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Math Help - Integral

  1. #1
    Super Member Random Variable's Avatar
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    Integral

    Challenge Question:


     \int^{\infty}_{0} \frac{\cos x - \cos 3x}{x^{2}} \ dx



    Moderator editor: Approved Challenge question.
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  2. #2
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    Quote Originally Posted by Random Variable View Post
    Challenge Question:


     \int^{\infty}_{0} \frac{\cos x - \cos 3x}{x^{2}} \ dx



    Moderator editor: Approved Challenge question.

     \int_0^{\infty} \frac{ \cos x - \cos{3x} }{ x^2}~dx

     = \int_0^{\infty} \int_1^3 \frac{\sin{xy}}{x}~dxdy

     = \frac{\pi}{2} \int_1^3 dy = \pi
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    Challenge Question:


     \int^{\infty}_{0} \frac{\cos x - \cos 3x}{x^{2}} \ dx



    Moderator editor: Approved Challenge question.
    Alternatively. Let J(\theta)=\int_0^{\infty}\frac{\cos(x)-\cos(\theta x)}{x^2}dx. Then, J'(\theta)=\int_0^{\infty}\frac{\partial}{\partial \theta}\frac{\cos(x)-\cos(\theta x)}{x^2}dx=\int_0^{\infty}\frac{\sin(\theta x)}{x}dx. Let \theta x=z to get \int_0^{\infty}\frac{\sin(z)}{z}dz but this is a routine calculation ( \frac{1}{z}=\int_0^{\infty}e^{-zt}dt) and it equals \frac{\pi}{2}. So, J'(\theta)=\frac{\pi}{2} and so J(\theta)=\frac{\pi\theta}{2}+C. But, 0=J(1)=\frac{\pi}{2}+C\implies C=-\frac{\pi}{2}. Thus, J(\theta)=\frac{\pi}{2}\left(\theta-1\right)
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  4. #4
    Super Member Random Variable's Avatar
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    or

     \int^{\infty}_{0} \frac{\cos x - \cos 3x}{x^{2}} \ dx = \int^{\infty}_{0} \int^{\infty}_{0} te^{-xt} (\cos x - \cos 3x) \ dt \  dx

     = \int^{\infty}_{0} t \Big(\mathcal{L} \{\cos x\} - \mathcal{L} \{\cos  3x\} \Big) dt = \int^{\infty}_{0} \Big(\frac{t^{2}}{t^{2}+1} - \frac{t^{2}}{t^{2}+9}  \Big) \ dt

    = \int^{\infty}_{0} \Big(\frac{9}{t^{2}+9}  - \frac{1}{1+t^{2}} \Big) \ dt = \Big(3 \tan (t/3) -\tan t\Big) \Big|^{\infty}_{0} = -\frac{\pi}{2} +  \frac{3 \pi}{2} = \pi
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