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Thread: \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:

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

    But first, I will try to prove that $\displaystyle I$ converge.

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


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


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

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

    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:

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

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

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

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

    and

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


    Question:

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





    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 $\displaystyle f(t)=\frac{sin(t)}{t}$ 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 $\displaystyle f(t)=\frac{sin(t)}{t}$ 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...

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

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Aug 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 $\displaystyle \frac{1}{\sin t}$,


    $\displaystyle \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; Aug 21st 2011 at 12:41 AM.
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