Convergence of $\int_0^\infty\frac{\sin x}{x}dx$

Question

I'm working on proving the convergence of the below integral, the value of which I know to be $\displaystyle \frac{\pi}{2}$. $$ \int_0^\infty\frac{\sin x}{x}dx $$ I'm also well aware that this question has been answer several times on this site. Yet, I'm writing this as a question because I've tried the below method which looks (to me) like it has some hope, yet I'm not able to see it.

Note: I'm trying to find a proof with the following constraints:

No using of vector calculus or multiple integrals
No using of techniques of infinite sequence and series
I've just been introduced to improper integrals and hence could only use the direct comparison or limit comparison tests or something elementary like that.

My Proof:
Consider $$ \begin{aligned} I &= \int_a^\infty\frac{\sin x}{x}dx \\ &= \lim_{b\rightarrow\infty}\int_a^b\frac{\sin x}{x}dx\\ &= -\lim_{b\rightarrow\infty}\left[\frac{\cos x}{x}\right]_a^b-\lim_{b\rightarrow\infty}\int_a^b\frac{\cos x}{x^2}dx \\ &= \frac{\cos a}{a}-\lim_{b\rightarrow\infty}\int_a^b\frac{\cos x}{x^2}dx \\ \end{aligned} $$ Now $$ \lim_{b\rightarrow\infty}\int_a^b\frac{\cos x}{x^2}dx \leq \lim_{b\rightarrow\infty}\int_a^b\frac{dx}{x^2}=\frac{1}{a} $$ Hence we have $$ \lim_{b\rightarrow\infty}\int_a^b\frac{\cos x}{x^2}dx =\frac{1}{a}-h \text{ | }h\geq0 $$ Substituting this result back to $I$, we get $$ I=\frac{\cos a}{a}-\left[\frac{1}{a}-h\right]=\frac{\cos a-1}{a}+h $$

Now $$ \begin{aligned} \int_0^\infty\frac{\sin x}{x}dx &= \lim_{a\rightarrow0}\int_a^\infty\frac{\sin x}{x}dx \\ &= \lim_{a\rightarrow0}\frac{\cos a-1}{a}+h \\ &= h \end{aligned} $$

This looks neat, but since $h$ is not bound on the other end, I don't see how I could proceed.

I would do it this way: $$ \mathop {\lim }\limits_{b \to + \infty } \int_a^b {\frac{{\sin x}}{x}dx} = \frac{{\cos a - 1}}{a} - \mathop {\lim }\limits_{b \to + \infty } \int_a^b {\frac{{\cos x - 1}}{{x^2 }}dx} . $$ Can you show that $$ \int_0^{ + \infty } {\frac{{\cos x - 1}}{{x^2 }}dx} $$ converges absolutely? — Gary, Nov 30 '21 at 13:59
Your idea of approaching the tail, from $a$ to $\infty$, using integration by parts is excellent, good job. There is a detail in your $\leq$ step you need to justify, because for now you only have an upper bound, but your left-hand side can be negative. — Jakob Streipel, Nov 30 '21 at 14:04
This question https://math.stackexchange.com/questions/67198/does-int-0-infty-frac-sin-xxdx-have-an-improper-riemann-integral-or?rq=1 has some more information about this function — user62498, Nov 30 '21 at 14:04
To reconcile the $a \to 0$ part: consider splitting your integral some place and do two improper integrals, one from $0$ to $a$, and one from $a$ to $\infty$. — Jakob Streipel, Nov 30 '21 at 14:05
Can you use the fact $\frac{\sin x}{x} \leq 1$ for all real $x \neq 0$? — aschepler, Nov 30 '21 at 14:07
@Koro I've already proven that the integral is $\leq \frac{1}{a}$ — Abhishek A Udupa, Nov 30 '21 at 14:10
Combine my hint with the one from @prets: $\int_0^\infty \frac{sin x}{x} dx = \int_0^1 \frac{sin x}{x} dx + \int_1^\infty \frac{sin x}{x} dx$. Use two different comparison tests to show each part converges. — aschepler, Nov 30 '21 at 14:27
@AbhishekAUdupa The integral $\int_a^b f(x)dx$ converges absolutely if and only if $\int_a^b |f(x)|dx$ converges. The definition is analogous to absolute convergence of series. Have you learnt about the latter? For the other question: $(\cos x-1)'=\sin x$. I just chose a different function to do the integration by parts. This is because $(\cos x-1)/x^2$ is finite as $x\to 0+$. — Gary, Nov 30 '21 at 22:37
@Gary I haven't learned about this 'absolute convergence' yet. I think I inadvertently ran into this concept. I've made a mistake in trying to apply direct comparison test to prove the convergence of $\lim_{b\rightarrow\infty}\int_a^b\frac{\cos x}{x^2}dx$ since the integrand is not positive at certain intervals. So, I've started to see how I can use $\left|\frac{\cos x}{x^2}\right|$ since I can use DCT to this function...in progress. — Abhishek A Udupa, Dec 01 '21 at 02:52
@AbhishekAUdupa How come you know about DCT but not about absolute convergence? DCT should come much later in one's studies of analysis. — Gary, Dec 01 '21 at 02:59
@Gary I'm studying from 'Thomas Calculus 14the Edition' and that's what I have... — Abhishek A Udupa, Dec 01 '21 at 04:27

am301 · Answer 1 · 2021-12-01T05:35:25.650

The integral $\int^\infty_0\frac{\sin x}xdx$ converges in the same fashion that the sum $\sum_{n=1}^\infty\frac{(-1)^n }n$ converges. This can be seen if we brake the infinte interval $[0,\infty]$ into successive intervals over half periods of the sine function - then the integrand over each half period roughly decreases as $\frac 1x$ and has alternating sign due to the sine function. This idea can be used to show that the integral converges. Firstly we note that the integral of a continous function ($\sin x/x$ is such at $x=0$ due to removability of the singulatrity) over a finite interval always exist. So we only need to show that the integral $\int_{2\pi}^{m\pi}\frac{\sin x}xdx$ exists as $m\to \infty$ for integers $m$. Indeed: $$ I(k)=\int_{2\pi}^{2\pi k }\frac{\sin x}xdx=\frac12\bigg\{\int_{2\pi}^{2\pi k }\frac{\sin x}xdx+\int_\pi^{2\pi k-\pi}\frac{\sin (x+\pi)}{x+\pi}dx\bigg\} $$ where the second term is obtained from the first by change of veariable $x'=x-\pi$. Taking the limit $k\to \infty$ and subtracting and adding finite integrals (over $[2\pi k,2\pi k-\pi]$ and $[2\pi,\pi]$, to the second integral on the right (calling the result now $I_1(k)$) we find: $$ I_1(\infty)=\frac12\lim_{k\to\infty}\int_{2\pi}^{2\pi k}\sin x\bigg[\frac1x-\frac1{x+\pi}\bigg]dx\le\frac{\pi}2\int_{2\pi}^\infty\frac{dx} {x^2} $$ and the last integral is known to exist.

Convergence of $\int_0^\infty\frac{\sin x}{x}dx$

1 Answers1