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This is the beginning of the proof that $\frac{\sin x}{x}, x\in (0,\infty)$ is improperly Riemann integrable. However, I don't understand how to attain the second identity below.

In the first inequality, I can see that $|\lim_{a\to \infty} \int_{N\pi}^a \frac{\sin x}{x}dx| \le \lim_{a\to \infty} \int_{N\pi}^{(N+1)\pi} |\frac{\sin x}{x}|dx$, but how do we get the same inequality for $N\to \infty$? What justifies this change in variables? Also, similarly, how does this enable us to conclude the second identity? To get the second identity, we would need $\lim_{a\to \infty} \int_0^{N(a)\pi} \frac{\sin x}{x}dx = \lim_{n\to \infty} \int_0^{n\pi} \frac{\sin x}{x}dx$, but how do we get this? I mean, I don't understand how we are able to switch the limit from $a\to \infty$ to $N \to \infty$. I can see this intuitively but I would appreciate if anyone could explain how this works rigorously.

For $a>0$ we can find $N=N(a)\in \mathbb{N}$ such that $N\pi \le a <(N+1)\pi$. Thus $$\int_0^\infty \frac{\sin x}{x}dx = \lim_{a\to \infty} \bigg( \int_0^{N\pi} \frac{\sin x}{x}dx + \int_{N\pi}^a \frac{\sin x}{x}dx\bigg) =\lim_{N\to \infty} \sum_{i=0}^{N-1} \int_{i\pi}^{(i+1)\pi} \frac{\sin x}{x}dx,$$ where we use $$\Bigg| \lim_{a\to \infty} \int_{N\pi}^a \frac{\sin x}{x}dx \Bigg| \le \lim_{N\to \infty} \int_{N\pi}^{(N+1)\pi} \Bigg| \frac{\sin x}{x}\Bigg| dx \le \lim_{N\to \infty} \frac{\pi}{N\pi}=0.$$

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    $\begingroup$ For those kind of oscillatory function, integrating by parts is always a good idea, as $ \int_1^\infty \frac{\cos x}{x^2}dx$ is easy to understand $\endgroup$ – reuns Oct 16 '17 at 21:05
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Perhaps it would be more clear if you maintained the explicit dependence of $N$ on $a$ throughout. Adding a couple extra steps for clarity, we have that $N(a)\pi\le a\lt(N(a)+1)\pi$ implies

$$\left|\int_{N(a)\pi}^a{\sin x\over x}dx\right|\le\int_{N(a)\pi}^a\left|\sin x\over x\right|dx\le\int_{N(a)\pi}^a\left|\sin x\over x\right|dx+\int_a^{(N(a)+1)\pi}\left|\sin x\over x\right|dx\\=\int_{N(a)\pi}^{(N(a)+1)\pi}\left|\sin x\over x\right|dx\le\int_{N(a)\pi}^{(N(a)+1)\pi}\left|1\over N(a)\pi\right|dx={\pi\over N(a)\pi}$$

Finally, it's clear that $N(a)\to\infty$ as $a\to\infty$ (and vice versa).

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  • $\begingroup$ Yes this is much clearer. But from this, how do we get $\lim_{a\to \infty} \int_0^{N\pi} \frac{\sin x}{x}dx = \lim_{N\to \infty} \sum_{i=0}^{N-1} \int_{i\pi}^{(i+1)\pi} \frac{\sin x}{x}dx$? Also, to split the limit of the sum of the first two integrals into the sum of the limit of the two integrals and use the above identity, don't we need the integrability of $\int_0^\infty \frac{\sin x}{x} dx$, which is what we want to prove in the first place? $\endgroup$ – takecare Oct 16 '17 at 21:09
  • $\begingroup$ @takecare, writing $$\int_0^{N\pi}f(x)dx=\sum_{i=0}^{(N-1)}\int_{i\pi}^{(i+1)\pi}f(x)dx$$ is a general property for definite integrals. Keep in mind, this is only the beginning of the proof that the improper integral converges. Presumably what comes next is an inequality of some sort that bounds the sum on the right hand side with a convergent infinite series. $\endgroup$ – Barry Cipra Oct 16 '17 at 22:32
  • $\begingroup$ Right I knew that. I was thinking why $\lim_{a\to \infty} \int_0^{N(a)\pi} f = \lim_{n \to \infty} \int_0^{n\pi} f$. Because the $N$ on the left hand side depends on $a$ but the right hand side doesn't. $\endgroup$ – takecare Oct 16 '17 at 22:39
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    $\begingroup$ @takecare, as $a$ runs through all (large) real values, $N(a)$ runs through all (large) integer values. If you like, think of it as $$\lim_{a\to\infty}\int_0^{N(a)\pi}f=\lim_{N(a)\to\infty}\int_0^{N(a)\pi}f=\lim_{n\to\infty}\int_0^{n\pi}f$$ $\endgroup$ – Barry Cipra Oct 16 '17 at 22:53
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A more correct chain of inequalities would be

$$\Bigg| \int_{N\pi}^a \frac{\sin x}{x}dx \Bigg| \le \int_{N\pi}^{a} \Bigg| \frac{\sin x}{x}\Bigg| dx \le \int_{N\pi}^{(N+1)\pi} \Bigg| \frac{\sin x}{x}\Bigg| dx \le \frac{1}{N\pi}\int_{N\pi}^{(N+1)\pi} | \sin x| dx\\= \frac{2}{N\pi}= \frac{2(N+1)}{N}\frac{1}{(N+1)\pi}\le \frac{4}{a},$$ where $1/(N+1)\pi <1/a \le 1/N\pi$. And now you send $a\to\infty$ to get $0$.

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  • $\begingroup$ Thanks that resolves the inequality problem. But how do we get $\lim_{a\to \infty} \int_0^{N\pi} \frac{\sin x}{x}dx = \lim_{N\to \infty} \sum_{i=0}^{N-1} \int_{i\pi}^{(i+1)\pi} \frac{\sin x}{x}dx$? In fact, to split the limit of the sum of the first two integrals into the sum of the limit of the two integrals and use the above identity, don't we need the integrability of $\lim_0^\infty \frac{\sin x}{x} dx$, which is what we want to prove in the first place? $\endgroup$ – takecare Oct 16 '17 at 21:03
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HINT:

You should estimate the integral on a given interval $[a,b]$ and see what happens when that interval moves to infinity. For this, use integration by parts:

$$\int_a^b \frac{\sin x}{x} d x = \int_ a^b (-\cos x)' \frac{1}{x} d x = = - \cos x \cdot \frac{1}{x} \mid_a^b - \int_a^b (-\cos x)(- \frac{1}{x^2})dx= \\=\frac{\cos a}{a} - \frac{\cos b}{b} - \int_a^b \frac{\cos x}{x^2} dx $$ We conclude that $$\left |\int_a^b \frac{\sin x}{x} d x \right |\le \frac{1}{a} + \frac{1}{b} + \frac{1}{a}- \frac{1}{b} = \frac{2}{a}$$

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