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0-cos-x-cos-3x-x-2-dx-




Question Number 132954 by LUFFY last updated on 17/Feb/21
∫_0 ^( ∞) ((cos(x)−cos(3x))/x^2 )dx
$$\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left({x}\right)−{cos}\left(\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$
Answered by Dwaipayan Shikari last updated on 17/Feb/21
I(α)=∫_0 ^∞ ((cos(αx)−cos(3x))/x^2 )dx    I′(α)=−∫_0 ^∞ ((sin(αx))/x)dx  I′(α)=−(π/2) ⇒I(α)=−(π/2)α+C  α=3 ,I(α)=0  C=((3π)/2)  I(α)=((3π)/2)−(π/2)α  ⇒I(1)=π
$${I}\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{cos}\left(\alpha{x}\right)−{cos}\left(\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$ \\ $$$${I}'\left(\alpha\right)=−\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left(\alpha{x}\right)}{{x}}{dx} \\ $$$${I}'\left(\alpha\right)=−\frac{\pi}{\mathrm{2}}\:\Rightarrow{I}\left(\alpha\right)=−\frac{\pi}{\mathrm{2}}\alpha+{C} \\ $$$$\alpha=\mathrm{3}\:,{I}\left(\alpha\right)=\mathrm{0}\:\:{C}=\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$${I}\left(\alpha\right)=\frac{\mathrm{3}\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}\alpha\:\:\Rightarrow{I}\left(\mathrm{1}\right)=\pi \\ $$

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