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(adsbygoogle = window.adsbygoogle || []).push({});

[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!

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# Integral of sinx/x

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