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Math Help - covergence

  1. #1
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    covergence

    show that int(sin^2(x)/ (x^2))dx from 0 to infinity converges and THEN use integration by parts and the identity sin(2x)=2sin(x)cos(x) to show that

    int(sin^2(x)/ (x^2))dx from 0 to infinity = int(sin(x) / x) dx
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  2. #2
    Moo
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    Hello,

    You can see here : http://www.mathhelpforum.com/math-he...-integral.html for something similar, proving it's convergent.

    (similar method as (1-cos)/x)

    As for the integration by parts, integrate 1/x and differentiate sin(x) (which will give 2sin(x)cos(x), and this is where you'll use the identity)
    And this would rather be int(sin(2x)/x), wouldn't it ?
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  3. #3
    Super Member girdav's Avatar
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    Hello.
    \ 0 \leq sin^2 x \leq 1 so  \int_1^{+\infty} \frac{sin^2 x}{x^2}dx converges and we can use \lim_{x\to 0} \dfrac{\sin x}x = 1 to show that
     \int_0^1 \frac{sin^2 x}{x^2}dx converges.
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