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Integral of sinx/x

  1. Oct 7, 2013 #1
    It is well known that [itex]\sin x / x[/itex] is not Lebesgue integrable on [itex][0, +\infty)[/itex] though it is (improper) Riemann Integrable. It is also fairly easily shown (integrating by parts) that
    [tex]\Bigg\lvert \int\limits_{a}^{b} \frac{\sin x}{x} dx\Bigg\rvert \leq 4[/tex]

    Since [itex][a,b][/itex] is compact, the Riemann and Lebesgue Integral of [itex]\sin x / x[/itex] coincide on this set. As [itex]b \to \infty[/itex] and [itex]a\to 0[/itex] the upper bound of 4 remains valid, though in the limit, the Lebesgue integral does not exist.


    I am reading a book that asks the reader to prove the above bound, and in the text, it uses this fact in the computation of this integral:

    [tex] \lim_{\epsilon \to 0}\int\limits_{-\infty}^{+\infty}f(y) \Bigg\lvert \int\limits_{\epsilon^{-1}\geq x \geq \epsilon} \frac{\sin xy}{x} dx \Bigg\rvert dy [/tex]

    If we let [itex]F_{k}(y) = f(y) \Bigg\lvert \int\limits_{k^{-1}\geq x \geq k} \frac{\sin xy}{x} dx \Bigg\rvert [/itex] and the bound of 4 above, we dominate [itex]F_k(y)[/itex] by [itex]f(y)4[/itex].
    If we assume [itex]f[/itex] is bounded and in [itex]L^1[/itex] then we can use Lebesgue Dominated Convergence Theorem to pass the limit inside the integral.


    Now, I get that [itex]F_k[/itex] is uniformly bounded in [itex]k[/itex]. However, if the integral is taken to be a Lebesgue integral (which, it is initially) then I don't see how [itex]\lim F_k [/itex] is even defined. So, what is going on?


    So, perhaps I am missing something obvious, but I am just a little confused by this. If I have missed something completely obvious, please don't hesitate to tell me!
     
  2. jcsd
  3. Oct 7, 2013 #2

    jbunniii

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    ##\sin(x)/x## fails to be strictly Lebesgue integrable on ##[0,\infty)## because the integrals of its positive and negative parts are both infinite.

    We may of course form an improper Lebesgue integral as ##\lim_{a \rightarrow 0, b \rightarrow \infty} \int_{a}^{b} \sin(x)/x dx##. Since the Lebesgue and Riemann integrals coincide for ##\int_{a}^{b} \sin(x)/x dx##, the limit as ##a \rightarrow 0, b \rightarrow \infty## exists and has the same value in both cases.

    Your ##\lim_{k \rightarrow 0} F_k(y)## exists because it only requires the improper integral to exist:

    $$\begin{align}
    \lim_{k \rightarrow 0} F_k(y) &= \lim_{k \rightarrow 0} f(y) \left|\int_{k}^{1/k} \sin(xy)/x dx\right|\\
    &= f(y) \lim_{k \rightarrow 0} \left|\int_{k}^{1/k} \sin(xy)/x dx\right|\\
    &= f(y) \left| \lim_{k \rightarrow 0} \int_{k}^{1/k} \sin(xy)/x dx\right|\\
    \end{align}$$
    where the last equality follows from continuity of the absolute value function. The object inside the absolute value in the last expression is simply the improper Lebesgue integral. Evaluating the limit does not require ##\int_{0}^{\infty} \sin(xy)/x dx## to exist.
     
  4. Oct 8, 2013 #3
    Ahhh, yes, I see. It is obvious. Just as we can evaluate [itex]\lim_{x\to 0}x/x[/itex] with out the function existing at 0.

    Thanks!
     
  5. Oct 8, 2013 #4

    jbunniii

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    Yes, or ##\lim_{x \rightarrow 0}\sin(x)/x## for that matter.
     
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