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Math Help - \int_0^{\infty}\frac{\sin x}{x}\ dx

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    \int_0^{\infty}\frac{\sin x}{x}\ dx

    Okay, I'm trying to compute the following integral:

    I=\int_0^{\infty}\frac{\sin x}{x}\ dx

    But first, I will try to prove that I converge.

    I=\int_0^{\infty}\frac{\sin x}{x}\ dx  =\int_0^{1}\frac{\sin x}{x}\ dx +\int_1^{\infty}\frac{\sin x}{x}\ dx.


    The integral \int_1^{\infty}\frac{\sin x}{x}\ dx <\infty (Dirichlet's test)


    But what about \int_0^{1}\frac{\sin x}{x}\ dx ?

    I know that x=0 is our problematic point and also I know that \lim_{x\to 0} \frac{\sin x}{x}=1.

    Can I say that because that limit exists the function is integrable?



    Now to the the second part-computation.


    It's from a book by G. M. Fikhtengoltz, The Fundamentals of Mathematical Analysis.

    (I'm translating this from Russian into Hebrew and then into English.)

    We will represent I in form of sum:

    I=\sum_{\nu =0}^{\infty}\int^{(\nu+1)\frac{\pi}{2}}_{\nu\frac{  \pi}{2}}.

    Putting \nu=2\mu or \nu=2\mu-1 and the he says the following:

    we substituting x=\mu\pi+t or x=\mu\pi-t, and we get:

    \int_{2\mu\frac{\pi}{2}}^{(2\mu+1)\frac{\pi}{2}} = (-1)^{\mu}\int_{0}^{\frac{\pi}{2}}\frac{\sin t}{\mu\pi+t}

    and

    \int_{(2\mu-1)\frac{\pi}{2}}^{2\mu\frac{\pi}{2}} =(-1)^{\mu - 1}\int_{0}^{\frac{\pi}{2}}\frac{\sin t}{\mu\pi-t}


    Question:

    Why we need the both substitutions, x=\mu\pi+t or x=\mu\pi-t ?





    Thank you.


    To be continued...
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  2. #2
    Super Member General's Avatar
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    Re: \int_0^{\infty}\frac{\sin x}{x}\ dx

    Your integral is just the laplace transformation of f(t)=\frac{sin(t)}{t} evaluated at s=0.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: \int_0^{\infty}\frac{\sin x}{x}\ dx

    Quote Originally Posted by General View Post
    Your integral is just the laplace transformation of f(t)=\frac{sin(t)}{t} evaluated at s=0.

    Without Laplace transformation please.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: \int_0^{\infty}\frac{\sin x}{x}\ dx

    A more general form of the indefinite integral is...

    \int_{0}^{\infty} \frac{\sin p x}{x}\ d x =\begin{cases}\frac{\pi}{2}&p>0\\0&p=0\\-\frac{\pi}{2}&p<0\end{cases} (1)

    This integral has great importance in several applcations, so that almost nobody remembers the 'theoretical bag' hidden in it: the definite integral (1) converges if it is processed as Riemann integral and doesn't converge if it is processed as Lebesgue integral, as for example in...

    Lebesgue integration - Wikipedia, the free encyclopedia

    All that is well known of course but, proper considering this example, I wonder how to justify motivations regarding the 'superiority' of the Lebesgue integral like that ...

    ... the Riemann integral is not defined for a wide range of functions and situations of importance in applications (and of interest in theory). For example, the Riemann integral can easily integrate density to find the mass of a steel beam, but cannot accommodate a steel ball resting on it. This motivates other definitions, under which a broader assortment of functions are integrable. The Lebesgue integral, in particular, achieves great flexibility by directing attention to the weights in the weighted sum...

    Kind regards

    \chi \sigma
    Last edited by chisigma; August 18th 2011 at 01:08 AM. Reason: forgotten the' dx' in integral...
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Re: \int_0^{\infty}\frac{\sin x}{x}\ dx

    Also sprach Zarathustra, have you tried double integration? It's a neat approach and the justification is easy.
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: \int_0^{\infty}\frac{\sin x}{x}\ dx

    Quote Originally Posted by Krizalid View Post
    Also sprach Zarathustra, have you tried double integration? It's a neat approach and the justification is easy.
    No, I didn't.

    I just want to understand that interesting method that the author presenting.

    By the way, he using in the computation, the interesting formula for \frac{1}{\sin t},


    \frac{1}{\sin t}=\frac{1}{t}+\sum^{\infty}_{1}(-1)^\mu \left (\frac{1}{t+\mu\pi}+\frac{1}{t-\mu\pi}  \right ) .
    Last edited by Also sprach Zarathustra; August 21st 2011 at 12:41 AM.
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