\mathrm{6i}\right)}{\mathrm{8}}\mathrm{Ci}\left(\mathrm{4x}-\mathrm{6i}\right)+\mathrm{C} \\ $$$$ \\ $$$$\mathrm{By}:\:\mathrm{Kin}\:\mathrm{Tom} \\ $$$$\mathrm{contact}\:\:\:\:\:\:\:\:\mathrm{www}.\mathrm{hppt}://\mathrm{kintom077}\partial\mathrm{gmail}.\mathrm{com} \\ $$

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