Solving the Integral of $\sin x/x$ with Lebesgue Dominated Convergence

[Robert1986]

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

Since $[a,b]$ is compact, the Riemann and Lebesgue Integral of $\sin x / x$ coincide on this set. As $b \to \infty$ and $a\to 0$ 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:

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

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


Now, I get that $F_k$ is uniformly bounded in $k$. However, if the integral is taken to be a Lebesgue integral (which, it is initially) then I don't see how $\lim F_k$ 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!

 

[jbunniii]

$\sin(x)/x$ 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.

 

[Robert1986]

$\sin(x)/x$ 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.

Ahhh, yes, I see. It is obvious. Just as we can evaluate $\lim_{x\to 0}x/x$ without the function existing at 0.

Thanks!

 

[jbunniii]

Ahhh, yes, I see. It is obvious. Just as we can evaluate $\lim_{x\to 0}x/x$ without the function existing at 0.

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

 

[blue_raver22]

First of all, it is important to note that the Lebesgue Dominated Convergence Theorem only applies to sequences of functions, not sequences of integrals. In this case, the sequence of integrals is being used to approximate the original integral, but the theorem cannot be directly applied to it.

That being said, the use of the bound of 4 is valid in this situation. The sequence of functions F_k is indeed uniformly bounded by f(y)4, which is integrable since f is assumed to be bounded and in L^1. This allows for the application of the Lebesgue Dominated Convergence Theorem to the sequence of functions F_k, which then justifies the interchange of limit and integral in the given expression.

It is also worth noting that the Lebesgue integral of \sin x / x does not exist, but this does not mean that the integral cannot be approximated by a sequence of Riemann integrals. This is exactly what is being done in the given computation. So while the Lebesgue integral may not exist, we can still use Riemann integrals to approximate it, as long as we have a bound on the error of the approximation.