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  1. #1
    Member Nacho's Avatar
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    Improper integral

    Diverge?

    <br />
\int_0^\infty  {\frac{{\sin x}}<br />
{x}dx} <br /> <br />

    thanks
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  2. #2
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    Krizalid's Avatar
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    It does converge. (Not absolutely, of course.)
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  3. #3
    Member Nacho's Avatar
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    Quote Originally Posted by Krizalid View Post
    It does converge. (Not absolutely, of course.)
    why?
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  4. #4
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    I'll try the proof I read a time ago...

    Let I_{a,b}=\int_a^b \frac{\sin(x)}{x} ~dx

    Integrate by parts with :
    u(x)=\frac 1x and v'(x)=\sin(x). So we have u'(x)=-\frac 1{x^2} and v(x)=-\cos(x)+c, and we'll take c=1. So v(x)=1-\cos(x)


    I_{a,b}=\left[\frac{1-\cos(x)}{x}\right]_a^b+\int_a^b \frac{1-\cos(x)}{x^2} ~dx

    But we know by Taylor series that 1-\cos(x) \approx \frac{x^2}{2} when x is near from 0.

    So we can make a=0 since \frac{1-\cos(x)}{x} and \frac{1-\cos(x)}{x^2} have finite limits when x \to 0

    I_b=\left[\frac{1-\cos(x)}{x}\right]_a^b+\int_0^b \frac{1-\cos(x)}{x^2} ~dx

    I_b=\frac{1-\cos(b)}{b}+\int_0^1 \frac{1-\cos(x)}{x^2} ~dx+\int_1^b \frac{1-\cos(x)}{x^2} ~dx



    I=\lim_{b \to + \infty} I_b=\lim_{b \to + \infty} \frac{1-\cos(b)}{b}+\int_0^1 \frac{1-\cos(x)}{x^2} ~dx+\lim_{b \to + \infty} \int_1^b \frac{1-\cos(x)}{x^2} ~dx

    _________________________________________________
    See \int_0^1 \frac{1-\cos(x)}{x^2} ~dx
    By extension, the integrand is continuous for all x in [0,1] and finite. Thus the integral exists and it is finite value. So it doesn't intervene in the convergence or divergence of I
    (actually, you have to check this explanation...it's the sort of parts in which I am always wrong)


    See \lim_{b \to + \infty} \frac{1-\cos(b)}{b}
    This obviously converges.


    See \lim_{b \to + \infty} \int_1^b \frac{1-\cos(x)}{x^2} ~dx
    This converges by comparison with the Riemann integral, since :
    \frac{1-\cos(x)}{x^2} \le \frac{2}{x^2}.
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  5. #5
    Member Nacho's Avatar
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    thanks!!
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