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Math Help - Integral

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

    Challenge Problem:

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

    Moderator Approved - CB
    Last edited by CaptainBlack; May 4th 2010 at 12:52 PM.
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  2. #2
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    Quote Originally Posted by Random Variable View Post
    Challenge Problem:

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

    Moderator Approved - CB
    starting with by parts we have:

    \int_0^{\infty} \frac{\sin^3x}{x^2} \ dx = 3 \int_0^{\infty} \frac{\sin^2 x \cos x}{x} \ dx = 3 \int_0^{\infty} \int_0^{\infty} \sin^2 x \cos x \ e^{-yx} \ dy \ dx

    = 3 \int_0^{\infty} \int_0^{\infty} \sin^2 x \cos x \ e^{-yx} \ dx \ dy = \frac{3}{4} \int_0^{\infty} \int_0^{\infty}[\cos x - \cos(3x)]e^{-yx} \ dx \ dy

    =\frac{3}{4} \int_0^{\infty} \left( \frac{y}{y^2+1} - \frac{y}{y^2+9} \right) \ dy = \frac{3}{8} \left [ \ln \left( \frac{y^2+1}{y^2+9} \right) \right]_0^{\infty} =\frac{3}{4} \ln 3.
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    starting with by parts we have:

    \int_0^{\infty} \frac{\sin^3x}{x^2} \ dx = 3 \int_0^{\infty} \frac{\sin^2 x \cos x}{x} \ dx = 3 \int_0^{\infty} \int_0^{\infty} \sin^2 x \cos x \ e^{-yx} \ dy \ dx

    = 3 \int_0^{\infty} \int_0^{\infty} \sin^2 x \cos x \ e^{-yx} \ dx \ dy = \frac{3}{4} \int_0^{\infty} \int_0^{\infty}[\cos x - \cos(3x)]e^{-yx} \ dx \ dy

    =\frac{3}{4} \int_0^{\infty} \left( \frac{y}{y^2+1} - \frac{y}{y^2+9} \right) \ dy = \frac{3}{8} \left [ \ln \left( \frac{y^2+1}{y^2+9} \right) \right]_0^{\infty} =\frac{3}{4} \ln 3.
    Nice.


    The following was my approach:

     \int^{\infty}_{0} \frac{\sin^{3} x}{x^{2}} \ dx = \frac{1}{4} \int^{\infty}_{0} \frac{3 \sin x - \sin 3x}{x^{2}} \ dx (triple-angle formula for  \sin )

    = \frac{1}{4} \lim_{\delta \to 0} \Big(3\int^{\infty}_{\delta}  \frac{\sin x}{x^{2}} \ dx - \int^{\infty}_{\delta} \frac{\sin 3x}  {x^{2}} \ dx \Big)

     = \frac{1}{4} \lim_{\delta \to 0} \Big(3\int^{\infty}_{\delta}  \frac{\sin x}{x^{2}} \ dx - \frac{1}{3}\int^{\infty}_{3 \delta}  \frac{\sin u} {(\frac{u}{3})^{2}} \ du \Big)

     = \frac{3}{4} \lim_{\delta \to 0} \Big(3\int^{\infty}_{\delta}  \frac{\sin x}{x^{2}} \ dx - \int^{\infty}_{3 \delta} \frac{\sin u}  {u^{2}} \ du \Big)

     =\frac{3}{4} \lim_{\delta \to 0} \int_{\delta}^{3 \delta} \frac{\sin  x}{x^{2}} \ dx

     = \frac{3}{4} \lim_{\delta \to 0} \int_{\delta}^{3 \delta} \frac{1}{x} \  dx (since we're integrating over x-values close to 0)

     = \frac{3}{4} \lim_{\delta \to 0} (\ln {3\delta} - \ln {\delta}) =  \frac{3}{4} \ln 3
    Last edited by Random Variable; May 4th 2010 at 08:38 PM.
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  4. #4
    Senior Member pankaj's Avatar
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    Quote Originally Posted by Random Variable View Post

     =\frac{3}{4} \lim_{\delta \to 0} \int_{\delta}^{3 \delta} \frac{\sin x}{x^{2}} \ dx

     = \frac{3}{4} \lim_{\delta \to 0} \int_{\delta}^{3 \delta} \frac{1}{x} \ dx (since we're integrating over x-values close to 0)
    Is this justified since
    \lim_{x\to 0}\frac{\sin x}{x^2}
    does not exist.

    Moreover,
    \lim_{x\to a}f(x).g(x)=\lim_{x\to a}f(x).\lim_{x\to a}g(x)

    only if both
    \lim_{x\to a}f(x) and \lim_{x\to a}g(x)
    exist and are finite.

    You may have obtained the answer but there are loopholes.

    The solution is pretty good otherwise.
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  5. #5
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    We can use the infinite series  \sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - ....
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